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Stella [2.4K]
3 years ago
8

The attendance at a family reunion was 160 people. This was 125% of last year's attendance. How many people attended the reunion

last year.
Mathematics
1 answer:
Alex787 [66]3 years ago
4 0
If you would like to know how many people attended the reunion last year, you can calculate this using the following steps:

125% of last year's attendance is 160 people
125% of x is 160
125% * x = 160
125/100 * x = 160     /*100/125
x = 160 * 100 / 125
x = 128 people (last year)

Result: 128 people attended the reunion last year.
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Answer:

A. 13.5

Step-by-step explanation:

60 ft. = 3 in.

20 ft. = 1 in.

270/20 = 13.5

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2 years ago
Hi i need help on trying to find out how to find the quotient od 3/4 divided by 1/8 can u please tell me the answer i need it in
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6

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4 years ago
e chairman of the statistics department in a certain college believes that 70% of the department’s graduate assistantships are g
8_murik_8 [283]

Answer:

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

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.

For a proportion p in a sample of size n, we have that \mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.7, n = 50

So

\mu = 0.7, \sigma = \sqrt{\frac{0.7*0.3}{50}} = 0.0648

What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

This is 1 subtracted by the probability that it is between 0.6 and 0.73.

Probability it is between 0.6 and 0.73

pvalue of Z when X = 0.73 subtracted by the pvalue of Z when X = 0.6. So

X = 0.73

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

Z = \frac{0.73 - 0.7}{0.0648}

Z = 0.46

Z = 0.46 has a pvalue of 0.6772

X = 0.6

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

Z = \frac{0.6 - 0.7}{0.0648}

Z = -1.54

Z = -1.54 has a pvalue of 0.0618

0.6772 - 0.0618 = 0.6154

NOT be between 0.60 and 0.73?

1 - 0.6154 = 0.3846

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

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