How i can answer this question

The table shows the percentage of students in each of three grade levels who list soccer as their favorite sport. Soccer Sophomo

Olegator [25]

The answer should be roughly 35.4%.

You can obtain this answer by looking at the percentage of each subgrouping. For instance, 33% of the class in juniors and 45% of them list soccer as their favorites. Thus showing that 14.85% of the entire school is made up of juniors that enjoy soccer.

If you do the totaling for all soccer lovers, you get a total of 41.95% of the school. By dividing the two numbers you get the answer above.

4 0

3 years ago

What is the sum of all of the odd numbers from 1 to 59? Answer 900

jasenka [17]

The sum of all the odd numbers from 1 to 59 is 900.

5 0

3 years ago

Read 2 more answers

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago

Solve for f(-1) f(x)=-3x +3 f(-1)=?

krok68 [10]

9 or -9 ...if I’m wrong sorry

6 0

3 years ago

The sum of reciprocals of rehman’s ages 3 years ago and 5 years from now is 1/3. Find his present age

Afina-wow [57]

Answer:

His age is 7 years.

Step-by-step explanation:

Let Rehman's present age be x years.

1 / (x - 3) + 1 /(x + 5) = 1/3

Multiply through by the LCD 3(x - 3)(x + 5):

3(x + 5) + 3(x - 3) = (x + 5)(x - 3)

3x + 15 + 3x - 9 = x^2 + 2x - 15

x^2 + 2x - 6x - 30 + 9 = 0

x^2 - 4x - 21 = 0

(x - 7)(x + 3) = 0

x = 7 (answer).

6 0

3 years ago