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2 years ago

Find a confidence interval for estimating the population μ "unknown s"

Mathematics

1 answer:

marysya [2.9K]2 years ago

5 0

Answer:the answer would be the unknown

Step-by-step explanation:

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Julia is selling bracelets (B) for each $3 each and necklaces (N) for $8 each. Write a inequality to represent all of the combin

trapecia [35]

$3b + 8n \geqslant 150$

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3 years ago

What is the length of the hypotenuse in the right triangle shown below?

Mrrafil [7]

Picture needed. not enough info

8 0

3 years ago

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Logarithms:

WITCHER [35]

Answer:

A) k = 18.75
B) R(0.8) = 161419
C) R(0.3) = 275

Step-by-step explanation:

Given expression

R(x) = 6*e^(kx)

A) Given

x = 0.04
R = 10

Solving for k

10 = 6*e^(0.04k)
e^(0.04k) = 10/6
ln (e^0.04k) = ln (1.6666)
0.04k = 0.51
k = 0.51/0.04
k = 12.75

B) Given

x = 0.8
R= ?

The value of R(0.8) is:

R(0.8) =
6*e^(0.8*12.75) =
6*e^(10.2) =

161419 rounded

C) Given

x = 0.3
R = ?

The value of R(0.3) is:

R(0.3) =
6*e^(0.3*12.75) =
6*e^(3.825) =

275 rounded

4 0

3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago

-7 2/3 + (-5 1/3) + 8 3/4=? A: 6 7/12 B: -4 5/12 C: -21 11/12 D: 4 2/3

balandron [24]

Its B

Hope this helps!

May I have brainliest please? :D

4 0

3 years ago

Read 2 more answers