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2 years ago

Mind lending a hand here?

Mathematics

2 answers:

sweet-ann [11.9K]2 years ago

8 0

Answer:

option B

Step-by-step explanation:

$75 is already saved plus X which is greater or equal to 200.

motikmotik2 years ago

7 0

Answer: B

Step-by-step explanation: 75 + the amount of money she still needs is equal to or greater than (atleast) 200. hope this helps

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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

PPLEASE HELP ME, I NEED HELP I CANT UNDERSTAND I COULD JUST GUESS BUT ITS IMPORTANT PLS

Korolek [52]

Answer: where’s the question??

8 0

1 year ago

Kate is arranging new furniture in her rectangular sitting room. She has mapped the arrangement on a coordinate plane. A, B, and

madreJ [45]

Answer:

Step-by-step explanation:

To find the distance between V1 and AQ use the distance formula

$d = \sqrt{(x2-x1)^2+(y2-y1)^2}$

First find the coordinates of both V1: (-6, 5) AQ: (5, 5)

Now plug it in and solve for d (11)

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2 years ago

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I need the answers to this guys please help

Anvisha [2.4K]

C. 660 ft
If you multiply 3 by 22, you get 66, and multiply 10 by 66 and get 660.

8 0

3 years ago

What is the length of AB ? Round to the nearest tenth.

Vinvika [58]

The only way you can find x here is to use a trig ratio. We have the reference angle of 75, CA is 10 which is the side adjacent to the reference angle, and we are looking for the hypotenuse. The ratio that relates the side adjacent to a reference angle to the hypotenuse is cosine. Therefore, $cos(75)= \frac{10}{x}$ . Solving for x we get $x= \frac{10}{cos(75)}$ . Do that on your calculator and find that the hypotenuse is 38.6 m long, last choice from above.

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2 years ago

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