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pashok25 [27]
3 years ago
15

Find the value for q for the following equation: 15q – 16 = 4(2q + 3)

Mathematics
2 answers:
Mars2501 [29]3 years ago
4 0
Answer:
Q=4
15q-16=8q+12
7q-16=12
7q=28
Q=4
TEA [102]3 years ago
3 0
Step 1: Multiply 4 by 2q and 3

Now your equation is 15q - 16 = 8q + 12

Step 2: Subtract 8q from both sides

Now your equation is 7q - 16 = 12

Step 3: Add 16 to both sides

Now the equation is 7q = 28

Step 4: Divid each side by 7

Q= 4
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2 years ago
Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points. Rachel scored 78 p
Artemon [7]

Answer:

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Step-by-step explanation:

Z-score:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.

This means that X = 80, \mu = 77, \sigma = 4.9

So

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

Z = \frac{80 - 77}{4.9}

Z = 0.61

Keenan's z-score was of 0.61.

Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.

This means that X = 78, \mu = 75, \sigma = 3.7. So

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

Z = \frac{75 - 78}{3.7}

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2 years ago
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3 0
3 years ago
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