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quester [9]
2 years ago
4

A study was conducted on the effect of daytime running lights on cars. Researchers gathered data from a random sample of 3,248 d

rivers and measured quite a few variables, in addition to the explanatory variable of daytime running lights and the response variable of less accidents. According to a newspaper article summarizing the study, those cars that had daytime running lights were more likely to be operated by drivers who were confident and attentive. Drivers whose cars did not have daytime running lights were about 22% more likely to have an accident. What conclusion can we draw from this study
Mathematics
1 answer:
HACTEHA [7]2 years ago
3 0

Answer:

We can infer a cause-and-effect relationship because multiple variables were tested in a experimental and control groups of a random sample.

The cause and effect relationships discovered from the study include

1) The presence (or absence) of daytime running lights and the type of drivers that'll drive the car.

2) The presence (or absence) of daytime running lights and whether or not the car was prone to accidents.

Step-by-step explanation:

From the results printed in the newspaper, it is evident that

- The sample were randomly selected.

- There was an experimental group of cars with daytime running lights.

- And a control group of cars with no daytime running lights.

- Then, different drivers were sampled, ones that were confident and attentive being one of the favourable groups.

- And the cars were investigated for which ones were more prone to accidents.

That the sample is of considerable size and is a random sample means that the results of this study can be generalized.

There were multiple variables tested, presence or absence of daytime running lights, the type of drivers, whether or not the car is prone to accidents.

Then the conclusion from the study, according to the newspaper, shows that there is a relationship between

- The presence (or absence) of daytime running lights and the type of drivers that'll drive the car.

- The presence (or absence) of daytime running lights on whether or not the car was prone to accidents.

Hope this Helps!!

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The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr
9966 [12]

Answer:

(a) Perimeter = 32.2\ cm

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

Area = 64.8

Required

Calculate the perimeter.

Area is calculated as:

Area = L * W

Where

L = Length and W = Width

Substitute 64.8 for Area

64.8 = L * W

Make L the subject:

L = \frac{64.8}{W}

Perimeter is calculated as:

P = 2 * (L + W)

Substitute 64.8/W for L

P = 2 * (\frac{64.8}{W} + W)

P = \frac{129.6}{W} + 2W

To solve further, we take the derivative of P and set it to 0, afterwards.

dP = -\frac{129.6}{W^2} + 2

Set to 0

0 = -\frac{129.6}{W^2} + 2

Collect Like Terms

\frac{129.6}{W^2} = 2

Cross Multiply:

2W^2= 129.6

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Take square roots

W = \sqrt{64.8

W = 8.05

Recall that:

L = \frac{64.8}{W}

So:

L= \frac{64.8}{8.05}\\

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Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

Tenths = \frac{1}{10}

So, the expression becomes:

((1 Group of 10 * \frac{1}{10}) and (1 group of 8 * \frac{1}{10}))/(6 groups of 3 * \frac{1}{10})

This gives:

((1 Group of \frac{10}{10}) and (1 group of \frac{8}{10}))/(6 groups of \frac{3}{10})

Group means product, so the expression becomes:

\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}

And, as used here means addition

\frac{(1 * \frac{10}{10} +  1 * \frac{8}{10})}{6 * \frac{3}{10}}

Simplify:

\frac{(1 * 1 +  1 * 0.80)}{6 *0.30}

\frac{(1 +  0.80)}{1.80}

\frac{1.80}{1.80}

= 1.00

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