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3 years ago

What is an expression for 5(9m+2n-6m)

Mathematics

2 answers:

VARVARA [1.3K]3 years ago

8 0

45m+10n-30m
15m+10n

What I got, let me know if you need something else, <3

Elis [28]3 years ago

6 0

45m+10n-30m
15m+10n

I believe this is right.

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Given that f(x) = x2 + 12x + 32 and g(x) = x+ 4, find (f - g)(x) and

frosja888 [35]

Answer:

x²+11x+28

Step-by-step explanation:

(f - g) (x) = f(x) - g(x)

x² + 12x + 32 - (x + 4) = x² + 11x + 28

6 0

3 years ago

A teacher can spend as much as $95.00 to take students to a movie. A movie theater charges a $10.00 group fee and a $5.50 per ti

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3 years ago

what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

kramer

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

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

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

$\bar x = 20$

$x = 21$

Calculate the z score

$x = \frac{x - \mu}{\sigma}$

$x = \frac{21 - 20}{2}$

$x = \frac{1}{2}$

$x = 0.50$

So, we have:

$P(x \le 21) = P(z \le 0.50)$

From the z table

$P(z \le 0.50) = 0.69146$

So:

$P(x \le 21) = 0.69146$

4 0

3 years ago

$7000 is compounded semiannually at a rate of 11% for 21years Find the total amount and round to the nearest cent

photoshop1234 [79]

$\text{Given that }\$7000 \text{ is compounded semiannualy at a rate of }11\% \text{ for 21 years}\\
\\
\text{we know that the amount after t year when compounded is given by}\\
\\
A=P\left ( 1+\frac{r}{n} \right )^{nt}\\
\\
\text{here P is the principal amoount, so }P=7000,\\
\\
\text{r is the interest rate, }r=11\%=0.11\\
\\
\text{n is the number of times in a year, here semiannulay, so }n=2,\\
\\
\text{and t is the time, so }t=21\\
\\
\text{so the amount after 21 years is}$

$A=7000\left ( 1+\frac{0.11}{2} \right )^{2(21)}\\
\\
\Rightarrow A=7000\left ( 1+0.055 \right )^{42}\\
\\
\Rightarrow A=7000\left ( 1.055 \right )^{42}\\
\\
\Rightarrow A\approx 66328.68$