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REY [17]
3 years ago
6

PLEASE HELPP!!! ITS REALLY IMPORTANT

Mathematics
2 answers:
victus00 [196]3 years ago
4 0

Answer:

70

Step-by-step explanation:

trust me

Fantom [35]3 years ago
3 0

Answer:

what do you need no one can help if you don't say what you need

Step-by-step explanation:

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14+6 :)) ILL GIVE BRAINLIEST ITS 100 POINTS :))
emmainna [20.7K]

Answer:

20

Step-by-step explanation:

lol

3 0
4 years ago
What is the ratio of the radius of a circle to its circumference
maksim [4K]
ANSWER:

r : 2 ( Pi ) r

Hope this helps! <3
5 0
3 years ago
What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and
e-lub [12.9K]
First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
  y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
  area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
  d(area)/dw = 0 = -kw +k
  w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
  area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
  h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
  triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
  16 square units

4 0
4 years ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
4 years ago
A study of 25 graduates of 4-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The s
Harman [31]

Answer:

Step 1

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

Step 2

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

5 0
2 years ago
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