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Andrew [12]
3 years ago
6

Dr. Betz, a vet, is running a free rabies clinic. He estimates that it will take him 12 minutes for each animal he treats. Dr. B

etz has already seen 20 animals, the last of which was a shaggy dog. Write an equation for how many minutes Dr. Betz worked. How long did Dr. Betz work at the rabies clinic if he saw 20 more animals after the shaggy dog?
Mathematics
1 answer:
Olin [163]3 years ago
7 0

Answer:

equation: y = 12x

Dr. Betz worked for 8 hours.

Step-by-step explanation:

To write the equation we'll use the form y = mx, where y represents the amount of time the Doctor worked, m is the amount of minutes it takes to treat each animal, and x is how many animals have been treated. We know the value of m is 12, so let's put that into the equation:

y = 12x

Now we have our equation. We can use it to see how long Dr. Betz worked. He treated a total of 40 animals - 20 including the shaggy dog, and 20 after the shaggy dog. let's replace x with 40:

y = 12(40)

Now solve:

y = 480

Dr. Betz worked for 480 minutes. There are 60 minutes in an hour. Divide 480 by 60 to find out how many hours Dr. Betz worked for:

480/60 = 8

Dr. Betz worked for 8 hours.

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yes

Step-by-step explanation:

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2 years ago
According to a survey, the average American person watches TV for 3 hours per week. To test if the amount of TV in New York City
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Answer:

Test statistic (t-value) of this one-mean hypothesis test is -2.422.

Step-by-step explanation:

We are given that according to a survey, the average American person watches TV for 3 hours per week. She surveys 19 New Yorkers randomly and asks them about their amount of TV each week, on average. From the data, the sample mean time is 2.5 hours per week, and the sample standard deviation (s) is 0.9 hours.

We have to test if the amount of TV in New York City is less than the national average.

Let Null Hypothesis, H_0 : \mu \leq 3  {means that the amount of TV in New York City is less than the national average}

Alternate Hypothesis, H_1 : \mu > 3   {means that the amount of TV in New York City is more than the national average}

The test statistics that will be used here is One-sample t-test statistics;

        T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean time = 2.5 hours per week

             s = sample standard deviation = 0.09 hours

             n = sample size = 19

So, <u>test statistics</u> = \frac{2.5 - 3}{\frac{0.9}{\sqrt{19} } } ~ t_1_8

                              = -2.422

Therefore, the test statistic (t-value) of this one-mean hypothesis test (with σ unknown) is -2.422.

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Answer:

There is not sufficient evidence to support the claim μ > 54.4.

Step-by-step explanation:

1) Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

2) Solution to the problem

On this case we want to test is \mu>54.4 and the system of hypothesi on this case are:

Null Hypothesis: \mu \leq 54.4

Alternative hypothesis:\mu >54.4

On this case is our decision is FAILS to reject the null hypothesis then we can conclude that we don't have enough evidence to support the claim at the significance level provided.  So the correct conclusion would be:

There is not sufficient evidence to support the claim μ > 54.4.

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