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2 years ago

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Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we

randomly sample 350 students, which of the following statements about the sampling distribution of the sample mean age is incorrect?
a. The shape of the sampling distribution is approximately normal.
b. The mean of the sampling distribution is approximately 24-years old.
c. The standard deviation of the sampling distribution is equal to 5 years.
d. All of the above statements are correct.

1 answer:

natta225 [31]2 years ago

8 0

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Mean of students' age = 24 years

Standard deviation of students' age = 3 years

Sample size = number of students = 350

So, according to options,

a. The shape of the sampling distribution is approximately normal.

It is true as n >30, we will use normal.

b. The mean of the sampling distribution is approximately 24-years old.

It is true as it is given.

c. The standard deviation of the sampling distribution is equal to 5 years.

It is not true as it is given 3 years.

Hence, Option 'c' is correct.

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r Flag For Review A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to h

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Answer:

$P = \frac{132000}{12}=11000$

So then P =11000 is the minimum that the least populated district could have.

Step-by-step explanation:

We have a big total of N = 132000 for the population.

And we know that we divide this population into 11 districts

And we have this info given "no district is to have a population that is more than 10 percent greater than the population of any other district"

Let's assume that P represent our minimum value for a district in the population. The range of possible values for the population of each district would be between P and 1.1 P

The interest on this case is find the minimum value for P and in order to do this we can assume that 1 district present the minimum and the other 10 the maximum value 1.1P in order to find which value of P satisfy this condition, and we have this:

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So then P =11000 is the minimum that the least populated district could have.

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2 years ago

a guy wire makes a 67 degree angle with the ground. walking out 32 ft further grom the tower,the angle of elevation to the top o

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Answer:

39.5 ft

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relation between angles and sides of a right triangle.

Tan = Opposite/Adjacent

This lets us write two equations in two unknowns:

tan(67°) = AD/CD . . . . . . . . . . angle at guy point

tan(39°) = AD/(CD+32) . . . . . .angle 32' farther

__

Solving the first equation for CD and using that in the second equation, we can get an equation for AD, the height of the tower.

CD = AD/tan(67°)

tan(39°)(CD +32) = AD . . . . eliminate fractions in the second equation

tan(39°)(AD/tan(67°) +32) = AD

32·tan(39°) = AD(1 -tan(39°)/tan(67°)) . . . simplify, subtract left-side AD term

32·tan(39°)tan(67°)/(tan(67°) -tan(39°)) = AD . . . . divide by AD coefficient

AD ≈ 39.486 . . . . feet

The tower is about 39.5 feet high.

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2 years ago

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Answer:

(a) $Perimeter = 32.2\ cm$

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

$Area = 64.8$

Required

Calculate the perimeter.

Area is calculated as:

$Area = L * W$

Where

$L = Length$ and $W = Width$

Substitute 64.8 for Area

$64.8 = L * W$

Make L the subject:

$L = \frac{64.8}{W}$

Perimeter is calculated as:

$P = 2 * (L + W)$

Substitute 64.8/W for L

$P = 2 * (\frac{64.8}{W} + W)$

$P = \frac{129.6}{W} + 2W$

To solve further, we take the derivative of P and set it to 0, afterwards.

$dP = -\frac{129.6}{W^2} + 2$

Set to 0

$0 = -\frac{129.6}{W^2} + 2$

Collect Like Terms

$\frac{129.6}{W^2} = 2$

Cross Multiply:

$2W^2= 129.6$

Divide through by 2

$W^2 = 64.8$

Take square roots

$W = \sqrt{64.8$

$W = 8.05$

Recall that:

$L = \frac{64.8}{W}$

So:

$L= \frac{64.8}{8.05}\\$

$L= 8.05$

The perimeter is:

$Perimeter = 2 * (8.05 + 8.05)$

$Perimeter = 2 * (16.10)$

$Perimeter = 32.2\ cm$

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

$Tenths = \frac{1}{10}$

So, the expression becomes:

((1 Group of $10 * \frac{1}{10}$ ) and (1 group of $8 * \frac{1}{10}$ ))/(6 groups of $3 * \frac{1}{10}$ )

This gives:

((1 Group of $\frac{10}{10}$ ) and (1 group of $\frac{8}{10}$ ))/(6 groups of $\frac{3}{10}$ )

Group means product, so the expression becomes:

$\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

And, as used here means addition

$\frac{(1 * \frac{10}{10} + 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

Simplify:

$\frac{(1 * 1 + 1 * 0.80)}{6 *0.30}$

$\frac{(1 + 0.80)}{1.80}$

$\frac{1.80}{1.80}$

$= 1.00$

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