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Mathematics

2 answers:

Ket [755]3 years ago

5 0

So 4x^2+-2x+-6 then multiply by 5 would be the Area of the entire rectangle using the formula A=bh

You'll get 4x^2 − 2x − 6

myrzilka [38]3 years ago

5 0

a = 5

b = 4x² - 2x - 6

The area is:

a · b = 5(4x² - 2x - 6) = 20x² - 10x - 30

Or you can add areas of shapes:

- blue: 5 · 4x² = 20x²

- pink: 5 · (-2x) = -10x

- green: 5 · (-6) = -30 <- a little strange, the length of a side shouldn't be negative

Sum of those areas is the area of entire rectangle: 20x² - 10x - 30

But the first option is easier here.

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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?

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