Answer:
[tex]\mathbf{V = 1296 \pi }[/tex]
Step-by-step explanation:
Given that :
Find the volume of the region enclosed by the cylinder [tex]x^2 + y^2 =36[/tex] and the plane z = 0 and y + z = 36
From y + z = 36
z = 36 - y
The volume of the region can be represented by the equation:
[tex]V = \int\limits \int\limits_D(36-y)dA[/tex]
In this case;
D is the region given by [tex]x^2 + y^2 = 36[/tex]
Relating this to polar coordinates
x = rcosθ y = rsinθ
x² + y² = r²
x² + y² = 36
r² = 36
r = [tex]\sqrt{36}[/tex]
r = 6
dA = rdrdθ
r → 0 to 6
θ to 0 to 2π
Therefore:
[tex]V = \int\limits^{2 \pi} _0 \int\limits ^6_0 (36-r sin \theta ) (rdrd \theta)[/tex]
[tex]V = \int\limits^{2 \pi} _0 \int\limits ^6_0 (36-r^2 sin \theta ) drd \theta[/tex]
[tex]V = \int\limits^{2 \pi} _0 [\dfrac{36r^2}{2}- \dfrac{r^3}{3}sin \theta]^6_0 \ d\theta[/tex]
[tex]V = \int\limits^{2 \pi} _0 [648- \dfrac{216}{3}sin \theta]d\theta[/tex]
[tex]V = \int\limits^{2 \pi} _0 [648+\dfrac{216}{3}cos \theta]d\theta[/tex]
[tex]V = [648+\dfrac{216}{3}cos \theta]^{2 \pi}_0[/tex]
[tex]V = [648(2 \pi -0)+\dfrac{216}{3}(1-1)][/tex]
[tex]V = [648(2 \pi )+\dfrac{216}{3}(0)][/tex]
[tex]V = 648(2 \pi )[/tex]
[tex]\mathbf{V = 1296 \pi }[/tex]
Losses covered by a flood insurance policy are uniformly distributed on the interval (0,2). The insurer pays the amount of the loss in excess of a deductible d. The probability that the insurer pays at least 1.20 on a random loss is 0.30. Calculate the probability that the insurer pays at least 1.44 on a random loss.
Answer:
The probability that the insurer pays at least 1.44 on a random loss is 0.18.
Step-by-step explanation:
Let the random variable X represent the losses covered by a flood insurance policy.
The random variable X follows a Uniform distribution with parameters a = 0 and b = 2.
The probability density function of X is:
[tex]f_{X}(x)=\frac{1}{b-a};\ a<X<b\\\\\Rightarrow f_{X}(x)=\frac{1}{2}[/tex]
It is provided, the probability that the insurer pays at least 1.20 on a random loss is 0.30.
That is:
[tex]P(X\geq 1.2+d)=0.30\\[/tex]
⇒
[tex]P(X\geq 1.2+d)=\int\limits^{2}_{1.2+d}{\frac{1}{2}}\, dx[/tex]
[tex]0.30=\frac{2-1.2-d}{2}\\\\0.60=0.80-d\\\\d=0.80-0.60\\\\d=0.20[/tex]
The deductible d is 0.20.
Compute the probability that the insurer pays at least 1.44 on a random loss as follows:
[tex]P(X\geq 1.44+d)=P(X\geq 1.64)[/tex]
[tex]=\int\limits^{2}_{1.64}{\frac{1}{2}}\, dx\\\\=|\frac{x}{2}|\limits^{2}_{1.64}\\\\=\frac{2-1.64}{2}\\\\=0.18[/tex]
Thus, the probability that the insurer pays at least 1.44 on a random loss is 0.18.
Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that 300 claimed they always buckle up.
We are interested in the population proportion of drivers who claim they always buckle up.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
(i) Enter an exact number as an integer, fraction, or decimal.
x =
(ii) Enter an exact number as an integer, fraction, or decimal.
n =
(iii) Round your answer to four decimal places.
p' =
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P' _ ( , )
Answer:
x = 300
n = 410
p' = 0.7317
[tex]\mathbf{P' \sim Normal (\mu = 0.7317, \sigma = 0.02188)}[/tex]
Step-by-step explanation:
From the given information;
the objective is to answer the following:
(i) Enter an exact number as an integer, fraction, or decimal.
Mean x = 300
(ii) Enter an exact number as an integer, fraction, or decimal.
Sample size n = 410
(iii) Round your answer to four decimal places.
Sample proportion p' of the drivers who always claimed they buckle up is :
p' = x/n
p' = 300/410
p' = 0.7317
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P' _ ( , )
The normal distribution is required to be used because we are interested in proportions and the sample size is large.
Let consider X to be the random variable that follows a normal distribution.
X represent the number of people that always claim they buckle up
∴
[tex]P' \sim Normal (\mu = p' , \sigma = \sqrt{\dfrac{p(1-p)}{n}})[/tex]
[tex]P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.7317(1-0.7317)}{410}})[/tex]
[tex]P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.7317(0.2683)}{410}})[/tex]
[tex]P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.19631511}{410}})[/tex]
[tex]P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{4.78817341*10^{-4}})[/tex]
[tex]\mathbf{P' \sim Normal (\mu = 0.7317, \sigma = 0.02188)}[/tex]
Solving exponential functions
Answer:
approximately 30Step-by-step explanation:
[tex]f(x) = 4 {e}^{x} [/tex]
[tex]f(2) = 4 {e}^{2} [/tex]
[tex]f(2) = 4 \times 7.389[/tex]
[tex]f(2) = 29.6[/tex]
( Approximately 30)
Hope this helps..
Good luck on your assignment..
Answer:
approximately 30
Step-by-step explanation:
[tex]f(x)=4e^x[/tex]
Put x as 2 and evaluate.
[tex]f(2)=4e^2[/tex]
[tex]f(2)=4(2.718282)^2[/tex]
[tex]f(2)= 29.556224 \approx 30[/tex]
Explain how the interquartile range of a data set can be used to identify outliers. The interquartile range (IQR) of a data set can be used to identify outliers because data values that are ▼ less than equal to greater than ▼ IQR Upper Q 3 minus 1.5 (IQR )Upper Q 3 plus IQR Upper Q 3 plus 1.5 (IQR )or ▼ less than equal to greater than ▼ IQR Upper Q 1 plus 1.5 (IQR )Upper Q 1 minus IQR Upper Q 1 minus 1.5 (IQR )are considered outliers.
Answer:
- greater than Upper Q 3 plus 1.5 (IQR)
- less than Upper Q 1 minus 1.5 (IQR)
Step-by-step explanation:
To identify outliers the interquartile range of the dataset can be used
Outliers can be identified as data values that are
- greater than Upper Q 3 plus 1.5 (IQR)
- less than Upper Q 1 minus 1.5 (IQR)
Using the interquartile range concept, it is found that:
The interquartile range (IQR) of a data set can be used to identify outliers because data values that are 1.5IQR less than Q1 and 1.5IQR more than Q3 and considered outliers.
----------------------------
The interquartile range of a data-set is composed by values between the 25th percentile(Q1) and the 75th percentile(Q3).It's length is: [tex]IQR = Q3 - Q1[/tex]Values that are more than 1.5IQR from the quartiles are considered outliers, that is:[tex]v < Q1 - 1.5IQR[/tex] or [tex]v > Q3 + 1.5IQR[/tex]
Thus:
The interquartile range (IQR) of a data set can be used to identify outliers because data values that are 1.5IQR less than Q1 and 1.5IQR more than Q3 and considered outliers.
A similar problem is given at https://brainly.com/question/14683936
Need Assistance With This
*Please Show Work*
Answer:
a =7.5
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2+ b^2 = c^2 where a and b are the legs and c is the hypotenuse
a^2 + 10 ^2 = 12.5^2
a^2 + 100 =156.25
Subtract 100 from each side
a^2 = 56.25
Take the square root of each side
sqrt(a^2) = sqrt( 56.25)
a =7.5
Find the slope on the graph. Write your answer as a fraction or a whole number, not a mixed number or decimal.
Explanation:
Two points on this line are (0,1) and (2,0)
Use the slope formula
m = (y2-y1)/(x2-x1)
m = (0-1)/(2-0)
m = -1/2
The negative slope means the line goes downhill as you move from left to right.
what is the volume of the specker below volume of a cuboid 50cm 0.4m 45cm
Answer:
50*0.4*45=900cm²
Find the area under the standard normal probability distribution between the following pairs of z-scores. a. z=0 and z=3.00 e. z=−3.00 and z=0 b. z=0 and z=1.00 f. z=−1.00 and z=0 c. z=0 and z=2.00 g. z=−1.58 and z=0 d. z=0 and z=0.79 h. z=−0.79 and z=0
Answer:
a. P(0 < z < 3.00) = 0.4987
b. P(0 < z < 1.00) = 0.3414
c. P(0 < z < 2.00) = 0.4773
d. P(0 < z < 0.79) = 0.2852
e. P(-3.00 < z < 0) = 0.4987
f. P(-1.00 < z < 0) = 0.3414
g. P(-1.58 < z < 0) = 0.4429
h. P(-0.79 < z < 0) = 0.2852
Step-by-step explanation:
Find the area under the standard normal probability distribution between the following pairs of z-scores.
a. z=0 and z=3.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 3.00) = 0.9987
Thus;
P(0 < z < 3.00) = 0.9987 - 0.5
P(0 < z < 3.00) = 0.4987
b. b. z=0 and z=1.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 1.00) = 0.8414
Thus;
P(0 < z < 1.00) = 0.8414 - 0.5
P(0 < z < 1.00) = 0.3414
c. z=0 and z=2.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 2.00) = 0.9773
Thus;
P(0 < z < 2.00) = 0.9773 - 0.5
P(0 < z < 2.00) = 0.4773
d. z=0 and z=0.79
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 0.79) = 0.7852
Thus;
P(0 < z < 0.79) = 0.7852- 0.5
P(0 < z < 0.79) = 0.2852
e. z=−3.00 and z=0
From the standard normal distribution tables,
P(Z< -3.00) = 0.0014 and P(Z< 0) = 0.5
Thus;
P(-3.00 < z < 0 ) = 0.5 - 0.0013
P(-3.00 < z < 0) = 0.4987
f. z=−1.00 and z=0
From the standard normal distribution tables,
P(Z< -1.00) = 0.1587 and P(Z< 0) = 0.5
Thus;
P(-1.00 < z < 0 ) = 0.5 - 0.1586
P(-1.00 < z < 0) = 0.3414
g. z=−1.58 and z=0
From the standard normal distribution tables,
P(Z< -1.58) = 0.0571 and P(Z< 0) = 0.5
Thus;
P(-1.58 < z < 0 ) = 0.5 - 0.0571
P(-1.58 < z < 0) = 0.4429
h. z=−0.79 and z=0
From the standard normal distribution tables,
P(Z< -0.79) = 0.2148 and P(Z< 0) = 0.5
Thus;
P(-0.79 < z < 0 ) = 0.5 - 0.2148
P(-0.79 < z < 0) = 0.2852
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard deviation of 2.1-cm. For shipment, 17 steel rods are bundled together. Find the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259-cm.
Answer:
The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.
Step-by-step explanation:
We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.
Let [tex]\bar X[/tex] = the average length of rods in a randomly selected bundle of steel rods
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean length of rods = 259.2 cm
[tex]\sigma[/tex] = standard deviaton = 2.1 cm
n = sample of steel rods = 17
Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P([tex]\bar X[/tex] > 259 cm)
P([tex]\bar X[/tex] > 259 cm) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{259-259.2}{\frac{2.1}{\sqrt{17} } }[/tex] ) = P(Z > -0.39) = P(Z < 0.39)
= 0.65173
The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.
Perform the indicated operation and write the result in standard form: (-3+2i)(-3-7i)
A. -5+27i
B. 23+15i
C. -5+15i
D. 23-15i
E-5-27I
Answer:
23+15i
Step-by-step explanation:
(-3+2i) (-3-7i)
multiply -3 w (-3+2i) and multiply -7i w (-3+2i)
9-6i+21i-14i^2
combine like terms
9+15i-14i^2
i squared is equal to -1 so
9+15i-(14x-1)
9+14+15i
23+15i
hope this helps :)
If x + 4 = 12, what is the value of x?
Answer:
8
Step-by-step explanation:
To find the answer to these problems you can work backwards
12-4=8
Evaluate the function y=1/2(x)-4 for each of the given domain values? PLZ HELP ME
Answer:
c. -13/4.
d. -13/3.
Step-by-step explanation:
c. f(3/2) = (1/2)(3/2) - 4
= 3 / 4 - 4
= 0.75 - 4
= -3.25
= -3 and 1/4
= -13/4.
d. f(-2/3) = (1/2)(-2/3) - 4
= -2/6 - 4
= -1/3 - 4
= -1/3 - 12/3
= -13/3.
Hope this helps!
Write your height in inches. Suppose it increases by 15%, what would your new height be? Now suppose your increased height decreases by 15% after the 15% increase; what is your new height?
Answer:
New height= 41.4 inches
Second new height= 36inches
Step-by-step explanation:
Height is assumed to be 36 inches
If it increases by 15%.
15%= 0.15
It's new height =( 36*0.15) +36
New height= 5.4+36
New height = 41.4 inches
This expression (36*0.15) is the expression of adding 15% to the height.
So if the 15% is taken away again , height= 41.4-(36*0.15)
Height= 41.4-5.4
Height= 36 inches
which of these shapes is congruent to given shape ?
Answer:
Step-by-step explanation:
shape D
Answer:
D.
Step-by-step explanation:
Well congruent means same size and same shape.
a) rectangle
This shape is a rectangle where as the given shape is a parallelogram.
This is not congruent to the given shape.
b) Parallelogram
This may be a parallelogram but it is too wide,
Hence, it is not congruent.
c) Rectangle
This is not a parallelogram,
Hence, this s not congruent
d) Parallelogram
This is a parallelogram with the same size just not in the same place but it is still congruent.
Thus, answer choices D. is the correct answer.
Solve the equation.
y + 3 = -y + 9
y= 1
y=3
y = 6
y = 9
Answer: y=3
Step-by-step explanation:
To solve the equation, we want to get the same terms onto the same side and solve.
y+3=-y+9 [add y on both sides]
2y+3=9 [subtract 3 on both sides]
2y=6 [divide 2 on both sides]
y=3
Answer:
y=3
Step-by-step explanation:
Please answer this correctly without making mistakes
Answer:
16 km
Step-by-step explanation:
Given:
Distance from Washington to Stamford = distance from Washington to Salem + distance from Salem to Stamford = 10.3 km + 11.9 km = 22.2 km
Distance from Washington to Oakdele = 6.2 km
Required: the difference between the distance from Washington to Stamford and from Washington to Oakdele
Solution:
Distance from Washington to Stamford = 22.2 km
Distance from Washington to Oakdele = 6.2 km
The difference = 22.2 km - 6.2 km = 16 km
Therefore, from Washington, it is 16 km farther to Stamford than to Oakdele.
find 10th term of a geometric sequence whose first two terms are 2 and -8. Please answer!!
Answer:
The 10th term is -524,288
Step-by-step explanation:
The general format of a geometric sequence is:
[tex]a_{n} = r*a_{n-1}[/tex]
In which r is the common ratio and [tex]a_{n+1}[/tex] is the previous term.
We can also use the following equation:
[tex]a_{n} = a_{1}*r^{n-1}[/tex]
In which [tex]a_{1}[/tex] is the first term.
The common ratio of a geometric sequence is the division of the term [tex]a_{n+1}[/tex] by the term [tex]a_{n}[/tex]
In this question:
[tex]a_{1} = 2, a_{2} = -8, r = \frac{-8}{2} = -4[/tex]
10th term:
[tex]a_{10} = 2*(-4)^{10-1} = -524288[/tex]
The 10th term is -524,288
An article reported that for a sample of 46 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 163.7.
Required:
a. Calculate and interpret a 9596 (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected.
b. Suppose the investigators had made a rough guess of 175 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 50 ppm for a confidence level of 95%?
Answer:
a) CI = ( 148,69 ; 243,31 )
b) n = 189
Step-by-step explanation:
a) If the Confidence Interval is 95 %
α = 5 % or α = 0,05 and α/2 = 0,025
citical value for α/2 = 0,025 is z(c) = 1,96
the MOE ( margin of error is )
1,96* s/√n
1,96* 163,7/ √46
MOE = 47,31
Then CI = 196 ± 47,31
CI = ( 148,69 ; 243,31 )
CI look very wide ( it sems that if sample size was too low )
b) Now if s (sample standard deviation) is 175, and we would like to have only 50 ppm width with Confidence level 95 %, we need to make
MOE = 25 = z(c) * s/√n
25*√n = z(c)* 175
√n = 1,96*175/25
√n = 13,72
n = 188,23
as n is an integer number we make n = 189
A manufacturer produces bolts of a fabric with a fixed width. The quantity q of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can write q = f(p). Then the total revenue earned with selling price p is R(p) = pf(p).
A. What does it mean to say that f(20)= 10,000 and f firstderivative (20)= -350?
B. Assuming the values in part a, find R first derivative (20).
Answer:
(A) the selling price is $20 per yards, and the expected yards to be sold is 10,000 yards
the derivative f'(20) is negative, which means the fabric producing company will sell 350 fewer yards when selling price is $20 per yard
(B) = R'(20) = $3000
∴the company will get extra $3000 revenue when selling price is $20 per yard
Step-by-step explanation:
A. given that
f(20)= 10,000
f'(20)= -350(first derivative)
the selling price is $20 per yards, and the expected yards to be sold is 10,000 yards
the derivative f'(20) is negative, which means the higher the price, it wil reduce the number of yards to be sold making it 350 fewer yards
(B) R(p) = p f(p)
f(20)= 10,000
f'(20)= -350(first derivative)
R(p) = p f(p)
differentiate with respect to p, using product rule
R'(p) = p f' (p) + f(p) (first derivative)
where p = 20
R'(20) = 20 f' (20) + f(20)
R'(20) = 20(-350) + 10,000
R'(20) = -7000 + 10,000
R'(20) = $3000
∴ the revenue is increasing by $3000 for every selling sold yard and increase in price per yard
If the triangle on the grid below is translated by using the rule (x, y) right-arrow (x + 5, y minus 2), what will be the coordinates of B prime?
Answer:
[tex]B'(x,y) = (0,-2)[/tex]
Step-by-step explanation:
Given
The attached grid
Translation rule: [tex](x,y) = (x + 5, y - 2)[/tex]
Required
Determine the coordinates of B'
First, we have to write out the coordinates of B
[tex]B(x,y) = (-5,0)[/tex]
Next is to apply the translation rule [tex](x,y) = (x + 5, y - 2)[/tex]
[tex]B(x,y) = (-5,0)[/tex] becomes
[tex]B'(x,y) = B(x+5,y-2)[/tex]
Substitute -5 for x and 0 for y
[tex]B'(x,y) = (-5+5,0-2)[/tex]
[tex]B'(x,y) = (0,-2)[/tex]
Pat is taking an economics course. Pat's exam strategy is to rely on luck for the next exam. The exam consists of 100 true-false questions. Pat plans to guess the answer to each question without reading it. If a grade on the exam is 60% or more, Pat will pass the exam. Find the probability that Pat will pass the exam.
Answer:
The probability that Pat will pass the exam is 0.02775.
Step-by-step explanation:
We are given that exam consists of 100 true-false questions. Pat plans to guess the answer to each question without reading it.
If a grade on the exam is 60% or more, Pat will pass the exam.
Let X = grade on the exam by Pat
The above situation can be represented through binomial distribution such that X ~ Binom(n = 100, p = 0.50).
Here the probability of success is 50% because there is a true-false question and there is a 50-50 chance of both being the correct answer.
Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).
So, the new mean of X, [tex]\mu[/tex] = [tex]n \times p[/tex] = [tex]100 \times 0.50[/tex] = 50
and the new standard deviation of X, [tex]\sigma[/tex] = [tex]\sqrt{n \times p \times (1-p)}[/tex]
= [tex]\sqrt{100 \times 0.50 \times (1-0.50)}[/tex]
= 5
So, X ~ Normal([tex]\mu=50, \sigma^{2} = 5^{2}[/tex])
Now, the probability that Pat will pass the exam is given by = P(X [tex]\geq[/tex] 60)
P(X [tex]\geq[/tex] 60) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{60-50}{5}[/tex] ) = P(Z [tex]\geq[/tex] 2) = 1 - P(Z < 2)
= 1 - 0.97725 = 0.02275
Hence, the probability that Pat will pass the exam is 0.02775.
10=12-x what would match this equation
Answer:
x=2
Step-by-step explanation:
12-10=2
Answer:
x=2
Step-by-step explanation:
10=12-x
Subtract 12 from each side
10-12 = 12-12-x
-2 =-x
Multiply by -1
2 = x
4.0.3x= 2.1 Equals what
Answer:
x= 1.75
Step-by-step explanation:
Answer:
1.75 = x?
Step-by-step explanation:
Select all the correct coordinate pairs and the correct graph. Select the correct zeros and the correct graph of the function below.
Answer:
(0, 0), (-1, 0), (2, 0), (3, 0) are the zeros.
First graph in top row is the answer.
Step-by-step explanation:
The given function is, f(x) = x⁴ - 4x³ + x² + 6x
For zeros of the given function, f(x) = 0
x⁴ - 4x³ + x² + 6x = 0
x(x³ - 4x² + x + 6) = 0
Therefore, x = 0 is the root.
Possible rational roots = [tex]\frac{\pm 1, \pm 2, \pm 3, \pm 6}{\pm1}[/tex]
= {±1. ±2, ±3, ±6}
By substituting x = -1 in the polynomial,
x⁴ - 4x³ + x² + 6x = (-1)⁴ - 4(-1)³+ (-1)² + 6(-1)
= 1 + 4 + 1 - 6
= 0
Therefore, x = -1 is also a root of this function.
For x = 2,
x⁴ - 4x³ + x² + 6x = (2)⁴ - 4(2)³+ (2)² + 6(2)
= 16 - 32 + 4 + 12
= 0
Therefore, x = 2 is a root of the function.
For x = 3,
x⁴ - 4x³ + x² + 6x = (3)⁴ - 4(3)³+ (3)² + 6(3)
= 81 - 108 + 9 + 18
= 0
Therefore, x = 3 is a root of the function.
x = 0, -1, 2, 3 are the roots of the given function.
In other words, (0, 0), (-1, 0), (2, 0), (3, 0) are the zeros.
From these points, first graph in top row is the answer.
Suppose that a forester wants to see if the average height of lodgepole pines in Yellowstone is different from the national average of 70 ft. The standard deviation lodgepole pine height is known to be 9.0 ft. The forester decides to measure the height of 19 trees in Yellowstone and use a one-sample z-test with a significance level of 0.01. She constructs the following null and alternative hypotheses, where mu is the mean height of lodgepole pines in Yellowstone.
H_0: mu = 70
H_1: mu notequalto 70
Use software to determine the power of the hypothesis test if the true mean height of lodgepole pines in Yellowstone is 62 ft. You may find one of these software manuals useful. Write your answer in decimal form and round to three decimal places.
Power =
Answer:
You can use your graphing calculator to find the answer.
Go to STAT, then TESTS, and hit "1: Z-Test..."
Make sure it is set to Stats, then for mu0, do 70; for standard deviation, do 9; for mean, you do 62; for sample size, you do 19. For mu, you do not equal to mu0. Then you hit "Calculate".
You then get a z-value (critical value) of -3.874576839, and a p-value of 0.00010685098.
This means that...
We reject the null hypothesis that the average height of lodgepole pines in Yellowstone is 70 feet because p = 0.0001 is less than the significance level of alpha = 0.01. There is sufficient evidence to suggest that the mean height of lodgepole pines in Yellowstone is NOT equal to 70 feet.
Hope this helps!
Isabel works as a tutor for $8 an hour and as a waitress for $9 an hour. This month, she worked a combined total of 93 hours at her two jobs. Let t be the number of hours Isabel worked as a tutor this month. Write an expression for the combined total dollar amount she earned this month.
We know that t is the number of hours Isabel worked as a tutor, and that she worked for a total of 93 hours. This means that the number of hours she worked as a waitress is 93 - t hours.
Now, for each of the t hours she worked as a tutor, she earned $8, so the total amount Isabel earned from that job is 8t dollars.
For each of the 93 - t hours Isabel worked as a waitress, she earned $9, so the amount she earned from her job as a waitress is 9(93 - t) dollars.
Therefore, the total amount she earned is 8t + 9(93 - t) dollars. This can be simplified to 837 - t dollars.
Determine by inspection whether the vectors are linearly independent. Justify your answer.
[4 1], [3 9], [1 5], [-1 7]
Choose the correct answer below.
A.The set is linearly dependent because at least one of the vectors is a multiple of another vector.
B. The set is linearly independent because at least one of the vectors is a multiple of another vector.
C. The set is linearly dependent because there are four vectors but only two entries in each vector.
D. The set is linearly independent because there are four vectors in the set but only two entries in each vector.
Answer:
B. The set is linearly independent because at least one of the vectors is a multiple of another vector.
Step-by-step explanation:
A set of n vector of length n is linearly independent if the matrix with these vectors as column has none of zero determinant. The set of vectors is dependent if the determinant is zero. In the given question the vectors have no zero determinants therefore it is linearly independent.
(a) The only even prime number is ....
Answer:
2 is the only even prime number
Answer:
2
Step-by-step explanation:
2 is the only even prime number. There is no even prime number other than 2. Prime numbers are the numbers which can only be divided by 1 and the number itself.
A simple random sample of 20 third-grade children from a certain school district is selected, and each is given a test to measure his/her reading ability. You are interested in calculating a 95% confidence interval for the population mean score. In the sample, the mean score is 64 points, and the standard deviation is 12 points. What is the margin of error associated with the confidence interval
Answer:
Margin of Error = ME =± 5.2592
Step-by-step explanation:
In the given question n= 20 < 30
Then according to the central limit theorem z test will be applied in which the standard error will be σ/√n.
Sample Mean = μ = 64
Standard Deviation= S= σ = 12
Confidence Interval = 95 %
α= 0.05
Critical Value for two tailed test for ∝= 0.05 = ±1.96
Margin of Error = ME = Standard Error *Critical Value
ME = 12/√20( ±1.96)=
ME = 2.6833*( ±1.96)= ± 5.2592
The standard error for this test is σ/√n
=12/√20
=2.6833
Emma words in a coffee shop where she is paid at the same hourly rate each day. She was paid $71.25 for working 7.5 hours on Monday. If she worked 6 hours on Tuesday, how much was she paid on Tuesday
Answer:
$57
Since $71.25 was paid for working 7.5 hours.
That means he was being paid $9.5 per hour.
Which is 71.25÷7.5.
And on tuesday that's 9.5×6 which is $57