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Bingel [31]
2 years ago
9

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and ob

serving the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met
Mathematics
1 answer:
klio [65]2 years ago
6 0

Answer:

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.

A.  The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.

C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.

E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake

The answer is E.

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In an article in Statistics and Computing [""An Iterative Monte Carlo Method for Nonconjugate Bayesian Analysis"" (1991, pp. 119
Marina CMI [18]

Answer:

Step-by-step explanation:

Hello!

a)

The dependent variable is

Y: length of a dugong

The explanatory variable is

X: age of a dugong

You need to estimate the linear regression of the length of the dugongs as a function of their age.

Using the given data I've estimated the regression using a statistic software:

The regression model is E(Yi)= α + βXi

The estimated model is ^Yi= a + bXi

Where a is the estimate of the intercept and b is the estimate of the slope:

a= 2.02

b= 0.03

And the estimate of the population variance of the error is Se²= 0.03

The estimated regression equation is ^Yi= 2.02 + 0.03Xi

b)

You have to estimate the length of a dugong when its age is 11 years using the model, for this all you have to do is replace X=11 in the regression line and calculate the corresponding ^Y value:

^Yi= 2.02 + 0.03*11= 2.35

The average length of an 11-year-old dugong should be 2.35.

I hope it helps!

5 0
2 years ago
I need help with number 5 please
steposvetlana [31]

Answer: Would be 55% I hope this helps!

7 0
2 years ago
Read 2 more answers
What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.
GalinKa [24]

Options

A. UV = 14 ft and m∠TUV = 45°

B. TU = 26 ft

C. m∠STU = 37° and m∠VTU = 37°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

E. m∠UST = 98° and m ∠TUV = 45°

Answer:

A. UV = 14 ft and m∠TUV = 45°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Step-by-step explanation:

Given

See attachment for triangle

Required

What proves that: ΔSTU ≅ ΔVTU using SAS

To prove their similarity, we must check the corresponding sides and angles of both triangles

First:

\angle UST must equal \angle UVT

So:

\angle UST = \angle UVT = 98

Next:

UV must equal US.

So:

UV = US = 14

Also:

ST must equal VT

So:

ST = VT = 20

Lastly

\angle TUV must equal \angle TUS

So:

\angle TUV = \angle TUS = 45

Hence: Options A and D are correct

4 0
2 years ago
How do you write 7.75 in expanded form?
Rashid [163]
7 + 0.70 + 0.05 =7.75
7 0
2 years ago
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A 22.5 ounce candle burns at the rate of one ounce every 5 hours. write an equation for the amount of candle left
maks197457 [2]

Answer:

y = 22.5 - 0.2t

Step-by-step explanation:

Given;

total number of candle, n = 22.5 ounce

Rate of candle burn, R = 1 ounce per 5 hours  = \frac{1 \ ounce}{5 \ hours} = 0.2 \ \frac{ounce}{hour}

The amount of candle left = total initial value - amount burnt

let the amount let = y

y = 22.5 - 0.2t

where;

t is the time in which the candle is burnt

Thus, the equation for the amount of candle left is given by;

y = 22.5 - 0.2t

7 0
2 years ago
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