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3 years ago

Gio is studying the rectangular pyramid below.

Mathematics

2 answers:

Irina18 [472]3 years ago

7 0

Answer:

a he used the wrong values as the bases of the lateral faces.

Step-by-step explanation:

3241004551 [841]3 years ago

7 0

Answer:

Step-by-step explanation:

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A linear set of points with a unique starting point and extending infinitely in one direction what term matches the definition?

bija089 [108]

Answer:- Option A "ray" is the right term which matches with the definition.

Explanation:-

A ray is a line that has one fixed endpoint, and extends infinitely along the line from the fixed endpoint.

Therefore, the term which matches with the given definition is "ray".

Thus A linear set of points with a unique starting point and extending infinitely in one direction is called a ray.

5 0

3 years ago

Read 2 more answers

Please help quick i have 10 minutes

zubka84 [21]

A=6
B= -1/2
C= 11
I had this one on my test and I got it right have a good day

3 0

3 years ago

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

Naddika [18.5K]

Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that $\sigma = 6$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$1.2 = 1.645*\frac{6}{\sqrt{n}}$

$1.2\sqrt{n} = 6*1.645$

$\sqrt{n} = \frac{6*1.645}{1.2}$

$(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}$

$n = 67.65$

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

$M = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.645*\frac{6}{\sqrt{n}}$

$\sqrt{n} = 6*1.645$

$(\sqrt{n})^{2} = (6*1.645)^2$

$n = 97.42$

Rounding up

A sample size of 98 should be used.

3 0

3 years ago

Last year, a large trucking company delivered 5.5*10^5 tons of goods with an average value of $23,000 per ton. What was the tota

Mademuasel [1]

It would be: 5.5 * 10⁵ * 23000
= 5.5 * 10⁵ * 2.3 * 10⁴
= 12.65 * 10⁹
= 1.265 * 10¹⁰

In short, Your Answer would be Option B

Hope this helps!

8 0

3 years ago

300 pounds of oranges were purchased at $0.24 per pound. The desired markup is 50% based on selling price, but 21% spoilage is e

andrezito [222]

1. The problem indicates that 300 pounds of oranges were purchased at $0.24 per pound, so the cost is:

300x$0.24=$72

2. A percent of 21% spoilage is expected, which means that 89% of 300 pounds of oranges left, is:

300x0.89=267 pounds of oranges

3. We want to calculate the selling price per pound of 267 pounds of oranges, so let's call this value "x":

Markup %=(Markup/cost)x100

Markup=267x-72

4. The desired markup, based on selling price, is 50%. So, when we substitute the values, we obtain:

(267x-72/72)x100=50

5. Let's clear the "x":

(26700x-7200/72)=50
26700x-7200=50x72
26700x-7200=3600
26700x=3600+7200
x=10800/26700
x=$0.40
x=40 cents

What should the selling price per pound be?

The answer is: 40 cents or $0.40

8 0

3 years ago