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2 years ago

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You roll a fair 666-sided die. What is \text{P(roll greater than 4})P(roll greater than 4)start text, P, left parenthesis, r, o,

l, l, space, g, r, e, a, t, e, r, space, t, h, a, n, space, 4, end text, right parenthesis?
If necessary, round your answer to 222 decimal places

1 answer:

andreev551 [17]2 years ago

6 0

Answer:

1/6 fraction is the answer

Step-by-step explanation:

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There are 20 children in the cast of a class play, and 8 of the children are boys. Of the boys, 4 have a speaking part in the pl

denpristay [2]

Answer:

A0 2/5

Step-by-step explanation:

First, find out how many girls there are by subtracting 8 (boys) from 20 (total). This means there are 12 girls. Since 8 girls aren't speaking, 4 are. 4+4=8. 8 out of the 20 children are speaking/ 8/20 simplified is 2/5

4 0

3 years ago

Read 2 more answers

What is the perimeter of a regular heptagon if one side measures 3 inches

svetlana [45]

What's a regular heptagon? "Hept-" means seven, so a "heptagon" is a seven-sided shape. A "regular" figure's sides are all the same length. So, each of the 7 sides are the same length.

Remember, perimeter is the sum of the lengths of all the sides in a polygon. We know one side measures three inches, so each of the others must also measure 3 inches. 3+3+3+3+3+3+3=3*7=21 inches.

Answer: 21 inches

7 0

3 years ago

Find y as a function of x if y"' + 9y' = 0, y(0) = 6, y'(0) = 9, y"(0) = 18. y(x) =

Andreas93 [3]

Answer:

y(x) = 8 - 2cos 3x + 3sin 3x

Step-by-step explanation:

Given the equation:

y''' + 9y' = 0

The characteristic equation :

⇒r³ + 9r =0

Solving for r, we get r = 0 and r = ± 3i

The general equation for such equation is :

<u>y = C₁ + C₂cos 3x + C₃sin 3x</u>

Given:

y(0) = 6

Thus, Applying in the above equation, we get

<u>6 = C₁ + C₂ .......................................................1</u>

Differentiating y , we get:

<u>y' = -3C₂sin 3x + 3C₃cos 3x</u>

Given:

y'(0) = 9

Thus, Applying in the above equation, we get

<u>9 = 3C</u><u>₃</u>

or,

<u>C</u><u>₃</u><u> = 3</u>

Differentiating y' , we get:

<u>y'' = -9C₂cos 3x - 9C₃sin 3x</u>

Given:

y''(0) = 18

Thus, Applying in the above equation, we get

<u>18 = -9C</u><u>₂</u>

or,

<u>C</u><u>₂</u><u> = -2</u>

Applying in equation 1 , we get:

<u>C₁ = 8</u>

Thus,

<u>y(x) = 8 - 2cos 3x + 3sin 3x</u>

4 0

3 years ago

Factor the polynomial x^2+7x+10. Your answer can written as (x+A) (x+B)

amm1812

(x+5)(x+2) is your answer

7 0

3 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

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

So

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

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

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

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

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

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

5 0

2 years ago