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ella [17]
3 years ago
10

Find the distance between (2, -75) and (10,235).

Mathematics
1 answer:
zmey [24]3 years ago
4 0

Answer:

310.1

Step-by-step explanation:

Use the distance formula:

\sqrt{(10-2)^2+(235-(-75))^2}

Simplify inside the parentheses

\sqrt{(8)^2+(310)^2}

Simplify exponents

\sqrt{64+96,100}

Simplify by adding

\sqrt{96,164}

Simplify

\sqrt{96,164} =310.1

Done. B-)

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The radius of a semicircle is 9.1 centimeters. What is the semicircle's perimeter?
Sloan [31]

Answer:

<u>46.77 cm</u>

Step-by-step explanation:

Perimeter of the semicircle = Circumference + 2r

  • πr + 2r = r (π + 2)
  • 9.1 (3.14 + 2)
  • 9.1 (5.14)
  • <u>46.77 cm</u>
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2 years ago
Is 398,468 divisible by 4?<br><br> YES or NO<br><br> HELP ME PLZ!!
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Step-by-step explanation:

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3 years ago
Which is the slope-intercept form of an equation for the line containing
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y=mx+b

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6 0
3 years ago
Assume that 8​% of people are​ left-handed. We select 6 people at random. ​a) How many lefties do you​ expect? ​b) With what sta
klemol [59]

Answer:

a) 0.48

b) 0.6645

c) 12.5

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are left-handed, or they are not. The probabilities of each person being left-handed are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

The number of trials expected to find r sucesses is given by

N = \frac{r}{p}

In this problem we have that:

Assume that 8​% of people are​ left-handed. We select 6 people at random. ​

This means that p = 0.08, n = 6

a) How many lefties do you​ expect?

E(X) = np = 6*0.08 = 0.48

​b) With what standard​ deviation?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6*0.08*0.92} = 0.6645

​c) If we keep picking people until we find a​ lefty, how long do you expect it will​ take?

Number of trials to find 1 success. So

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3 years ago
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