Look at the picture graphs. How many fewer students walk to school in Class A than in Class B?

A picture graph is titled

Enter your answer in the box.

fewer students

a. The mean of the binomial distribution is  4.7

b. The variance of the binomial distribution is 1.1

c. The standard deviation of the binomial distribution is 1.0

Probability of success p = 0.78

Number of trials n = 6

(a) Mean of the binomial distribution:

The mean or expected value of a binomial distribution is given by np.

Therefore, the mean of this binomial distribution is:

μ = np = 6 x 0.78 = 4.68

So, the mean is 4.7 (rounded to one decimal place).

(b) Variance of the binomial distribution:

The variance of a binomial distribution is given by np(1 - p).

Therefore, the variance of this binomial distribution is:

σ^2 = np(1 - p) = 6 x 0.78 x 0.22 = 1.0608

So, the variance is 1.1 (rounded to one decimal place).

(c) Standard deviation of the binomial distribution:

The standard deviation of a binomial distribution is the square root of the variance.

Therefore, the standard deviation of this binomial distribution is:

σ = √(np(1 - p)) = √(1.0608) = 1.03

So, the standard deviation is 1.0 (rounded to one decimal place).

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The integral ∫ ln(x^17)/x^2 dx converges, and its value is given by -ln(x^17)/x + 17/x + C.Therefore, the integral converges.

To determine if the integral ∫ ln(x^17)/x^2 dx converges or diverges, we need to use the integral test. The integral test states that if f(x) is a continuous, positive, and decreasing function on the interval [a, infinity), then the infinite series ∑f(n) and the improper integral ∫f(x) dx both converge or both diverge.

In this case, we have f(x) = ln(x^17)/x^2, which is continuous, positive, and decreasing on the interval [1, infinity). Therefore, we can use the integral test to determine if the improper integral ∫ ln(x^17)/x^2 dx converges or diverges.

To evaluate the integral, we can use integration by parts with u = ln(x^17) and dv = 1/x^2 dx, which gives us:

∫ ln(x^17)/x^2 dx = -ln(x^17)/x - ∫ -17/x^3 dx
= -ln(x^17)/x + 17/x + C

where C is the constant of integration.

Now, to determine if the integral converges or diverges, we need to evaluate the limit as x approaches infinity of the integrand. We have:

lim x→∞ [ln(x^17)/x^2] = lim x→∞ [(17/x) ln(x)]/x
= lim x→∞ (17/x) * lim x→∞ ln(x)/x

Since lim x→∞ ln(x)/x = 0 (which can be shown using L'Hopital's rule), the limit of the integrand is 0.

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for this question the required one is the distance b/n the light house and the boat (x) so in this case we are gonna use tan= opposite / hypotenuse :

- tan 20 = 89/ x (tan 20 is equivalent to 0.3640)

- 0.3640 = 89 / x

-x = 89 / 0.3640

- x = 244.5 ft. and when estimated x= 245 ft.

The real variables among the given options are: Bushels of Wheat and Numbers of bees per hectare used to pollinate farm crops. The nominal values are: Price of a pen and Price of a computer.

Real variables are measurable quantities that can take on any value within a given range. In this case, "Bushels of Wheat" and "Numbers of bees per hectare used to pollinate farm crops" are real variables. These variables can be measured and expressed as continuous quantities, such as the amount of wheat harvested or the number of bees present.

On the other hand, nominal values are categories or labels that cannot be measured on a numerical scale. They represent different groups or classes. The "Price of a pen" and "Price of a computer" are nominal values as they represent different price categories or labels for pens and computers, but they do not have a numerical measurement scale associated with them.

In summary, "Bushels of Wheat" and "Numbers of bees per hectare used to pollinate farm crops" are real variables, while "Price of a pen" and "Price of a computer" are nominal values.

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The price-demand equation is [tex]p(x) = 77.88^(x/5)[/tex] rounded to two decimal places.

We know that the marginal price dp/dx at x units of demand per week is proportional to the price p. Let k be the constant of proportionality. Then, we have:

dp/dx = kp

Separating variables and integrating, we get:

∫dp/p = ∫kdx

ln|p| = kx + C

Taking exponential on both sides and using the initial condition p(0) = 100, we get:

p(x) = 100e^(kx)

Using the second condition, p(5) = 77.88, we get:

[tex]77.88 = 100e^(5k)[/tex]

Solving for k, we get:

k = ln(77.88/100)/5

Substituting this value of k in the price-demand equation, we get:

[tex]p(x) = 100e^[(ln(77.88/100)/5)x] = 100(77.88/100)^(x/5)[/tex]

Therefore, the price-demand equation is [tex]p(x) = 77.88^(x/5)[/tex] rounded to two decimal places.

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The approximate value of P(z ≤ -1.25) will be; option (c) 11%

Since probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

Given P(z ≤ -1.25)

In the table on the first row and column represents values of z

For -1.25 we can see value of z at -1.2 from column and 0.05 from row and where they meet will be our value of P(z ≤ -1.25)

Here the value is 0.1056

If we round off this value will get 0.11 so it will be; 11%

Hence for P(z ≤ -1.25) from standard normal distribution the approximate value will be 11% that is option (c).

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1) 8 +/- 4 heads in 16 tosses is about as likely as 32 +/- 16 heads in 64 tosses. 2) the answer is 50% +/- 0.63% heads.

1. To find the equivalent number of heads for 64 tosses, you need to scale the given range of heads in proportion to the number of tosses. In the first question, there are 16 tosses with a range of 8 +/- 4 heads. The second question has 64 tosses, which is 4 times the number of tosses in the first question. Therefore, you'll need to multiply the range of heads by 4:

32 +/- (4 * 4) heads = 32 +/- 16 heads

So, 8 +/- 4 heads in 16 tosses is about as likely as 32 +/- 16 heads in 64 tosses.

2. In this question, you are given a percentage range instead of a specific number of heads. First, you need to find the equivalent percentage range for 512 tosses:

50% +/- 10% heads in 32 tosses

To find the new range, divide the original number of tosses (32) by the new number of tosses (512):

32/512 = 0.0625

Now, multiply the original percentage range (10%) by this ratio:

10% * 0.0625 = 0.625%

So, 50% +/- 10% heads in 32 tosses is about as likely as 50% +/- 0.625% heads in 512 tosses. Rounded to 2 decimal places, the answer is 50% +/- 0.63% heads.

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Dr. Howard's sample does vary from the general population, as the average grade in his classes is higher than the overall average grade.

Based on the information provided, it appears that Dr. Howard's sample has a higher average grade than the general population. The average grade on a statistics final exam for the general population is 85, while in Dr. Howard's classes, the average grade is 93. This suggests that Dr. Howard's students performed better on the final exam than the average statistics student. However, without additional information about the sample size or characteristics of Dr. Howard's students, it is difficult to draw definitive conclusions about the variation between the sample and the general population.
Dr. Howard's sample varies from the general population. Here's a step-by-step explanation:

1. The average grade of the general population on the statistics final exam is 85.
2. In Dr. Howard's classes, the average grade is 93.
3. Compare the two averages: 93 (Dr. Howard's classes) vs. 85 (general population).
4. Since 93 is higher than 85, it indicates that Dr. Howard's sample (his classes) has a higher average grade than the general population.

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The probability sampling method in which you randomly select one of the first k elements and then select every k element thereafter is known as c. systematic sampling. Therefore, option c. systematic sampling is correct.

Systematic sampling is a probability sampling technique where the sample is chosen by selecting every kth element from the population, where k is a constant. This method is often used when the population is large and the complete list of elements is not easily available.

Stratified random sampling is a technique where the population is divided into strata or subgroups based on certain characteristics and a random sample is chosen from each stratum.

Cluster sampling involves dividing the population into clusters or groups and then selecting a random sample of clusters. The elements within each selected cluster are then included in the sample.

Convenience sampling is a non-probability sampling method where the sample is chosen based on convenience and availability. This method is often used in situations where it is difficult or expensive to obtain a random sample.

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The equation that can be used to find the number of quarters q in any number of rolls of quarters r is: q = 40r

This equation simply states that the total number of quarters is equal to the number of rolls multiplied by 40, since each roll contains 40 quarters.

An equation is a mathematical statement that expresses the equality of two expressions using an equal sign (=). It contains one or more variables (letters or symbols that represent an unknown value) and constants (numbers that have a fixed value), as well as mathematical operations such as addition, subtraction, multiplication, division, and exponentiation. Equations are used to solve problems and find solutions for unknown values by manipulating the expressions using algebraic rules and properties.

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The assessed value of the property with a market value of $60,000 and an assessment rate of 45% is $27,000. T

To compute the assessed value of a property, you'll need to consider both the market value and the assessment rate. In this case, the market value is $60,000, and the assessment rate is 45%.

To find the assessed value, you can multiply the market value by the assessment rate. Using the provided values, you can calculate the assessed value as follows:

Assessed Value = Market Value x Assessment Rate
Assessed Value = $60,000 x 0.45

Assessed Value = $27,000

In this case, the assessed value of the property with a market value of $60,000 and an assessment rate of 45% is $27,000. This value represents the taxable amount on which property taxes will be calculated and is an essential factor for both property owners and tax authorities.

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The exact length of the curve. y = 3 + 4x^3/2, 0 ≤ x ≤ 1 is L = ∫√(9+108x-144x^(5/2)) / √(9-16x^3) dx, from 0 to 1.

To find the length of the curve, we need to use the arc length formula:

L = ∫√(1+(dy/dx)^2) dx, where y = 3 + 4x^(3/2) and 0 ≤ x ≤ 1.

First, we need to find dy/dx:

dy/dx = (12x^(1/2))/2√(3 + 4x^(3/2))

dy/dx = 6x^(1/2)/√(3 + 4x^(3/2))

Now, we can substitute this into the arc length formula:

L = ∫√(1+(6x^(1/2)/√(3 + 4x^(3/2)))^2) dx, from 0 to 1.

Simplifying the inside of the square root, we get:

L = ∫√(1+(36x)/(3 + 4x^(3/2))) dx, from 0 to 1.

We can simplify this further by multiplying the numerator and denominator of the fraction by (3 - 4x^(3/2)):

L = ∫√(1+36x(3-4x^(3/2))/(9-16x^3)) dx, from 0 to 1.

Expanding the numerator, we get:

L = ∫√((9+108x-144x^(5/2))/(9-16x^3)) dx, from 0 to 1.

Simplifying the expression under the square root, we get:

L = ∫√(9+108x-144x^(5/2)) / √(9-16x^3) dx, from 0 to 1.

We can evaluate this integral using numerical methods, such as Simpson's rule or the trapezoidal rule, to get an approximation of the length of the curve. The exact length of the curve cannot be expressed in a finite number of terms, but it can be approximated to any desired degree of accuracy using numerical methods.

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The equation of the tangent plane to the surface [tex]x^2yz[/tex] = 18 at the point (4, 1, 2) is: (8x-3y-16z) = (84 - 31 - 16*2)

8x - 3y - 16z = -21

The region $E$ is defined by the paraboloid [tex]$z = 24 - 2x^2 - 2y^2$[/tex]and the cone [tex]$z = 2\sqrt{x^2 + y^2}$[/tex]. To find the volume of this region, we integrate over [tex]$E$[/tex]using cylindrical coordinates:

[tex]\iiint_E dV &= \int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \int_{2r^2}^{24-2r^2} r , dz , dr , d\theta \[/tex]

[tex]&= \int_{0}^{2\pi} \int_{0}^{\sqrt{2}} (22r^3 - r^5) , dr , d\theta \&= \frac{64}{3} \pi.\end{align*}[/tex]

Therefore, the volume of the region[tex]$E$ is $\frac{64}{3} \pi$[/tex]

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From the first integration we have,

[tex]\frac{1}{81} (\sqrt{(1-\frac{81}{x^{4} }) }[/tex] +c

From the second integration we have,

[tex]1152[\frac{x (x^{2} + 24)^{3/2} }{2304} - \frac{1}{192} x\sqrt{x^{2} +24} - \frac{1}{8} ln|\frac{x\sqrt{x^{2} + 24} }{\sqrt{24} } | ][/tex] +c

What is integration?

Integration which is the opposite of differentiation is  is the calculation of an integral. Integrals in mathematics are used to find many useful mathematical as well physical quantities such as areas, volumes, displacement, etc.

The first integral given is,

[tex]\int\limits {\frac{2}{x^{3}\sqrt{x^{4} -81} } } \, dx[/tex]

This can be rewritten as,

[tex]\int\limits {\frac{2}{x^{3}\sqrt{x^{4}(1 -\frac{81}{x^{4} } )} } } \, dx[/tex]

This equals to,

[tex]\int\limits {\frac{2}{x^{5}\sqrt{(1 -\frac{81}{x^{4} } )} } }} \, dx[/tex]  --------(1)

Let us take 1- (81/x⁴)= z²

differentiating both sides we get,

2zdz= 162/x⁵ dx

zdz/162= (1/x⁵)dx

Putting these values in equation (1) we get,

2 ∫ (z/162z) dz

= (2/162)∫ dz

= (1/81) z + c where c is the integrating constant.

Hence from the given integration we have,

[tex]\frac{1}{81} (\sqrt{(1-\frac{81}{x^{4} }) }[/tex] + c

For the second integral,

[tex]\int\limits {x^{2} \sqrt{(96+4x^{2} }) } \, dx[/tex]

= [tex]4\int\limits {x^{2} \sqrt{(24+x^{2} }) } \, dx[/tex]

let us take x= √24 tanα

                dx= √24 sec²α dα

putting these values in integration we get,

[tex]4\int\limit 24tan^{2} \alpha \sqrt{24(1+tan^{2}\alpha } \, \sqrt{24} sec^{2}\alpha d\alpha[/tex]

=2304∫ tan²α sec³α dα

= 2304∫ sec³α( sec²α -1) dα

= 2304 [ ∫sec⁵α dα - ∫ sec³α dα] ---------(2)

Now at first we integrate ∫sec⁵α dα

Let I₁ = ∫sec⁵α dα

        = ∫ sec³α sec²α dα

    Integrating by parts we get,

      = sec³α tanα - 3∫ sec³α tan²α dα

      = sec³α tanα - 3∫ sec³α (sec²α - 1)dα

      = sec³α tanα - 3∫ sec⁵α dα + 3 ∫ sec³α dα

     = sec³α tanα - 3I₁ + 3 ∫ sec³α dα

4I₁= sec³α tanα  + 3 ∫ sec³α dα

I₁= ( sec³α tanα  + 3 ∫ sec³α dα)/4 ----------- (3)

Here we have to solve I₂=∫ sec³α dα

                                    = ∫ sec α  sec²α dα

Integrating by parts we get,

                        I₂= secα tan α- ∫ secα tan²α dα

                           = secα tan α- ∫ secα ( sec²α -1) dα

                           = secα tanα - ∫ secα(sec²α -1) dα

                          =  secα tanα - ∫ sec³α dα + ∫ secα dα

                           =  secα tanα -  I₂ + ln| secα + tanα |

               2I₂= secα tanα  + ln| secα + tanα |

               I₂= ( secα tanα  + ln| secα + tanα |)/2

Now at first putting the values in equation (3) and from that calculating and deriving the value of equation (1) we get,

2304×[tex]\frac{1}{4} [ \frac{sec^{3} \alpha tan\alpha }{4} - \frac{sec\alpha tan\alpha }{8} - \frac{ln|sec\alpha + tan\alpha| }{8} ][/tex]+ c

Now using x= √24 tanα we get the value of the above integral in terms of x and that is,

[tex]1152[\frac{x (x^{2} + 24)^{3/2} }{2304} - \frac{1}{192} x\sqrt{x^{2} +24} - \frac{1}{8} ln|\frac{x\sqrt{x^{2} + 24} }{\sqrt{24} } | ][/tex] +c

Hence, from the given integration we have,

[tex]1152[\frac{x (x^{2} + 24)^{3/2} }{2304} - \frac{1}{192} x\sqrt{x^{2} +24} - \frac{1}{8} ln|\frac{x\sqrt{x^{2} + 24} }{\sqrt{24} } | ][/tex] +c

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Answer:

20%

Step-by-step explanation:

1200 is 120% of 1000, therefor the profit percentage is 20.

The given function is S(t) = t.e^(2t) on the interval (0, ∞). To compute S(0), we substitute t = 0 in the expression for S(t) to obtain: S(0) = 0.e^(2(0)) = 0.

Therefore, the value of S(0) is 0.

The surge function S(t) = 1 - e^(-kt) is commonly used to model drug-loading curves.

It reflects an initial surge in the drug level, followed by a gradual decline. In this case, we were given a function that models the drug concentration in the bloodstream on the interval (0, ∞) as S(t) = t.e^(2t), with unrealistic values for the constants A, p, and k. We were asked to compute the value of S(0), which is simply 0.

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We can use this formula: N = N0 * 2^(t/T) to solve the problem. We know that N0 = 10000, N = 67500, and T = 5.5 hours. We want to find t, the time elapsed.

Now, let's solve the problem step-by-step:

Step 1: Identify the given information.
- Doubling time: 5.5 hours
- Initial population: 10,000 bacteria
- Final population: 67,500 bacteria

Step 2: Use the formula for exponential growth:
Final population = Initial population * (2 ^ (time elapsed / doubling time))

Step 3: Solve for the time elapsed (t).
67,500 = 10,000 * (2 ^ (t / 5.5))

Step 4: Divide both sides by the initial population to isolate the exponential term.
6.75 = 2 ^ (t / 5.5)

Step 5: Take the logarithm of both sides (base 2) to solve for t.
log2(6.75) = log2(2 ^ (t / 5.5))

Step 6: Use the logarithm property to simplify the equation.
log2(6.75) = t / 5.5

Step 7: Solve for t.
t = 5.5 * log2(6.75)

Step 8: Calculate the value of t.
t ≈ 13.08 hours

Step 9: Round the answer to two decimal places.
t ≈ 13.08 hours

So, it will take approximately 13.08 hours for an initial population of 10,000 bacteria to reach 67,500 in the petri dish.

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(a) Yes, the given series is an Alternating Series because the terms alternate in sign, switching between positive and negative values. Both conditions are satisfied, so the series is convergent according to the Alternating Series Test.

(b) The series is convergent because the terms alternate and decrease in magnitude. Specifically, the terms approach zero as n increases and alternate between positive and negative values. This satisfies the alternating series test, which states that if a series is alternating, decreasing in magnitude, and approaching zero, then the series is convergent. Therefore, we can conclude that the given series converges.

(a) The given series is:

Σ (-1)^n / n, for n = 1 to infinity

This series is an alternating series because the term (-1)^n alternates the signs of the terms in the series. Specifically, when n is even, the term is positive, and when n is odd, the term is negative.

(b) To determine if the series converges or diverges, we can use the Alternating Series Test. The test has two conditions:

1. The absolute value of the terms is decreasing: |a_(n+1)| <= |a_n| for all n.
2. The limit of the terms as n approaches infinity is 0: lim (n->∞) a_n = 0.

For our series, a_n = (-1)^n / n, let's check both conditions:

1. |a_(n+1)| = |(-1)^(n+1) / (n+1)| and |a_n| = |(-1)^n / n|.
Since n + 1 > n, we have |(-1)^(n+1) / (n+1)| <= |(-1)^n / n|, so the terms are decreasing in absolute value.

2. The limit of a_n as n approaches infinity is: lim (n->∞) (-1)^n / n. Since the limit of (-1)^n is oscillating and the limit of 1/n as n->∞ is 0, the overall limit is 0.

Both conditions are satisfied, so the series is convergent according to the Alternating Series Test.

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X = 3. 5y is an equation that shows the number of teaspoons of sugar, y, needed to make x tarts where a recipe 3.5 teaspoons of sugar are added to make a tart.

An equation is a mathematical sentence where we equalize two expressions using an equal sign. An expression refers to a phrase with two or more variables or numbers with any mathematical operation.

The situation given is a recipe 3.5 teaspoons of sugar is required to make a tart. Thus to calculate the number of teaspoons of sugar needed to make tarts, we have to multiply 3.5 by the number of teaspoons of sugar to make tarts

Thus if x is the number of tarts

y is the number of teaspoons of sugar

The equation is given by x = 3.5y

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An expression for 2n-1 is: 2n-1 = (2n+1 - 1)/(2n(2n-1)), for n = 1, 2, ....  The general solution is: [tex]y(x) = c1exp(x^2) + c2exp(-2x^2)[/tex], where c1 and c2 are constants. The power series solution of the initial value problem is:[tex]y(x) = 4 - 6x - 12x^2 + 16x^3 - 16x^4 + 256/15 x^5 - 1024/96 x^6 + 2048/315 x^7 - 32768/460[/tex]

2.1 Using the recurrence relation, we can obtain an expression for 2n-1 as follows:

Cn+2 = (n+1)Cn+1 + 22Cn

For n=0, we have:

C2 = C1 + 22C0

Substituting C1 = 1.3.5 and C0 = c, we get:

C2 = 1.3.5 + 22c

For n=1, we have:

C3 = 2C2 + 22C1

Substituting C2 = 1.3.5 + 22c and C1 = 1, we get:

C3 = 2(1.3.5 + 22c) + 22

Simplifying, we get:

C3 = 1.3.5.7 + 2.2.3.5c + 22

Comparing with the general expression for Cn+2, we get:

2n+1 - 1 = 2.2n(2n-1)cn

Solving for 2n-1, we get:

2n-1 = (2n+1 - 1)/(2n(2n-1))

Hence, an expression for 2n-1 is:

2n-1 = (2n+1 - 1)/(2n(2n-1)), for n = 1, 2, ...

2.2 The general solution of the differential equation y" - 2xy' - 4y = 0 can be written as a linear combination of the two linearly independent solutions:

[tex]y1(x) = c1exp(x^2)\\y2(x) = c2exp(-2x^2)[/tex]

Hence, the general solution is:

[tex]y(x) = c1exp(x^2) + c2exp(-2x^2)[/tex]

where c1 and c2 are constants.

2.3 To find the power series solution of the initial value problem y" - 2xy' - 4y = 0, y(0) = 4, y'(0) = -6, we first need to find the coefficients of the power series solution y(x).

Substituting y = Σn=0∞ anxn into the differential equation, we get:

Σn=0∞ [(n+2)(n+1)an+2 - 2n an - 4an]xn = 0

Equating the coefficients of xn, we get:

(n+2)(n+1)an+2 - 2n an - 4an = 0

Simplifying, we get:

an+2 = (2n/(n+2))an

Using the initial conditions y(0) = 4 and y'(0) = -6, we get:

a0 = 4

a1 = -6

Substituting the recurrence relation, we get:

a2 = -12

a3 = 48/3 = 16

a4 = -128/8 = -16

a5 = 256/15

a6 = -1024/96

a7 = 2048/315

a8 = -32768/4608

Hence, the power series solution of the initial value problem is:

[tex]y(x) = 4 - 6x - 12x^2 + 16x^3 - 16x^4 + 256/15 x^5 - 1024/96 x^6 + 2048/315 x^7 - 32768/460[/tex]

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The midpoint of the points A and B is (7/2, 5/2)

From the question, we have the following parameters that can be used in our computation:

A has the coordinates [0,0] and B has coordinates [7, 5].

The midpoint is calculated as

Midpoint = 1/2(A + B)

Substitute the known values in the above equation, so, we have the following representation

Midpoint = 1/2(0 + 7, 0 + 5)

Evaluate

Midpoint = (7/2, 5/2)

Hence, the Midpoint is (7/2, 5/2)

Also, the complete statement is

To find where the median from (5, 0) to that midpoint intersects the other medians, you need to find 1/2 of point (5,0) and 1/2 of the ordered pair for the midpoint. After you then add the new x-values together and the new y-values together, you find that the medians intersect at point (7/2, 5/2).

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The best instrument to determine the predictive validity of a metric scale would be a correlation-coefficient.

This measure assesses the strength of the relationship between two variables, in this case, the metric scale scores and the predicted outcome. A high correlation would indicate that the metric scale is a good predictor of the outcome, whereas a low correlation would indicate that the metric scale is not a reliable predictor.

Cronbach's alpha is a measure of internal consistency and would not be appropriate for determining predictive validity. Fisher's r-to-z test is used to compare the strength of two correlations and is not necessary in this scenario. Therefore, the answer is B, a correlation-coefficient.

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A) The probability is approximately 0.257. B) The expected number of customers is 5, and the standard deviation is approximately 1.71.

A) The number of customers who qualify for the membership follows a binomial distribution with parameters n = 20 and p = 0.1. The probability of having between 2 and 5 (inclusive) customers who qualify for the membership can be calculated by summing the probabilities of having exactly 2, 3, 4, or 5 customers who qualify for the membership.

Using the binomial probability formula or a binomial probability table, we find that this probability is approximately 0.257.

B) The expected number of customers who qualify for the membership in a randomly selected sample of 50 customers can be calculated using the formula E(X) = np, where X is the number of customers who qualify for the membership.

Thus, E(X) = 50 × 0.1 = 5. The variance of X is given by Var(X) = np(1-p), so the standard deviation is given by the square root of the variance, which is approximately 1.71.

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The points where tangents are horizontal are only at point A (0,-40). The answer is only at point A and the word count is 307.

In order to find the points where tangents are horizontal, we need to first understand what a tangent is. A tangent is a straight line that touches a curve at only one point. When we say that a tangent is horizontal, it means that it is parallel to the x-axis or has a slope of 0.
To find the points where tangents are horizontal, we need to differentiate the given function and set it equal to 0. This is because the slope of the tangent at a point is given by the derivative of the function at that point. When the derivative is 0, the slope of the tangent is also 0, which means it is a horizontal line.
Let's differentiate the given function:
f(x) = -5x^4 + 30x^2 - 40
f'(x) = -20x^3 + 60x
Now, we set f'(x) = 0 to find the critical points where the slope is 0:
-20x^3 + 60x = 0
-20x(x^2 - 3) = 0
x = 0, √3, -√3
These are the critical points where the slope is 0. To determine whether the tangents at these points are horizontal, we need to look at the second derivative:
f''(x) = -60x^2 + 60
At x = 0, f''(x) = 60 > 0, which means the point (0,-40) is a minimum point and the tangent is horizontal.
At x = √3 and x = -√3, f''(x) = -60 < 0, which means the points (±√3, -28) are maximum points and the tangents are not horizontal.

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Check the picture below.

[tex]\cfrac{1}{6}+\cfrac{1}{4}\implies \cfrac{(2)1+(3)1}{\underset{\textit{using this LCD}}{12}}\implies \cfrac{5}{12}~\hfill {\Large \begin{array}{llll} > \end{array}} ~\hfill \cfrac{1}{6}+\cfrac{1}{6}\implies \cfrac{2}{6}\implies \cfrac{1}{3}[/tex]

notice, 1/4 is really larger than 1/6 of the same whole.

There are 61872 integer solutions to the equation x1 + x2 + x3 + x4 = 30 that obey the given conditions.

To find the number of integer solutions to the equation x1 + x2 + x3 + x4 = 30 that obey the condition -10 ≤ xi ≤ 20 for i = 1, ..., 4, we can use the Principle of Inclusion-Exclusion.

First, let's change variables to make it easier to count. Define yi = xi + 10, so now 0 ≤ yi ≤ 30 for i = 1, ..., 4. Then, the equation becomes y1 + y2 + y3 + y4 = 70. Now, consider the number of non-negative integer solutions to y1 + y2 + y3 + y4 = 70 without any restrictions.

Using the stars and bars method, there are C(70+4-1, 4-1) = C(73, 3) = 62196 solutions. Now we must account for the restriction that 0 ≤ yi ≤ 30. Let Ai be the event that yi > 30.

For each Ai, if yi > 30, let zi = yi - 31. Then, z1 + z2 + z3 + z4 = 70 - 4*31 = 6. There are C(6+4-1, 4-1) = C(9, 3) = 84 solutions for each Ai. Using the Principle of Inclusion-Exclusion, we find the number of integer solutions to the equation that meet the given conditions: Total solutions = 62196 - (4 * 84) + (6 * C(2,1)) - 0 Total solutions = 62196 - 336 + 12 - 0 Total solutions = 61872

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The calculated value of the area of the figure is 89 sq meters

From the question, we have the following parameters that can be used in our computation:

Composite figure

The shapes in the composite figure are

This means that

Area = Rectangle - Triangle

Using the area formulas on the dimensions of the individual figures, we have

Area = 13 * 8 - 1/2 * (13 - 8) * 6

Evaluate

Area = 89

Hence, the area of the figure  is 89 sq meters

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Answer:

-1x, also can be re-written as -x

Step-by-step explanation:

You can use the formula as follows:

[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

Find two points on the graph.

We will use (3,1) and (4,0)

We now plug in the values into the formula.

[tex]\frac{0-1}{4-3} = -\frac{1}{1} = -1[/tex]

The slope is equal to -1x, which can be re-written as -x.

Research with larger sample sizes may be necessary to determine the true impact of industrial waste on fish mercury levels.

Based on the results of the correctly conducted test of significance, the researcher can conclude that there is no statistically significant difference in the average mercury level between the two groups of fish. Therefore, it is likely that industrial waste does not cause higher levels of mercury in fish. However, it is important to note that this conclusion is based on the assumption that the test was conducted correctly and that the sample size was sufficient to detect a significant difference if it existed. The researcher can conclude that the sample size may be too small to detect a statistically significant difference in average mercury levels between fish caught near industrial sites and those caught away from industrial sites.

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