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2 years ago

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Mathematics

2 answers:

levacccp [35]2 years ago

8 0

Answer:

The formula for the volume of a right cylinder is: V = A h, where A is the area of the base, or πr2. Therefore, the total formula for the volume of the cylinder is: V = πr2h. First, we must solve for r by using the formula for a circumference (c = 2πr): 25π = 2πr; r = 12.5.

Step-by-step explanation:

Explanation: The formula for the volume of a right cylinder is: V = A * h, where A is the area of the base, or πr2. Therefore, the total formula for the volume of the cylinder is: V = πr2h. First, we must solve for r by using the formula for a circumference (c = 2πr): 25π = 2πr; r = 12.5.

Lostsunrise [7]2 years ago

5 0

Answer:

The answer is C

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Yuri has 12 counters He put them in 2 equal group How many counters are in each group

Andrej [43]

Answer: 6

Step-by-step explanation:

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8 0

3 years ago

Determine the point (x,y) ( x , y ) on the unit circle associated with the following real number s. Write the exact answer as an

Harlamova29_29 [7]

The required coordinates of the unit circle for s = -11π/6 is (√3/2, 1/2). Using trigonometric functions, the required answer is calculated.

<h3>What are the trigonometric functions?</h3>

There are six trigonometric functions.

They are sin θ, cos θ, tan θ, sec θ, cosec θ, and cot θ.

Where θ - 0 to 360° or 0 to 2π

<h3>Calculation:</h3>

For a unit circle, the coordinates are given in the form of trigonometric functions as below:

x = cos θ and y = sin θ

Here it is given that s = θ = -11π/6

So, on substituting,

x = cos θ = cos (-11π/6)

⇒ x = √3/2

y = sin θ = sin (-11π/6)

⇒ y = 1/2

Thus, the required coordinate pair is (√3/2, 1/2).

Learn more about trigonometric functions here:

brainly.com/question/2291993

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7 0

1 year ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

The surface area of a square pyramid is 189 in.². The side length of the base is 7 inches. What is the value of X?

alina1380 [7]

Not enough information is given to put an absolute answer it there could be a million reason like d332 nut5 for example

8 0

3 years ago

You have a bag that contains 20 quarters, 20 dimes, and 15 nickels. When randomly selecting two coins with no replacement find t

iogann1982 [59]

Answer:

P(2 quarters) = 19/ (13*16) = 19/208 = 0.0913

Step-by-step explanation:

probability of getting 2 quarters?

first draw: there are 20 quarters to choose from

second draw: there are 19 quarters left to choose

so Total coins are 65 coins

P(2 quarters) = P( 1st quarter)* P(2nd quarter)

P(2 quarters) = 20/65 * (19/64)

P(2 quarters) = 4/13 * 19/64

P(2 quarters) = 19/ (13*16) = 19/208

4 0

3 years ago