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3 years ago

Solve for 20-u=3u simplify answer

2 answers:

6 0

<h3>Step-by-step explanation: Solve for <em>u</em> by simplifying both sides of the equation, then isolating the variable . u=5</h3>

postnew [5]3 years ago

5 0

$20 - u = 3u$
simplify both sides of the equation by adding u to both sides and then you have
$20 = 4u$
simplify again by dividing both sides by 4 and then you get
$u = 5$

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Alex recently bought some new video games for the following prices: 24.00,36.00,50.00,44.00 If the sales tax is 8% , find the me

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3 years ago

I put $500 in a saving account that warns 7% interest compounded quarterly. How much will be in the account after 1 year? Rounde

Fantom [35]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 7\%\to \frac{7}{100}\to &0.07\\
n=
\begin{array}{llll}
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\end{array}\to &4\\
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\end{cases}
\\\\\\
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3 years ago

Rationalize the denominators.

Ronch [10]

Step-by-step explanation:

that probably should be your answer

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3 years ago

According to the US Bureau of Labor Statistics publication News, self-employed persons with home-based businesses work a mean of

worty [1.4K]

Answer:

The answer is below

Step-by-step explanation:

Given that mean (μ) = 23 hours, standard deviation (σ) = 10 hours

a) The population is a group of self employed home based workers while the variable is the number of hours worked per week.

b) The mean of the distribution of sample means (also known as the Expected value of M) is equal to the population mean μ.

$\mu_x=\mu=23\ hours$

The standard deviation of the distribution of sample means is called the Standard Error of M, it is given by:

$\sigma_x=\sigma/\sqrt{n} =\frac{10}{\sqrt{100} }=1$

c) $\mu_x=\mu=23\ hours$

$\sigma_x=\sigma/\sqrt{n} =\frac{10}{\sqrt{1000} }=0.32$

d) The sample size has no effect on the mean, hence increasing the sample size does not change the mean.

The square root of sample size is inversely proportional to the standard deviation therefore increasing the sample size reduces the standard deviation.

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3 years ago