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vredina [299]
3 years ago
8

Frank and his family are going to the grand opening of the circus there is a special price on tickets this weekend tickets cost

$36 each this is of 60% of the cost of a regular price ticket what is the cause of a regular price ticket
Mathematics
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

$60

Step-by-step explanation:

36/0.6 = 60

60 x 0.6 = 36

I hope that's correct? :)

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Jesse enjoys watching ‘Rudolph the Red-Nosed Reindeer’ during the holidays. If Jesse watches the movie three times every year, h
pogonyaev

Answer:

<em>72</em>

Step-by-step explanation:

Since every year she watches it 3 times, she would watch 3·24 times when she is 24. 3·24=3·20+3·4=60+12=72

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Solve this inequality.<br> -35 – 2 &gt;7<br> A.<br> B.<br> C.<br> D. &gt; -3
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Answer:

-2> 7+35

-2>42

0<21

the answer is 0 is less than 21

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SOMEONE PLSSSS HELP ME WITH THIS QUESTION!!!
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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

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Answer:

1&a 2&b 3&c 4&d

Step-by-step explanation:

...

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3 years ago
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