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s344n2d4d5 [400]
3 years ago
9

100 POINT AND BRAINIEST TO THE CORRECT ANSWER WITH THE BEST EXPLANATION.

Mathematics
2 answers:
Naddika [18.5K]3 years ago
4 0

If 2/5 of the population are boys, then that means 3/5 of the population is girls

The proportion for girls is 3/5

Total students, divide total boys by fraction of boys:

450 / 2/5 = 1125 total students.

Number of girls = 1125 - 450 = 675 girls.

Vitek1552 [10]3 years ago
3 0

Answer:

⅗ of total;

675 girls

Step-by-step explanation:

⅖ × Total = 450

Total = 1125

p(Girls) = 1 - p(Boys)

= 1 - ⅖

= ⅗ × 1125

= 675

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OlgaM077 [116]

Answer: d = ±10

<u>Step-by-step explanation:</u>

169 = (\frac{24}{2})^{2} + (\frac{d}{2})^{2}

169 = (12)² + \frac{d^{2}}{4}

169 = 144  + \frac{d^{2}}{4}

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6 0
3 years ago
A set of data items is normally distributed with a mean of 400 and a standard deviation of 60. Find the data item in this distri
Softa [21]

Answer:

The data item is X = 580

Step-by-step explanation:

Normal Probability Distribution:

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.

Mean of 400 and a standard deviation of 60.

This means that \mu = 400, \sigma = 60

z=3

We have to find X when Z = 3. So

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

3 = \frac{X - 400}{60}

X - 400 = 3*60

X = 580

The data item is X = 580

6 0
3 years ago
the average rainfall in Clearview is 38 inches . this year 29.777 inches fell . How much less rain fell this year than falls in
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38 (average) - 29.777 (this year) = 8.223 inches less this year than the average year.
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