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OLga [1]
2 years ago
5

In 2014, there were 16,400 students at College A, with a projected enrollment increase of 640 students per year. In the same yea

r, there were 15,168 students at College B, with a projected enrollment increase of 948 students per year. According to these projections, how many years will it take for colleges to have the same enrollment
Mathematics
1 answer:
zhenek [66]2 years ago
8 0

Answer:

4 Years

Step-by-step explanation:

I just continually added the numbers til' I got 18,960 on Both Colleges.

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Find Each Difference.<br> 1) 2-(-2)=<br> 2) (-1)-10=<br> 3) 8-7=<br> 4) (-8)-(-6)
Afina-wow [57]

Answer:

1) 4

2) -11

3) 1

4) -2

Step-by-step explanation:

5 0
3 years ago
Is 2/13 a repeating decimal
Cloud [144]

The denominator 13 cannot be factored so that only 2's and 5's show up, so this means that 2/13 is a non-terminating decimal. Therefore, this decimal repeats itself

Use a calculator to see that: 2/13 = 0.153846  153846  153846 ....

The spaces are put in to help make the number more readable. Note how the "153846" keeps repeating forever

8 0
3 years ago
Charlotte and Edgar's ages add up to 22. Charlotte is one year
nadya68 [22]

Answer: Charlotte's 15 and Edgar's 7

Step-by-step explanation:

7*2=14

14+1 = 15

15 + 7 = 22

8 0
2 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
Find the area of the given triangle.
earnstyle [38]

Answer:

75 squared feet

Step-by-step explanation:

Area=1/2×base ×height

4 0
2 years ago
Read 2 more answers
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