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Natali [406]
2 years ago
5

The number of water bottles used during a teams football practice varies directly with the temperature. If a team uses 75 water

bottles when the temperature is 60 degrees, what is the temperature if they use 120 bottles?

Mathematics
1 answer:
AysviL [449]2 years ago
6 0
To find the cost of 1 water bottle you divide 60 degrees into 75 (75/60) which gives you 1.25 degrees for each water bottle. To find 120 water bottles temperature multiply the 1.25 degrees for 1 water bottle times 120 (120x1.25). 120x1.25=150 The answer would be 150degrees
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Answer:

9/8 or 1.125

Step-by-step explanation:

10 1/2 ÷ 9 2/6

= 21/2 ÷ 56/6

= 9/8

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=》 See the attachment.

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A weight clinic recorded the weight lost (in pounds) by each client of a weight control clinic during the last year, and got the
borishaifa [10]

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Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

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51 - 60 ___________ 1 _________ 16

Step-by-step explanation:

Given the data :

35, 26, 31, 17, 46, 30, 28, 21, 26, 34, 15, 27, 7, 18, 16, 57

Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

31 - 40 ___________3 _________ 14

41 - 50 ___________ 1 _________ 15

51 - 60 ___________ 1 _________ 16

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7 0
3 years ago
The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o
frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 36, \sigma = 5

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40 - 36}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0
3 years ago
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