Calculate the perimeter of each figure <br> How do you find the perimeter of a figure

Triangle FGH is equilateral. The midpoints of the sides are connected to form triangle XYZ. Line segment XY is parallel to line

Hatshy [7]

Answer:

Isosceles trapezoid

Step-by-step explanation:

The figure will look like one large triangle on the outside and four smaller triangles on the inside. you are only concerned with the inside triangle that is pointing to the line FH. Since XY and FH are parallel and the distance between FX and YH are equivalent (the midpoints of each side should be the same for all three sides of the larger triangle) the shape FXYH is an isosceles trapezoid.

3 0

3 years ago

A tetrahedron has 4 sides. If the numbers 1, 2, 3, and 4 are on the faces of the tetrahedron, what

Savatey [412]

Probability of randomly rolling a 3 is 1/4

Step-by-step explanation:

Step 1:

It is given that a tetrahedron has 4 sides with 1,2,3 and 4 and is randomly rolled. We need to compute the probability of randomly rolling a 3.

Step 2:

Probability of an event = (Favorable outcomes) / (Total outcomes).

There are 4 total outcomes when the tetrahedron is rolled, these are the values 1 to 4.

The only favorable outcome is getting a 3

Probability of getting a 3 when the tetrahedron is randomly rolled = 1/4

Step 3:

Answer:

Probability of randomly rolling a 3 is 1/4

7 0

4 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.

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

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