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3 years ago

Plssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss help

Mathematics

2 answers:

KengaRu [80]3 years ago

8 0

Answer:

10 and 5

6 and 9

11 and 4

Step-by-step explanation:

I hope this helps!

almond37 [142]3 years ago

3 0

Answer:

5, 6, 4

Step-by-step explanation:

10+(?)=15

10 and 15 have a difference of 5

10+5=15

(?)+9=15

9 and 15 have a difference of 6

6+9=15

11+(?)=15

11 and 15 have a difference of 4

11+4=15

I hope this helps!

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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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Help find the answer for problem number 4

Olin [163]

Adult ticket (a) = $5
Child ticket (c) = $2

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Translate into English please

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4 years ago

Sorry my brain hurts... Most of u love the points anyway

Amanda [17]

I think the answer's C: $7000/$8000
That's a surprisingly easy question for high school

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dolphi86 [110]

Answer:

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here,

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