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2 years ago

12

Given the right triangle ABC, if segment AC is 3V6, then AB = _ 3.14 3,13 443 37

1 answer:

Luda [366]2 years ago

4 0

The answer would be 37, i’m pretty sure.

:D

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Is it legal to keep monkeys in a refrigerator in the United States?

Mice21 [21]

Answer: No unless the fbi cathches up to you mr.gieco

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What’s the answer to this important problem it is very idk

White raven [17]

1) 4/(1/5)

=4 ÷ 1/5
to divide fractions, multiply by the reciprocal/inverse of 1/5

=4 * 5/1
=20

4) (7/9)/7

=7/9 ÷ 7
to divide fractions, multiply by the reciprocal/inverse of 7

=7/9 * 1/7
multiply numerators; multiply denominators

=(7*1)/(9*7)
=7/63
simplify by 7
=1/9

7) (9/10)/6

= 9/10 ÷ 6
to divide fractions, multiply by the reciprocal/inverse of 6

= 9/10 * 1/6
multiply numerators; multiply denominators

= (9*1)/(10*6)
= 9/60
simplify by 3
= 3/20

ANSWER: 1) 20; 4) 1/9; 7) 3/20

Hope this helps! :)

3 0

3 years ago

Find the probability that the next student will select hamburger for lunch. You

adell [148]

Answer=3/10

Step wise =

Total number of student=40

student who eat Hamburger=12

now,

Probability of eating hamburger student = 12/40

=3/10

6 0

2 years ago

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of <em>z</em> for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

2 years ago

Brainlyyyyyyyyyyyyyy :)

Rashid [163]

Wow, that IS really BRAINLY! I’m so jealous

6 0

3 years ago