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3 years ago

15

A square pyramid has a base of 6 inches. The height of each triangle face is 11 inches. Which expression correctly calculate the

surface area of the pyramid

1 answer:

stira [4]3 years ago

6 0

Step-by-step explanation:

see the image for explanation.

please ask if something not clear.

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Which statements are true? check all that apply.the line x = 0 is perpendicular to the line y = –3.all lines that are parallel t

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The first and the second statements are correct
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three times the sum of a number and -5 equals -6 times the sum of the number and -2. Find the number.

Rashid [163]

3(x-5)=-6(x-2)

3x-15=-6x+12

9x-15=12

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Para las celebraciones del barrio de Santiago se junto cierta cantidad de dinero que se distribuirá de la siguiente forma:?

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4,767

Step-by-step explanation:

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2 years ago

What is the measure of what is the measure of (either 50,90,95, or 115)

Anuta_ua [19.1K]

Answer:

65 & 115

Step-by-step explanation:

<EGF = (180 - 50) /2 = 65 degree

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3 years ago

Franchise Business Review stated over 50% of all food franchises earn a profit of less than $50,000 a year. In a sample of 130 c

nydimaria [60]

Answer:

We need a sample size of 564.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = \frac{81}{130} = 0.6231$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.

We need a sample size of n

n is found when $M = 0.04$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.6231*0.3769}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.6231*0.3769}$

$\sqrt{n} = \frac{1.96\sqrt{0.6231*0.3769}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.6231*0.3769}}{0.04})^{2}$

$n = 563.8$

Rounding up

We need a sample size of 564.

4 0

3 years ago