Simplify: -3152 – 10(4.32 + 5 = 5)]

My number has 2 hundreds, the tens digit has 9 more than the ones digit

Svetach [21]

The number is two hundred ninety- one.

6 0

3 years ago

Read 2 more answers

Is 64 divided by 4 16?

IrinaK [193]

Answer:

Yes, 64 divided by 4 is 16.

Step-by-step explanation:

64/4 = 16.

16*2 = 32

32+32 = 64

6 0

3 years ago

you have a party and 5 dozen people will be coming if you figure 2 ears of corn per person how many ears of corn will you need

torisob [31]

A dozen is 12 so 12 • 5 = 60
2 ears of corn per person would be 120 not counting you, counting you it would be 122

Hope this helped.

5 0

3 years ago

Jeremy has a stand in the farmer's market where he sells strawberries by the pound. The table shows the weights of the strawberr

34kurt

C because it’s proportional

3 0

2 years ago

In a large local high school, 19% of freshmen have had their wisdom teeth removed and 24% of seniors have had their wisdom teeth

soldi70 [24.7K]

Answer:

0.2643 = 26.43% probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors.

Step-by-step explanation:

To solve this question, we need to understand the normal distribution, the central limit theorem, and subtraction of normal variables.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction of normal variables:

When we subtract normal variables, the mean of the subtraction will be the subtraction of the means, while the standard deviation will be the square root of the sum of the variances.

In a large local high school, 19% of freshmen have had their wisdom teeth removed. Sample of 60 freshmen:

So, by the central limit theorem:

$\mu_f = 0.19, s_f = \sqrt{\frac{0.19*0.81}{60}} = 0.0506$

24% of seniors have had their wisdom teeth removed. Sample of 50 seniors.

So, by the central limit theorem:

$\mu_s = 0.24, s_s = \sqrt{\frac{0.24*0.76}{50}} = 0.0604$

What is the probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors?

This is the probability that the subtraction of the proportion of freshmen by the proportion of seniors is larger than 0. For this distribution, we have that:

$\mu = \mu_f - \mu_s = 0.19 - 0.24 = -0.05$