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bulgar [2K]
3 years ago
10

Please help as soon as possible I WILL MARK YOU AS BRAINLIEST ​

Mathematics
2 answers:
bulgar [2K]3 years ago
6 0

Answer:

The answer is c

Step-by-step explanation:

Because she has two more nickles than dimes so the variable has to be c and the two more refers to (c+2)

den301095 [7]3 years ago
3 0
The answer is the third option
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Students conducted a survey to see which cola was preferred most. People preferred cola A to cola B by a ratio of 9:7. If 108 pe
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Answer:

c:84

Step-by-step explanation:

set your equasion to 9/7 = 108/x .solve the equasion and you get 84 for x

4 0
3 years ago
A committee of four students is being randomly selected from a group of twenty students. How many combinations of four students
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Step-by-step explanation:

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7 0
2 years ago
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard
bulgar [2K]

We have been given that the lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard  deviation is 2.4 years. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.

First of all, we will find the z-score corresponding to sample score 10.1.

z=\frac{x-\mu}{\sigma}, where,

z = z-score,

x = Random sample score,

\mu = Mean

\sigma = Standard deviation.

z=\frac{10.1-12.5}{2.4}

z=\frac{-2.4}{2.4}

z=-1

Since z-score of 10.1 is -1. Now we need to find area under curve that is below one standard deviation from mean.

We know that approximately 68% of data points lie between one standard deviation from mean.

We also know that 50% of data points are above mean and 50% of data points are below mean.

To find the probability of a data point with z-score -1, we will subtract half of 68% from 50%.

\frac{68\%}{2}=34\%

50\%-34\%=16\%

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.

3 0
3 years ago
In 4 hours, the temperature steadily fell from 0°F to −12°F. What was the average change in temperature per hour? The average ch
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