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2 years ago

Solve for each variable. SHOW YOUR WORK!

Mathematics

1 answer:

vampirchik [111]2 years ago

4 0

Answer:

I DONT KNOW DO YOU WANT TO BE MY FRIEND

Step-by-step explanation:

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If a dog weighs 240 ounces, how many grams does the dog weigh?

agasfer [191]

Answer: If the dog weighed 240 ounces it would weigh 6803.89 grams

Step-by-step explanation:

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3 years ago

I need help asap.......

Vilka [71]

What do you need help with I will respond

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3 years ago

Read 2 more answers

Simplify the complex fraction.

lilavasa [31]

Given:

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}$

To find:

The simplified fraction.

Solution:

Step 1: Simplify the numerator

$$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}$

Step 2: Simplify the denominator

$$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}$

Step 3: Using step 1 and step 2

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}$

Step 4: Using fraction rule:

$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}$

$$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}$

Cancel the common factor r and t², we get

$$=\frac{64 r^{2} \cdot 9 }{16 \cdot 15 t^2 }$

Cancel the common factors 16 and 3 on both numerator and denominator.

$$=\frac{4 r^{2} \cdot 3 }{ 5 t^2 }$

$$=\frac{12 r^{2} }{ 5 t^2 }$

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2} }{ 5 t^2 }$

The simplified fraction is $\frac{12 r^{2} }{ 5 t^2 }$ .

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3 years ago

Which descriptions from the list below accurately describe the relationship between QRS and TUV

Nata [24]

Answer is A. similar

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3 years ago

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A simple random sample of 100 observations was taken from a large population. The sample mean and the standard deviation were de

olga2289 [7]

Answer:

Se=1.2

Step-by-step explanation:

The standard error is the standard deviation of a sample population. "It measures the accuracy with which a sample represents a population".

The central limit theorem (CLT) states "that the distribution of sample means approximates a normal distribution, as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population distribution shape"

The sample mean is defined as:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

And the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

Let X denotes the random variable that measures the particular characteristic of interest. Let, X1, X2, …, Xn be the values of the random variable for the n units of the sample.

As the sample size is large,(>30) it can be assumed that the distribution is normal. The standard error of the sample mean X bar is given by:

$Se=\frac{\sigma}{\sqrt{n}}$

If we replace the values given we have:

$Se=\frac{12}{\sqrt{100}}=1.2$

So then the distribution for the sample mean $\bar X$ is:

$\bar X \sim N(\mu=80,Se=1.2)$

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3 years ago