27y-12y^3 polynomial

The box plots show the average gas mileage, in miles per gallon, of cars and minivans tested by a certain company.

mr_godi [17]

Answer:

b.The interquartile range for cars is about 7 mpg, and the interquartile range for minivans is about 3 mpg.

Step-by-step explanation:

i got it right

4 0

3 years ago

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Tina is the top seller of the store. She earns a salary plus 2%. Tina's salary is $30,000 a year. This month her sales were $48,

Lilit [14]

I'm assuming she gets 2% commission on sales and that she gets paid monthly from her yearly salary.

If the above is true, her monthly pay is calculated as follows:

Commission: 48500 * 0.02 = $970
Monthly salary: 30000 ÷ 12 = $2500

So monthly pay is $2500 + $970 = $3470

7 0

3 years ago

+ 96

Brut [27]

Answer:

90°

Step-by-step explanation:

It looks slightly obvious, but here's the complicated explanation.

Converse of the Consecutive Interior Angles Theorem. Solve.

x+96+x+96=180

Simplify.

2x+192=180

Subtraction Property of Equality (subtract 192 from both sides to isolate 2x).

2x=-12

Division Property of Equality (divide both sides by 2 to isolate x).

x=-6

Now that you have your x value, plug it in into your bolded equation.

x+96

-6+96

Simplify.

90°

8 0

3 years ago

Is the output for this correct? (See full pic)

Burka [1]

It is correct if you're trying to add 0 for the first one
1 for the second one
and so on

4 0

4 years ago

Read 2 more answers

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of

11111nata11111 [884]

Answer:

We conclude that the average time someone spends in the gym is different from 56 minutes.

Step-by-step explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let $\mu$ = population average time someone spends in the gym

SO, Null Hypothesis, $H_0$ : $\mu$ = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average time someone takes in the gym = 50 min

$\sigma$ = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics = $\frac{50-56}{\frac{8}{\sqrt{32} } }$

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.

4 0

3 years ago