30 POINTS AND I'LL MARK YOU BRAINLIST IF YOU ANSWER!!

Troy made a scale drawing of the Statue of Liberty which has an actual height of 305 feet. He decides to use a scale in which 1 inch represents 25 feet. What is the height in inches of Troy's drawing?

The random variable is 3×sigma²/4.

Since, X1, X2 & X3 are independent & identically distributed.

so we have,

var(X1) = var(X2) = var(X3) = \sigma².

And,[tex]cov(X1 , X2) = cov(X2 , X3) = cov(X1 , X3) = 0.[/tex]

Now,  [tex]var(Y1) = var(X2 + X3) = var(X2) + var(X3) + 2.cov(X2 , X3) = \sigma² + \sigma² + 0 = 2.\sigma²[/tex]

[tex]var[(Y1 + Y2 + Y3)/2] = (1/4). [var(Y1) + var(Y2) + var(Y3) + 2.cov(Y1 , Y2) + 2.cov(Y1 , Y3) + 2.cov(Y2 , Y3)] = (1/4). (\sigma² + \sigma² + \sigma² + 0 + 0 + 0) = 3.\sigma²/4[/tex]

The collection of potential outcomes for a random event is referred to as the sample space, which is the domain of a random variable. When a coin is tossed, for instance, there are only two possible outcomes: heads or tails. Read Related Links as well. Probability.

Different from an algebraic variable is a random variable. An algebraic equation's variable is an unknowable but calculable value. The formula [tex]10 + x = 13[/tex] demonstrates that we can determine the precise value of x, which is [tex]3.[/tex] As seen in the example of the dice above, a random variable has a set of values, and any of those values could be the outcome.

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The number 1.092×[tex]10^{1}[/tex] is 15.6 times larger than 7×[tex]10^{-1}[/tex]

Given,

The first number = 1.092×[tex]10^{1}[/tex]  

The second number =7×[tex]10^{-1}[/tex]

Here,

Convert the given numbers to the simplest form

The first number 1.092×[tex]10^{1}[/tex] =1.092×10

=10.92

The second numbers 7×[tex]10^{-1}[/tex]= 7×0.1

=0.7

We know the division is one the basic arithmetic operations in the mathematics. It is the process of the splitting a particular amount or number into equal parts.

To find how many time larger 1.092×[tex]10^{1}[/tex] than 7×[tex]10^{-1}[/tex], we have to divide the first number by second number.

10.92 ÷ 0.7 =15.6

Hence, the number 1.092×[tex]10^{1}[/tex] is 15.6 times larger than 7×[tex]10^{-1}[/tex]

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Solving a quadratic equation, it is found that the largest possible value of a is of a = 5.

A quadratic equation is modeled by the following rule:

y = ax^2 + bx + c

The solutions of the equations are the values of x for which y = 0.

For this problem, the equation is:

a² - 6a - 9 = b.

b = -9 - a, hence we replace in the equation as follows;

a² - 6a - 9 = -9 - a

a² - 5a = 0.

Traditionally, the solutions are given using Bhaskara's formula, however, since the coefficient c is zero, we can solve the equation by elimination as follows:

a² - 5a = 0.

a(a - 5) = 0.

Then the solutions are:

5 > 0, hence the largest possible value of a is of a = 5.

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the value y = 0 does not belong to the range, that is why we need to use parenthesis next to the range.

When we write in interval notation, the parenthesis are used to denote that the "ends" of the interval don't belong to the interval.

This means that in (a, b).

a and b don't belong to that interval.

While if instead we wrote [a, b]

In that case, a and b belong.

For the case of the exponential function:

y = a^x

As x tends to really large negative values, the value of y will tend to zero asymptotically. This means that y is really close to zero, but it is never equal to zero.

Then the value y = 0 does not belong to the range, that is why we need to use parenthesis next to the range.

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The critical number of the given function {g(t) = 3t ㏑t} is t = e⁻¹

Assume f(x) is a function with an open interval (a,b). If c is in the interval (a,b) and either f'(c) =0 or f'(c) does not exist, we say c is a critical number of f(x).

Each local minimum as well as maximum of f(x) must correspond to a critical number of f(x). The opposite is not truly the case: not all critical number of f(x) must correspond to a local maximum as well as minimum of f(x).

Now, as per the question;

It should be noted that the broadest possible domain of {g(t) = 3t ㏑t} is given in the set {t : t > 0} as ㏑t can not be defined for negative numbers of zero.

Now, differentiate the given function with respect to t.

[tex]g^{\prime}(t)=\frac{d}{d t}(3 t \ln t)[/tex]

Apply the product rule of differentiation;

[tex]\begin{aligned}&=\left[\frac{d}{d t}(3 t)\right] \ln t+3 t \frac{d}{d t}(\ln t) \\&=3 \ln t+3 t \frac{1}{t} \\&=3 \ln t+3 \\&=3(\ln t+1)\end{aligned}[/tex]

Now, put this value of g'(t) = 0 for finding the critical points.

0 = g'(t)

  = 3(㏑t + 1)

Divide equation by 3.

0 = ㏑t + 1

㏑t = -1

Taking anti log;

t = e⁻¹ (critical value)

Therefore, the critical value obtained for the given function is  t = e⁻¹.

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The complete question is-

how many critical numbers does {g(t) = 3t ㏑t} have? note: critical numbers must be within the domain of the function.

The rope's diameter is 10⁻³ inches greater than that of the floss, for example.

The diameter of a stand of rope is 1.2 × 10⁻³ inch.

The diameter of a strand of floss is 2 × 10⁻⁴ inches.

Subtract the diameter of the floss strand from the diameter of the rope to determine how much longer the diameter of the rope is than the diameter of the floss.

Therefore, the length by which the diameter of the strand of rope is longer than the diameter of the strand of floss will be:

= 1.2 × 10⁻³ inch - 2 × 10⁻⁴ inches

Now,

= 12 × 10⁻⁴ inch - 2 × 10⁻⁴ inch

= 10 × 10⁻⁴ inch

= 10⁻³ inch

The diameter of the strand of rope is 10⁻³ inches longer than the diameter of the strand of floss.

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The interval to find the true average tip is  (14.62,13.78).

Average is a number that is calculated by adding quantities together and dividing the total by the number of quantities.

The Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range

Given:

We know that the confidence interval = sample mean ± standard deviation substitute the values in it then confidence interval = 14.2  ±2.5/(35)^1/2

confidence interval = 14.2 ± 2.5/5.91

confidence interval = 14.2 ± 0.42

confidence interval = 14.2 + 0.42 ,

confidence interval = 14.2 - 0.42

confidence interval = (14.62,13.78)

So, an interval to estimate the true average tip is  (14.62,13.78).

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Answer: 3.23

Step-by-step explanation:

Add 1/6 + 1/7 = 13/42 Take the reciprocal = 42/13 hrs ≈ 3.23 hrs

The interest rate of the account is 2%

From the question, we are to determine the interest rate of the account

From the compound interest formula, we have that

A = P(1 + r/n)^{nt}

Where A is the amount

P is the principal

r is the rate

n is the number of times compounded

t is the time in years

From the given information,

A = $1,325.71

P = $1,130

n = 6

t = 8 years

Thus,

1325.71 = 1130(1 + r/6)⁶⁽⁸⁾

1325.71 = 1130( 1 + r/6)⁴⁸

1325.71/1130 = ( 1 + r/6)⁴⁸

Take the 48th root of both sides

1.003333 = 1 + r/6

1.003333 - 1 = r/6

0.003333 = r/6

r = 6 × 0.003333

r = 0.019998

r ≈ 0.02

r ≈ 2%

Hence, the interest rate of the account is 2%.

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Step 1) : We know that,

[tex]\begin{cases}P = 7000& \\r = 5.6 \% & \\ n = 4 & \\ t = 4 & \end{cases}[/tex]

Step 2) : Formulate and Substitute:

[tex]Substitute\begin{cases}P = 7000& \\ r = 5.6 \% & into \: formula \\ n = 4 & \\ t = 4 & \end{cases}[/tex]

[tex]\bf\longrightarrow{F = P * (1 + \frac{r}{n} )^{(n*t)}} [/tex]

Put the values according to the formula:

[tex]\bf\longrightarrow{F = 7000 * (1 + \frac{ \frac{5.67}{100} }{4} )^{(4*4)}}[/tex]

Step 3) : Evaluate the equation/expression:

[tex]\bf\longrightarrow{8768.1}[/tex]

15 protrusions were counted by her.

Clearly, there are multiple ways to treat verruca. For this type of patient, we would try a prescription for Drysol to apply topically once daily. In many of these patients, the skin is too moist. In our practice, we write a prescription for physical therapy for ultrasound application to the verruca. There is no guarantee with any treatment for verruca. Vitamin A adjunct treatment is often instituted: approx 7500 IU /day for 3 weeks.

Let no. of protrusions be x

3 * x = -4 * (-x) - 15

3x-4x = -15

x = 15

Therefore 15 protrusions were counted by her.

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Answer: (-4-[tex]\sqrt{11}[/tex], -4+[tex]\sqrt{11}[/tex]) ==> B

Step-by-step explanation:

x^2+8x+5<0

x^2+8x+16-11<0

(x+4)^2-11<0

(x+4)^2<11

x+4<[tex]\sqrt{11}[/tex]

x<-4+[tex]\sqrt{11}[/tex]

x+4>-[tex]\sqrt{11}[/tex]

x>-4-[tex]\sqrt{11}[/tex]

(-4-[tex]\sqrt{11}[/tex], -4+[tex]\sqrt{11}[/tex]) ==> B

Remember, the solution doesn't include the x values -4-[tex]\sqrt{11}[/tex] and -4+[tex]\sqrt{11}[/tex]  since if they were plugged in x^2+8x+5, the expression would equal 0. The expression is supposed to be LESS than 0, not equal to 0.

Answer:

Answer: (-4-\sqrt{11}11 , -4+\sqrt{11}11 )

x^2+8x+5<0

x^2+8x+16-11<0

(x+4)^2-11<0

(x+4)^2<11

x+4<\sqrt{11}11

x<-4+\sqrt{11}11

x+4>-\sqrt{11}11

x>-4-\sqrt{11}11

(-4-\sqrt{11}11 , -4+\sqrt{11}11 ) 

The radius of 8th sphere is 3/2.

The 8th sphere's center and the hexagon's two opposing ends should form an isosceles triangle. Then put up another triangles between the point of tangency of the 7th and 8th spheres, and the point of tangency between the 7th sphere and 2 of the original spheres on different sides of the hexagon. Each triangle's side length should be expressed in terms of r (the sphere's 8 radius) and h. (the height of the first triangle). The values of h and r may then be determined by setting up two equations for the two triangles using the Pythagorean Theorem.

(1+r)² = [tex]2^{2}[/tex] + [tex]h^{2}[/tex]

(3√2)² =3² + (h+r)²

r= 3/2

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The value of x from the given expression is 1

If the point M is on the line LN, then the equivalent addition postulate is expressed as;

LM+MN = LN

Given the following

LN=9,

MN =4x

LM=5x

Substitute

5x + 4x = 9

9x = 9

x =9/9

x = 1

Hence the value of x from the given expression is 1

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The equation of the line passing through (14, 25) and (9, -10) is given by y = 7x - 73

An equation is an expression that shows the relationship between two or more numbers and variables.

The standard equation of a line is:

y = mx + b

Where m is the rate of change and b is the y intercept

The equation of the line passing through the point (14, 25) and (9, -10) is:

y - 25 = [(-10 - 25)/(9 - 14)](x - 14)

y = 7x - 73

The equation of the line passing through (14, 25) and (9, -10) is given by y = 7x - 73

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Answer:

The "different" answer is BC-AC and AB

The "same" answer is AC+CB and CA+BC

Step-by-step explanation:

AC+CB and CA+BC if are rearranged wold be the same CA becomes AC and CB becomes BC

Answer:

[tex]27^{\frac{1}{3}} \quad 125^{\frac{2}{3}} \quad 9^{\frac{3}{2}}[/tex]

Step-by-step explanation:

Rewrite 9 as 3²:

[tex]\implies 9^{\frac{3}{2}}=\left(3^2\right)^{\frac{3}{2}}[/tex]

[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]

[tex]\implies 3^{(2 \cdot \frac{3}{2})}=3^3[/tex]

Therefore:

[tex]\implies 3^3= 3 \cdot 3 \cdot 3=27[/tex]

---------------------------------------------------------

Rewrite 27 as 3³:

[tex]\implies 27^{\frac{1}{3}}=\left(3^3\right)^{\frac{1}{3}}[/tex]

[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]

[tex]\implies 3^{(3 \cdot \frac{1}{3})}=3^1[/tex]

Therefore:

[tex]\implies 3^1=3[/tex]

---------------------------------------------------------

Rewrite 125 as 5³:

[tex]\implies 125^{\frac{2}{3}}=\left(5^3\right)^{\frac{2}{3}}[/tex]

[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]

[tex]\implies 5^{(3 \cdot \frac{2}{3})}=5^2[/tex]

Therefore:

[tex]\implies 5^2=5 \cdot 5=25[/tex]

---------------------------------------------------------

Solution

In order, from smallest to largest:

[tex]27^{\frac{1}{3}} \quad 125^{\frac{2}{3}} \quad 9^{\frac{3}{2}}[/tex]

Answer:

addingadddddddddddddddddd

The expression of Mia's height in cm is 5 + x

The given parameters are

Mia's height = 5 cm taller than her sister

Her sister's height  = x

The parameter "Mia's height = 5 cm taller than her sister" means that

Mia's height = 5 cm + her sister's height

Substitute Her sister's height  = x in Mia's height = 5 cm + her sister's height

Mia's height = 5 cm + x

This means that the expression of Mia's height in cm is 5 + x

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Answer:

A=2.5×x

10cm=2.5×x

10÷2.5=2.5÷2.5×x

x=4

x+12+4x-16= 6x-13

5x-4=6x-13

-4=x-13

x=9

9+12+4(9)-16

21+36-16

57-16= 41

JL=41 units

The required subtraction problem that is represented on the number line is 3 -(-5). Option B is correct.

Given that,
To determine the subtraction problem that is represented on the number line.

A number line is defined as the number marked on the line calibrated into an equal number of units. For example -1, 0, 1, and so on.

Here,
The line marked above the number line from -5 to 3
The subtraction problem would be given as 3 - (-5)

Thus, the required subtraction problem that is represented on the number line is 3 -(-5). Option B is correct.

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Since questions seem to be incomplete,

The given solution is based on the question given below,
Which subtraction problem is represented by the number line?

A: 3-5

B: 3 - (-5)

C: 3-8

D: 3-(-8)

Answer:

16a

Step-by-step explanation:

14a+2=14+2+a=16a.

Answer:

There are 98 terms in this sequence.

Answer:

he has $6 left.

Step-by-step explanation:

Answer:

$18.99 for each student

Step-by-step explanation:

First we need to find out how much he spent on each student for the calculators.

We can do that by dividing 84.14, the total, by the number of students, 14:

84.14/14 = 6.01

He spent 6.01 on each student and has 25 for each student, so we can subtract that:

25 - 6.01 = 18.99

Pattern uses 6233.76 inches² fabric.

Firstly we will convert units in required form. Therefore, converting equivalent of yard² in inches². As per the given information, 1 yard = 3 feet and 1 foot = 12 inches.

As 1 foot = 12 inches, so, 3 feet = 36 inches, which is further equal to 1 yard.

Now, 1 yard² = (36 inches)²

1 yard² = 1296 inches²

Coming back to the problem, we will now calculate inches² from fabric measurement in yard². So, relating the units for conversion.

1 yard² = 1296 inches²

4.81 yard² = 1296 × 4.81

Performing multiplication

Measurement of 4.81 yard² in inches² = 6233.76

Hence, the pattern uses 6233.76 inches² fabric.

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