Determine the equation of the quadratic function shown below.

*WILL GIVE BRAINLIEST ANSWER* Analyze the graph. Determine what, if any, association there is. Explain. On a graph, points are g

Gnesinka [82]

Answer:

bro I thought someone else answered your question

4 0

2 years ago

What is 82,119 rounded to the nearest ten thousand?

Stels [109]

80,000.
To round up to the nearest ten thousand, we would see if the next lower place value has a amount either 5 or higher or 4 or lower.
If it’s five or higher, we round up.
If it’s four or lower, we round down.
We can see that the number in the ten thousands place is 8. The number in the place value after 8 is 2.
2<5 so we round down.
The answer is 80,000.
Hope this helps!

8 0

3 years ago

Please help on question b

Viktor [21]

Answer:

Step-by-step explanation:

Number of males who exclusively drink coffee: 21.

Total number of people: 100

P(MdC) = 21/100 * 100% = 21%

8 0

3 years ago

Solve for X and then pick a answer at the bottom. This is basically angles in transversal if I'm correct.

Sonbull [250]

The correct answer would be c but im not sure im really right

5 0

3 years ago

Read 2 more answers

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

3 years ago