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fenix001 [56]
3 years ago
14

A rectangular garden has length twice as great as its width. A second rectangular garden has the same width as the first garden

and length that is 4 meters greater than the length of the first garden. The second garden has area of 70 square meters.
What is the width of the two gardens?

Enter your answer in the box.


m
Mathematics
1 answer:
stiks02 [169]3 years ago
6 0
Use the information to write the dimensions of each rectangle in terms of w, the width of the 1st one.

1st rectangle;
l = 2w
w = 2

2nd rectangle:
w = w 
l = 2w + 4

If the area of the 2nd rectangle is 70 square meters, you will use the area formula to write an equation that you will solve using the factoring.

A = lw
70 = w(2w + 4)
70 = 2w^2 + 4w
0 = 2w^2 + 4w -70
0 = 2 (w^2 +2w - 35)
0 = 2 (w + 7) (w - 5)

To get zero, the width would need to be -7 or 5.  Because it is a distance, it has to be 5 meters.

The width of both rectangles is 5 meters.

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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
3 years ago
Which equation has no solution?
ololo11 [35]

Answer:

|5 - 3x| = -8

Step-by-step explanation:

The definition of absolute value is the distance from zero. Therefore, you cannot get a negative answer.

That is why this is wrong...

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7 0
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Given(x) = -x-2 find g(-3)
andreev551 [17]

Step-by-step explanation:

If you mean g(x) =-x-2 the answer is:

g(-3)=-(-3)-2

g(-3)=+3-2

g(-3)=+1

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