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3 years ago

How do I simplify the expression

Mathematics

1 answer:

Serggg [28]3 years ago

3 0

Answer:

Like this. -2x+10

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The ratio of boys to girls in a school are 2:3 there are 1100 pupils in the school how many are boys ?

uranmaximum [27]

First of all you do 2+3=5 then 1100/5 is 220
so 2*220=440
3*220= 660
so that means 440 boys and 660 girls
it depends on which order the boys and girls are put

6 0

3 years ago

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Solve the equation.<br> S + (-20) = -19

natima [27]

Answer:

S = 1

Step-by-step explanation:

If you add a positive 1 to -20, you would get -19

8 0

3 years ago

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assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an avera

aleksley [76]

Let
n = number of data
s = standard deviation (sample)
S = standard deviation (population)

The working equations is

$\frac{(n-1) s^{2} }{ x^{2}_{right} } \ \textless \ S^{2} \ \textless \ \frac{(n-1) s^{2} }{ x^{2}_{left} }$

To find $x^{2}_{right}$ , : (1 - 0.90)/2 = 0.05

To find $x^{2}_{left}$ , : 1 - 0.05 = 0.95

Degrees of freedom = n-1 = 24 - 1 = 23

This is shown in the figure attached. Since there is no row for df=23, we interpolate. Thus,

$x^{2}_{left} = 13.093$

$x^{2}_{right} = 35.17$

Substitute all values,

$\frac{(24-1) 5.6^{2} }{ 35.17} \ \textless \ S^{2} \ \textless \ \frac{(24-1) 5.6^{2} }{ 13.093} }$

Thus the answer is,

$20.51\ \textless \ S^{2} \ \textless \ 55.09$

5 0

3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

2 years ago

"Five times a number" can be written as 5 x. <br><br><br> True or false?

mars1129 [50]

5x = five times x.
Not to be confused with X^5 (sometimes accidentally written as x5) which means x to the fifth power.
Hope I helped!
~ Zoe

8 0

4 years ago