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Vladimir79 [104]
3 years ago
5

How to solve q =r/2(s+t) for t

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
4 0
<span>q =r/2(s+t) for t
</span><span>2q =r(s+t) 
2q = rs + rt
rt = 2q - rs
 t = 2q/r - s
hope it helps</span>
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Find the perimeter of this figure to the nearest hundreth use 3.14 as an approximate for pi​
maks197457 [2]

Answer: 40.56ft.

Step-by-step explanation:

First, lets find the perimeter of the half circle: c = 2pi(r)

pi = 3.14

r = 4

3.14 * 4 = 12.56

12.56 * 2 = 25.12

Then, because it is only have a circle, divide by 2.

25.12/2 = 12.56

10 + 10 + 8 + 12.56 = 40.56

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If you are given odds of 4 to 5 in favor of winning a​ bet, what is the probability of winning the​ bet?
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Answer:

Step-by-step explanation:

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John weighs 200 pounds. He lost 10% of his weight. Then he gained back 10% of his weight. How much does John weigh now?
tia_tia [17]

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Simplify 3√125. Thanks!
ivann1987 [24]
If you mean
3 times square root of 125 then go to AAA
if you mean ∛125 then go to BBBB


AAAAAA
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BBBBBBB
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5 0
4 years ago
Read 2 more answers
Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea
Talja [164]

Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c) P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d) P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

E(X)= n\frac{M}{N}

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

4 0
3 years ago
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