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yanalaym [24]
3 years ago
9

Anna's clothing store was open for 24 hours last weekend. If the store made $2,496 in sales over the weekend, what is the averag

e amount of sales for each hour?
Mathematics
1 answer:
Alexxx [7]3 years ago
6 0

Answer:

maybe im just dumb but isnt it just 2496 divided by 24? if so the answer is 104. sorry if im wrong :/

Step-by-step explanation:

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gizmo_the_mogwai [7]

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1/4

25 percent

Step-by-step explanation:

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3 years ago
A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.
krek1111 [17]

Answer:

a) \hat p=\frac{1627}{5120}=0.3178

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

X=1627 number of permanent dwellings on the entire reservation that are traditional hogans.

n=5120 random sample taken

\hat p=\frac{1627}{5120}=0.318 estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

p true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

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}})

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

\hat p=\frac{1627}{5120}=0.3178

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:

z_{\alpha/2}=-2.58, z_{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:

0.318 - 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.301

0.318 + 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.335

The 99% confidence interval would be given by (0.301;0.355)

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Step-by-step explanation:

(7x+3) (10x-14)

70x^{2}-98+30x-42

70x^{2}+30x-42-98

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