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skelet666 [1.2K]
3 years ago
14

Add. Simplify your answer. 3y^2/y+1 + 3y/y+1

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0
\dfrac{3y^2}{y+1}+\dfrac{3y}{y+1}=\dfrac{3y^2+3y}{y+1}=\dfrac{3y(y+1)}{y+1}=3y\to\fbox{B.}\\\\\\\dfrac{c+5}{3c+3}-\dfrac{4}{3c+3}=\dfrac{c+5-4}{3c+3}=\dfrac{c+1}{3c+3}=\dfrac{c+1}{3(c+1)}=\dfrac{1}{3}\to\fbox{D.}
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The owner of a computer repair shop has determined that their dailyrevenue has mean $7200 and standard deviation $1200. The dail
meriva

Answer:

The required probability is 0.0855

Step-by-step explanation:

Consider the provided information.

The daily revenue has mean $7200 and standard deviation $1200.

\mu_{\bar x}=7200

\sigma=1200

The daily revenue totals for the next 30 days will be monitored.

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

\sigma_{\bar x}=\frac{1200}{\sqrt{30}}=219.089

As we know Z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}

Substitute \bar x=7500, \mu_{\bar x}=7200\ and\ \sigma_{\bar x}=219.089 in above formula.

Z=\frac{7500-7200}{219.089}=1.3693

From the standard normal table P( Z >1.3693) = 0.0855

Hence, the required probability is 0.0855

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