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2 years ago

D is Midpoint of CE C = (-5,?) D = (3,10) E = (?,8)

1 answer:

3 0

Answer:
C = (-5,12) E = (11,8)
Step-by-step explanation:
The difference between -5 and 3 is 8, so add 8 to the midpoint (3), and it's 11
The difference between 10 and 8 is 2, so add 2 to the midpoint (10), and it's 12

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(-6) x 8 What is this ?????

torisob [31]

Answer:

-(6) x 8

ans: - 48

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2 years ago

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If a vsco girl walks 1.5 miles how many yards is that???? pls answer quickly pls

tekilochka [14]

One mile is equal to 1760 yards. Therefore, 1.5 miles is equal to 1.5 x 1760 = 2640 yards.

1.5 mile = 2640 yards

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3 years ago

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What percent of 82 is 7

katovenus [111]

Answer: 8.5%

7/82=0.0853

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3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

4 years ago

-20y + 15 = 2 - 16y + 11

shtirl [24]

In order to answer this, we will first need to combine like terms on each side. on the left, you can leave them alone. however on the right, we will need to combine 2 and 11. this is 13. the right side becomes 13-16y. after that, we can add 20y to both sides. that equals 15=13+4y now we can subtract 13 from both sides. 2=4y. then we divide by 4 on both sides to find y. y=.5

7 0

4 years ago