Does anyone know this ?

IF 24% of the sixth grade class is in Mrs Whites class and Mrs White has 30 students how many sixth graders are there in all?

Alexeev081 [22]

Given

0.24 × (number of 6th grade students) = 30

Find

(number of 6th grade students

Solution

Divide by 0.24.

... (number of 6th grade students) = 30/0.24 = 125

4 0

3 years ago

Read 2 more answers

6279÷4= could you help me with this problem

Feliz [49]

Answer: 1,569.75

Step-by-step explanation:

6279÷4=1,569.75

:)

7 0

3 years ago

marissa [1.9K]

L = 50 because 35 + 95 = 130. 180 - 130 means the missing angle is 50 degrees.

7 0

4 years ago

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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand 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.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago

Keri is making a doll clothes for a holiday craft show. The wholesale cost of the materials for one outfit is $9.38. If she sell

irinina [24]

Answer:

60%

Step-by-step explanation:

If I get you right, you're asking for the percentage gain to the nearest percentage, right?

Now to do this, we need to get the amount of profit she made. That's quite easy, we only need to subtract the cost price from the selling price. And that equals $15 - $9.38 that equals $5.62.

Now, we divide this value by the cost price and multiply by 100% to get the percentage gain.

I.e 5.62/9.38 × 100%.

That gives 59.9% , which equals 60% when approximated to the nearest percentage.

5 0

3 years ago