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3 years ago

The double number line shows that 111 pound of chocolate costs \$4$4dollar sign, 4.

Mathematics

2 answers:

alexandr1967 [171]3 years ago

8 0

In Khan Acadmy the answer is A

Yuliya22 [10]3 years ago

3 0

hey thanks for deleting my answer

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Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

3 years ago

Solve for x<br><br>-3x + 5= -8 + 10

IRINA_888 [86]

Answer:

Step-by-step explanation:

you should remove 8 out of 10 then 2 will remain

then add negative 5 to both sides then we will left with -3x = -3

divide -3 on both sides we find x = 1

5 0

3 years ago

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Customers generally find waiting in line to be an unpleasant experience. In addition to reducing the length of the wait itself,

garri49 [273]

Answer:

OOP

Step-by-step explanation:

5 0

3 years ago

Marco and Drew stacked boxes on a shelf. Marco lifted 9 boxes and Drew lifted 14 boxes. The boxes that Drew lifted each weighed

Ghella [55]

Marco and Drew stacked boxes on a shelf. Marco lifted 9 boxes and Drew lifted 14 boxes. The boxes that Drew lifted each weighed 8 lb less than the boxes Marco lifted.

m − 8

14(m − 8) **

9(m + 8)

112m

hope this helps

7 0

3 years ago

Read 2 more answers

Match the coordinates with the quadrMatch the coordinates with the quadrants they lie in. || Quadrant I || Quadrant II || Quadra

Dmitry_Shevchenko [17]

Answer:

Points B,E,F Quadrant 1

Points A,C,D Quadrant 2

Points G,I,K Quadrant 3

Points H,L,M Quadrant 4

Step-by-step explanation:

I've labeled the quadrants for you, take a look.

If you ever need more help, don't be afraid to reach out! I hope this helps!

5 0

3 years ago