12 more than 8.2 times n

(NO TROLLS 3 TIMES THE CHARM) In △ABC, AB = BC = 20 and DE ≈ 9.28. Approximate BD.

andreev551 [17]

Answer:

≈ 5.36

Step-by-step explanation:

If we have a triangle in which the angle measure of it being broken down are the same, that means that the legs in which the angle measures are the same will have the same length.

If we already know that DE ≈ 9.28, then we can subtract this from 20 and divide by two to get BD, which is equal to EC.

$20-9.28=10.72\\10.72\div2=5.36$

However, I'm not 100% sure about this answer.

Hope this helped (and I hope I'm right)

3 0

3 years ago

Read 2 more answers

Find the missing number in the proportion.

Soloha48 [4]

Answer: 7

Step-by-step explanation: Cross multiply, so 15 times x and 21 times 5. 21 times 5 is 105 then you divide by 15 which gives you 7.

8 0

2 years ago

Select the correct answer.

ValentinkaMS [17]

Answer:

The answer is A.

$g(x) = 2( {x}^{2} - 3x - 28)$

7 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where