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2 years ago

WILL MARK BRAINLIEST. Which of the following statements about the polynomial function given is true?

Mathematics

2 answers:

Katyanochek1 [597]2 years ago

4 0

Answer:

Step-by-step explanation:

because as X approaches infinity, the negative sign will cause it to go opposite of X.

andreyandreev [35.5K]2 years ago

3 0

Option A is true.

The graph will rise to the left and descend to the right and have 2 relative turning points. A minimum and a maximum.

It is option A that is true.

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Simplify. Assume that a is a positive integer and a>2. (a-2)!/a!

Paul [167]

Answer:

C. 1 / (a(a - 1))

Step-by-step explanation:

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Just know that:

n! = n(n-1)!

= n(n-1)(n-2)!

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3 years ago

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Sam and Annika ride their bike to school. Sam rides his bike (k) kilometers. Annika rides her bike 2 less than 4 times as many k

pav-90 [236]

Because he lives farther from the school than Anneika more than likely

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3 years ago

PLEASE HELP I really need help with it:/

Luda [366]

Answer:

q = 14

Step-by-step explanation:

39 + (4q - 5) = 90°

<u>Solution:</u>

34 + 4q = 90°
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=> q = 56/4
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Hence, the value of q is 14.

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2 years ago

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Which table has a constant of proportionality between Y and X of 1/6?

dezoksy [38]

Answer:

Step-by-step explanation:

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3 years ago

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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?

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

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

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3 years ago