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sladkih [1.3K]
2 years ago
5

A group of hikers buys 8 bags of mixed nuts. Each bag contains 3 and 1/2 cups of mixed nuts. the mixed nuts are shared evenly am

ong 12 hikers. How many cups of mixed nuts will each hiker receive? Explain your answer
Mathematics
1 answer:
nika2105 [10]2 years ago
4 0

Answer:

2 1/3 cups for each hiker

Step-by-step explanation:

8bags x 3.5 cups in each bag =28 cups in total
28 cups shared among 12 hikers = 28/12 = 2 1/3 cups for each hiker

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Which pair of numbers is relatively prime?
Vladimir [108]
I think it is 21 and 50. The only common factor they have is 1.


All the others are divisible by 1 & 3.
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Enter the numerator and the denominator of the fraction in simplest form that is equal to 0.998¯¯¯¯¯¯¯¯.
Nookie1986 [14]

Answer:

998/1000

or  

499/500 simplified

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The table below shows the illuminated percentage of the moon on a lunar cycle of 28 days. The relationship can be described usin
natulia [17]

Answer:

The amplitude of the function is 48

The period of the function is 28

The graph has a vertical shift of   48  units.

Step-by-step explanation:

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7 0
3 years ago
PLEASE HELP IM STRUGGLING!!!​
Mnenie [13.5K]

Answer:

  • 10. 0.99
  • 11. -60
  • 12. 5.5

Step-by-step explanation:

<u>Use the slope formula:</u>

  • m = (y2 - y1)/(x2 - x1)

10.

  • m = (6.24 - 3.27)/(5 - 2) = 2.97/3 = 0.99

11.

  • m = (240 - 360)/(3 - 1) = -120/2 = -60

12.

  • m = (8.84 - 6.09)/(7 - 2) = 2.75/5 = 5.5
6 0
3 years ago
10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0
2 years ago
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