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aniked [119]
3 years ago
13

Brad is testing whether school is more enjoyable when students are making high grades. He asked 110 students if they enjoyed sch

ool and whether their GPA was above or below 3.5. He found that 38 of the 45 students with a GPA above 3.5 reported that they enjoyed school, and 7 of the 65 students with a GPA below 3.5 reported that they enjoyed school. What is the probability that a student with a GPA below 3.5 does not enjoy school?
89%
92%
85%
75%
Mathematics
2 answers:
sattari [20]3 years ago
6 0
Answer is 75% hopefully I mean I don't really know that much
Nimfa-mama [501]3 years ago
5 0

Answer:

89%

Step-by-step explanation:

This exercise requires a simple probability calculation. The question concerns only students with GPA less than 3.5.

The universe of researched nulls whose GPA is less than 3.5 consists of 65 students.

Of these students, in the survey, 7 answered that they like school. So the likelihood that a student with a GPA less than 3.5 will like the choice will be found with a simple proportion account: (students who dislike / total students) * 100

 (7/65) * 100 = 11%

To calculate the probability that a student with a GPA below 3.5 doesn't like the school, we can use two ways:

1) Decrease from the sample universe the number of students who do not like school (7) to find the number of students who do not like school and make a simple proportion calculation, which consists of dividing the number of students who do not like by the total number of students and multiply by 100.

65 - 7 = 58

(58/65) * 100 = 89%

2) If you know the percentage of students who like school, you can only decrease the total to find the ones you don't like.

100% - 11% = 89%

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Answer:

\large\boxed{1.\ (4)^{-3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}}

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Step-by-step explanation:

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\\------------\\\\(4)^{-3x^2}=\left[(4)^{-1}\right]^{3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\ and\ (a^n)^m=a^{nm}\\--------------------\\\\ab^{-3x}=a\cdot b^{-3x}=a\left[(b)^{-1}\right]^{3x}=a\left(\dfrac{1}{b}\right)^{3x}\\\\ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x

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Step-by-step explanation:

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3 years ago
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Oksanka [162]

Answer:

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Step-by-step explanation:

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