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Delicious77 [7]

3 years ago

7

Adam has game and education apps on his tablet. He noticed that he has 4 game apps for every 3 education apps. Which of the foll

owing is another way to write this ratio?
1) 1:3

2) 3:4

3) 4:3

4) 7:3

1 answer:

brilliants [131]3 years ago

7 0

Answer: 4:3

Step-by-step explanation: Because there are 4 game app for every 3 educational app.

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Which of the following correctly shows clustering for the problem 24 x 14?

lbvjy [14]

Answer:

the answer is d

Step-by-step explanation:

because Let's see if you understood the definition of an arithmetic sequence. Which of the following is *not* an arithmetic sequence?

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4, 4, 4, 4, 4, 4, 4, 4, 4...

2.13, 3.34, 4.55, 5.76, 6.97, 8.18, 9.39, 10.6, 11.81...

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3 years ago

Ples help me i woud be aprishated

worty [1.4K]

Answer:

Number 2 is A

Step-by-step explanation:

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5 0

3 years ago

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mixas84 [53]

Answer:

X = 9

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Step-by-step explanation:

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8 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

Is y=2x^2 a linear function

Savatey [412]

Answer:

No

Step-by-step explanation:

Since x is squared it is a parabola. A linear function is a line. Therefore it is not linear.

8 0

3 years ago