A yard is equal in length to three feet. The function f(x) takes a measurement

Stuck on two. Plz help. Pre-all is hard

algol [13]

Gncgbvcrvbshj hnctvvdf

7 0

2 years ago

Solve for w.<br> = w-2/3=18 - 3w+18/2<br> Simplify your answer as much as possible.

mart [117]

I have a screenshot of the answer

8 0

2 years ago

The weight of strawberries follows a normal distribution with a mean weight of 12 grams and a standard deviation of 2.5 grams. I

Viefleur [7K]

This is a problem of Standard Normal distribution.

We have mean= 12 grams
Standard Deviation = 2.5 grams

First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:

$z= \frac{8.5-12}{2.5} =-1.4$

So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808

Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>

6 0

2 years ago

A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc

marysya [2.9K]

Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So 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.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that $p = \frac{3}{5} = 0.6$

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

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

$P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6$

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

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

$P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4$

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

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

$P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296$

12.96% probability that all 4 students selected took online courses.

3 0

2 years ago

If you help me I'll make as brilliant

Rus_ich [418]

Answer:

36 im pretty sure! wait for someone else cus im not that sure, hope u get it right !

Step-by-step explanation:

5 0

2 years ago