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Alex787 [66]
3 years ago
15

In a survey, 15 high school students said they could drive and 15 said they could not. Out of 60 college students surveyed, 30 s

aid they could drive. Micah concluded that knowing that a person is in college means they are more likely to drive. Is Micah’s conclusion correct? Explain.
Mathematics
2 answers:
luda_lava [24]3 years ago
3 0
Micas conclusion is incorrect. There were 30 high school students surveyed and 15 of them said that they could drive, and out of the 60 college students half of them also said they could drive. In both situations, half of the groups were able to drive which shows that his conclusion of knowing that a person is in college means they are more likely to drive is incorrect.

velikii [3]3 years ago
3 0

Sample Response: No, Micah is not correct. Since there are equal numbers in the group that drive and the group that doesn’t drive, the relative frequency for each is 50%. Because the relative frequencies are the same, there is no association between the variables.

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What’s the answer for this?
fgiga [73]

Answer:

1.98

Step-by-step explanation:

Use SOH-CAH-TOA:

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

Here, we're given the hypotenuse is 4.5, and we want to find the side opposite of the 26° angle.  So we should use sine.

sin 26° = x / 4.5

0.44 = x / 4.5

x = 1.98

7 0
3 years ago
Hey yall what is 2⋅2⋅2⋅n⋅n?
Svetach [21]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Carly has 15 stuffed animals. How many ways are there to organize her top 3 favorites?
Ivanshal [37]
Well, the answer will depend on whether the order will count or not (based on Permutations and Combinations). <em>If the order counts</em>, then we would use the formula for Permutations, which is:
\frac{n!}{(n-r)!}
Where n is the number of items you have, and r is the number of times you choose from the items.
\frac{15!}{(15-3)!}
Which simplifies to
\frac{15!}{(12)!}
Which simplifies to 15*14*13 (because all the numbers 1-12 in the factorial canceled out), which gets us the answer 2730.

Now, if you wanted to find the number of ways to order the toys without replacement (<em>order doesn't count</em>), you would use the formula:
\frac{n!}{r!(n-r)!}
The  variables are still the same, but you are now multiplying by r!.
\frac{15!}{3!(15-3)!}
Simplifies to
\frac{15!}{3!(12)!}
Which simplifies to (using the same cancellation method above)
\frac{2730}{3!}
Dividing 2730 by 3! will get us an answer of 455.

Really, it depends on whether they are ordered or not. In this case (since you didn't specify whether the order mattered), the answer would be 455 or 2730.

:)
4 0
3 years ago
The best player on a basketball team makes 85% of all free throws. The second-best player makes
xxTIMURxx [149]

Answer:

i. Estimated number of free throws of the best player = 0.85 * 40

ii. 34 free throws

Step-by-step explanation:

Percentage of free throws made by the best player = 85%

Percentage of free throws made by the second best player = 75%

Percentage of free throws made by the third best player = 70%

Therefore, for 40 attempts;

i. Estimated number of free throws made by the best player = 85% x 40

                                                         = 0.85 x 40

                                                         = 34

ii. Estimated number of free throws made by the second best player = 75%x 40

                                                                 = 0.75 x 40

                                                                 = 30

iii. Estimated number of free throws made by the third best player = 70% x 40

                                                                   = 0.70 x 40

                                                                   = 28

Thus, the equation that gives the estimated number of free throws is 0.85 * 40.

The best player will make about 34 free throws.

8 0
3 years ago
PLEASE HELP
balu736 [363]

Answer:

slope = 1

Step-by-step explanation:

-3 | -3

5  | 5

_____

8  | 8

8/8= 1

7 0
3 years ago
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