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3 years ago

Can someone solve this and explain?

Mathematics

1 answer:

Anarel [89]3 years ago

7 0

Answer: (y-c)/m
Reasoning:
Subtract c from both sides [ y-c=mx+c-c ]
Divide m from both sides [ (y-c)/m=mx/m ]

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15 POINTS PLEASE HELP IVE ONLY GOT 10 MINS

frosja888 [35]

C is what I think it is

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3 years ago

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A dog breeder would like to know how many Dalmatian puppies are typically born in a litter. He conducts some research and select

podryga [215]

Answer:

c. 6.2 ± 2.626(0.21)

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 101 - 1 = 100

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.626

The confidence interval is:

$\overline{x} \pm M$

In which $\overline{x}$ is the sample mean while M is the margin of error.

The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter.

This means that $\overline{x} = 6.2$

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.626\frac{2.1}{\sqrt{101}} = 2.626(0.21)$

In which s is the standard deviation of the sample and n is the size of the sample.

Thus, the confidence interval is:

$\overline{x} \pm M = 6.2 \pm 2.626(0.21)$

And the correct answer is given by option c.

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3 years ago

Tell me about what you know about "Rounding"

user100 [1]

It is finding the closest estimate to number. Rounding is used in everyday calculations and in many different fields of work

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3 years ago

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The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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