I need help !!!!!!!!!!!!!

Calcular el volumen en m3 de la esfera en el que el área de uno de sus círculos maximos es 36pim2

Strike441 [17]

Answer:

The volume of the sphere is $V=288\pi\ m^3$

Step-by-step explanation:

<u><em>The question in English is</em></u>

Calculate the volume in m^3 of the sphere in which the area of one of its maximum circles is 36pi m^2

we know that

The radius of the maximum circle in the sphere is equal to the radius of the sphere

Step 1

Find the radius of the maximum circle

The area of the circle is

$A=\pi r^{2}$

we have

$A=36\pi\ m^2$

substitute and solve for r

$36\pi=\pi r^{2}$

Simplify

$36=r^{2}$

take the square root both sides

$r=6\ m$

Step 2

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

substitute the value of r

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ m^3$

6 0

3 years ago

A coin has heads on one side and tails on the other the coin is tossed 12 times and lands heads up four times which best describ

Rufina [12.5K]

Step-by-step explanation:

According to the empirical

Definition of probability, by continuing to increase the number of trails, it will ultimately get very close to a number which is the probability of obtaining a head in a single toss of the coin. The this is known to be 1/2 or 0.5

5 0

4 years ago

Read 2 more answers

What is the product of (7.28 x 10-2). (9.1 x 10-2)?

tangare [24]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So

X = 35.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{35.5 - 26.751}{5.08}$

$Z = 1.72$

$Z = 1.72$ has a pvalue of 0.9573

X = 24.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24.5 - 26.751}{5.08}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 26.751}{5.08}$

$Z = 2.71$

$Z = 2.71$ has a pvalue of 0.9966

1 - 0.9966 = 0.0034

0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

6 0

3 years ago

Find the value of x. 2(x-3)-17=13-3(x+2)

Nonamiya [84]

Solve your equation step-by-step.

2(x−3)−17=13−3(x+2)

Simplify both sides of the equation.

2(x−3)−17=13−3(x+2)

Simplify

2x−23=−3x+7

Add 3x to both sides.

2x−23+3x=−3x+7+3x

5x−23=7

Add 23 to both sides.

5x−23+23=7+23

5x=30

Divide both sides by 5.

5x/5 = 30/5

x=6

5 0

4 years ago

Read 2 more answers