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Vinvika [58]
3 years ago
15

HELP!!!!! What is the length plz help me besties

Mathematics
1 answer:
Nesterboy [21]3 years ago
8 0

Answer: 30 inches

Step-by-step explanation:

20/8 = x/12

20(12) = 8x

240 = 8x = 30

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Rectangle ABCD translates 4 units down and 2 units to the right to form rectangle A'B'C'D'. The vertices of rectangle ABCD are l
AlekseyPX

The length of B'C' in the rectangle A'B'C'D' = 9 units.

<u>Step-by-step explanation</u>:

step 1 :

Draw a rectangle with vertices ABCD in clockwise direction.

where, AB and DC are width of the rectangle ABCD.

            AD and BC are length of the rectangle ABCD.

step 2 :

Now,

The length of the rectangle is AD = 5 units and

The width of the rectangle is AB = 3 units.

step 3 :

Draw another rectangle with vertices A'B'C'D' extended from vertices of the previous rectangle ABCD.

step 3 :

The length of the new rectangle is A'D' which is 4 units down from AD.

∴ The length of A'D' = length of AD + 4 units = 5+4 = 9 units

step 4 :

Since B'C' is also the length of the rectangle A'B'C'D', then the measure of B'C' is 9 units.

3 0
3 years ago
25+5 Times blank equals 35
mezya [45]
Let x = blank

25 + 5x = 35

5x = 35 - 25

5x = 10

x = 10/5

x = 2

Done!
6 0
3 years ago
Help me please I have no idea on how to do this. Thanks
andriy [413]
1) P=2w +2l
46=2w +14
32=2w
16= w or width

2) A=lw
A=7 x 12
A= 84

3) A= 12 x 21
A= 252
6 0
3 years ago
2(10.8 k+ 2.9) What's the answer
motikmotik
21.6k + 11.8

I believe is the answer good luck!
3 0
3 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

Answer:

A sample of 1068 is needed.

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}.

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.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

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

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

8 0
3 years ago
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