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3 years ago

1. Jackson spends 15 hours over 3 weeks watching movies. How many hours

Mathematics

1 answer:

Kamila [148]3 years ago

6 0

Answer:

I believe its 120 hours

Step-by-step explanation:

3 weeks = 15 hours

4 weeks = 15 x 2 = 30

4 + 4 = 8

8 x 15 = 120

Hope that helped

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Given a circle with an area of 25 pi in^2 <br> What is the exact circumference?

Taya2010 [7]

Answer:

31.4

Step-by-step explanation:

C = 2πr OR C = πd

But, we need to find r first.

$A = \pi r^2\\25\pi = \pi r^2\\\sqrt{25} = \sqrt{r^2} \\r=5\\\\C=2\pi r\\C=2\pi (5)\\C= 10\pi \\C= 31.4$

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4 years ago

Factor the following polynomial. 6x 2 + 14x

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Answer:

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Step-by-step explanation:

factor 2x out of 6x^2+14x

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3 years ago

Which number line model shows 8 x 1/2

77julia77 [94]

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3 years ago

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Identify the terms, coefficients, and constants in the expression 20x+13.

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Step-by-step explanation:

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2 years ago

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

3 years ago