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3 years ago

Sue bought 7 notebooks for $8.75. Jolie bought 4 notebooks for $5.08. Who got the better bargain? By how much?

Mathematics

2 answers:

statuscvo [17]3 years ago

7 0

Answer:

sue.

Step-by-step explanation:

sue paid 1.25 per book while Joliet paid 1.27. hope that helps

Art [367]3 years ago

3 0

Answer:

235

Step-by-step explanation:

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In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

4) Part d

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

And rounded up we have that n=649

5 0

3 years ago

Emily read 24 fiction books and 2 nonfiction books. What is the ratio of the number of fiction books she read to the total numbe

Tems11 [23]

The ratio of number of fiction books to total number of books that she read 12:13.

Given :

Emily reads 24 friction books and 2 nonfiction books.

Number of fiction books she read =24

Number of nofiction boos she read = 2

Total number of books = 26

To Find :

What is the ratio of the number of fiction books she read to the total number of books that she read ?

Solution :

Total number of books that she read=24+2=26

number of fiction books she read=24

Thus ratio of number of fiction books to total number of books that she read

⇒ $\frac{24}{26}$

⇒ $\frac{12}{13}$

⇒ 12: 13

Hence ratio of number of fiction books to total number of books that she read 12 : 13

To know more about total number of books visit :

brainly.com/question/1179123

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4 0

1 year ago

The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate,

Veronika [31]

Um not to sure about that

7 0

3 years ago

Can someone check this for me thanks so much

Zinaida [17]

Answer:

that looks like it is right

3 0

3 years ago

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Math HELP PLEASEE URGENT NOWWWWWW MATH!!!!

Viefleur [7K]

Answer:

Step-by-step explanation:

Answer is the first choice. you can tell by the notation.

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3 years ago

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