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Minchanka [31]
3 years ago
11

Use a graphing calculator to determine which equation for the line of regression, Pearson product-moment correlation value (r),

and predicted value match the data in the chart. The chart below shows the number of hours a student spent studying for a math test and the score the student earned on the test. Hours Spent Studying | 0.5|1.25| 1.5 |1.75 |2 | 2.5 |3 |3.5 |4 |
Test Score,,,................ | 90 | 75| 60 | 83 |75| 80| 99| 97| ? |

A) y = 6.4x + 69.5, r = 0.48
Predicted test score for 4 hours of study equals 95.

B) y = 6.1x + 74.5, r = 0.48
Predicted test score for 4 hours of study equals 99.

C) y = 5.1x + 74.5, r = 0.88
Predicted test score for 4 hours of study equals 95.

D) y = 7.4x + 69.5, r = 0.88
Predicted test score for 4 hours of study equals 99.

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
6 0
The best answer is:
B) y = 6.1x + 74.5, r = 0.48
<span>Predicted test score for 4 hours of study equals 99.
See the attachment for graph
</span>

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Divide 12 by 15 and then you should get 0.8
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Step-by-step explanation:

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An adult stingray can measure about 350 cm long. how many meters is 350 cm?
Liono4ka [1.6K]

Answer:

<u>3.5 meters</u>

Step-by-step explanation:

1 meter = 100 centimeters

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8 0
1 year ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
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