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o-na [289]
3 years ago
15

Estimate each percent. 19% of 51 69% of 203

Mathematics
2 answers:
lyudmila [28]3 years ago
7 0
The answer for
1 - 9.69
2 - 140.07
slamgirl [31]3 years ago
7 0

Answer:

9.69 & 140.07

Step-by-step explanation:

0.19*51=9.69

0.69*203=140.07

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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

3 0
3 years ago
1) Gena’s new outfit originally cost $75. She received 20% off. How much did Gena’s outfit cost?
Paha777 [63]

Answer:

1. $60

2. 600 Calories were burned

3. D

Step-by-step explanation:

1. $75 x .20=$15

75-15=$60

2. What I started with is 200/40=5 calories burned per min

60 plus 60=120

5x120=600

3. It's the only option that has something decreasing. and there is a -5 which means it's getting subtracted or its decreasing.  

6 0
3 years ago
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PLZ HELP ASAP A candy machine contains spherical candies. Each candy is solid and has a diameter of 0.25 in.
NikAS [45]
Each candy has a volume of about 8.18 * 10^-3 which is simplified to .00818
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3 years ago
A queen ant lays 1.83 x 10^6 eggs over a period of 30 days. Assuming she lays the same number of eggs each day, about how many e
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The answer is 61,000
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PLZS HELP NEED BY TODAY!!
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X=20. Because that line is 180 and u subtract 40 and u do 140/7= 20
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2 years ago
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