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MAXImum [283]
3 years ago
5

A scientist measures the water temperature in the Gulf at Gulfport on the fifteenth of each month. Her data is shown in the tabl

e. What is the average rate of change between March 15 and June 15?
A. 2.6°F per month
B. 3.9°F per month
C. 5.2°F per month
D. 7.8°F per month
Please explain!

Mathematics
2 answers:
andrew-mc [135]3 years ago
6 0
The average rate of change between March 15 and June 15 will be given by:
(82.4-66.8)/(6-3)
=15.6/3
=5.2° F per month

Answer: C
Zina [86]3 years ago
3 0

Answer:

<h2>C. 5.2°F per month</h2>

Step-by-step explanation:

The average rate of change refers to the relation between these variables.

Specifically, this problem can be modeled by a linear function, where the slope is the average rate of change.

So, such rate can be defined as

r=\frac{y_{2}-y_{1}  }{x_{2} -x_{1} }

Where (x_{1} ,y_{1} ) is the first pair and (x_{2} ,y_{2} ) is the second one.

In this case, the first pair is (3,66.8) and the second pair is (6,82.4). Replacing these coordinates, we have

r=\frac{y_{2}-y_{1}  }{x_{2} -x_{1} }=\frac{82.4-66.8}{6-3}=\frac{15.6}{3}=5.2

Which is expressed in Fahrenheit.

Therefore, the average rate of change is 5.2 °F.

This can be interpreted as "The temperature changes +5.2 °F each month between March 15 and June 15".

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Read 2 more answers
A corporation has 11 manufacturing plants. Of these, seven are domestic and four are outside the United States. Each year a perf
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Answer:

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Step-by-step explanation:

Total plants = 11

Domestic plants = 7

Outside the US plants = 4

Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure

Here we have n=4, p=4/11 and q=7/11

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰

          = 1 - 0.16399

P(X≥1) = 0.836

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

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