2 years ago

Enter the equation of the circle described below. Center (-3,0), radius = sqrt5

Mathematics

1 answer:

Marianna [84]2 years ago

5 0

Answer: $(x+3)^{2}+y^{2}=5$

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Using y = mx + b (-3, 2) and (-5, 4)

skelet666 [1.2K]

Answer: y=-1x-1

Step-by-step explanation:

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2 years ago

If f(x) = −2 + 8, what is f(−1.8)? 4 6 10 12

Oksanka [162]

F(x) = -2 + 8

f(x) = 6

No matter what x is, it does not change the value of f(x).

f(-1,8) = 6

Answer: 6

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2 years ago

Read 2 more answers

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$