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3 years ago

Please halp me in the question?

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

5 0

Answer: a. (x-6) + 5x = 90 b. Angle One = 10° Angle Two = 80°

Step-by-step explanation:

A. Since angle 1 plus angle 2 is a right angle, then their sum is 90 degrees. The equation form angle 1 is x-6 and the equation for angle 2 is 5x, so (x-6) + 5x = 90

B. To find x,

x - 6 + 5x = 90

x + 5x = 96

6x = 96

x = 16

So, angle one = (16) - 6 = 10° and angle two = 16 * 5 = 80°

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Find the value of x in each case.

ExtremeBDS [4]

62 is the awnser for this

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3 years ago

9 ten thousands / 10 in unit form

Phoenix [80]

First write nine ten thousands in standard form which is 90,000. Next divide 90,000 by 10 and you should get 9,000 by crossing out one zero in both numbers. To get 9,000 in unit form all you have to write is 9 thousands. And your answer is 9 thousands

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3 years ago

You can make a tent by throwing a piece of cloth over a clothesline, then securing the edge to the ground with stakes so that an

4vir4ik [10]

Based on the required width and height of the tent, the length of the cloth would have to be 14.4 ft.

<h3>How long should the length of the cloth be?</h3>

The opening of the tent would be an isosceles triangle but if it is divided in two from the top, a right-angled triangle will be formed and the hypotenuse will be the length of the cloth on one side.

That length - according to the Pythagoras theorem - would be:

= √(6² + (8 /2)²)

= √(36 + 16)

= 7.2 ft

As this is the length of one side of the cloth, the length of the entire cloth is:

= 7.2 x 2

= 14.4 ft

Find out more on the Pythagoras theorem at brainly.com/question/343682.

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2 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

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3 years ago

Which of these is an algebraic expression? A. 3-14 B. 14-3 C. 14-n=5 D. 14-n

Evgesh-ka [11]

The answer is C because for algebraic expression you need a coefficient, variable and a constant

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3 years ago