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3 years ago

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the ages ( in years) of the 5 employees at a particular computer store are 32,23,21,38,26 . assuming that theses ages constitute

an entire population, find the standard deviation of the popultion

1 answer:

jolli1 [7]3 years ago

3 0

Step 1: Add up the ages.

32+23+21+38+26=140.

Step 2: Divide answer by the amount of used numbers (5, in this case.)

140/5=28.

Your answer is 28.

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

s2008m [1.1K]

Answer: 912

===================================

Work Shown:

The starting term is a1 = 3. The common difference is d = 5 (since we add 5 to each term to get the next term). The nth term formula is

an = a1+d(n-1)

an = 3+5(n-1)

an = 3+5n-5

an = 5n-2

Plug n = 19 into the formula to find the 19th term

an = 5n-2

a19 = 5*19-2

a19 = 95-2

a19 = 93

Add the first and nineteenth terms (a1 = 3 and a19 = 93) to get a1+a19 = 3+93 = 96

Multiply this by n/2 = 19/2 = 9.5 to get the final answer

96*9.5 = 912

I used the formula

Sn = (n/2)*(a1 + an)

where you add the first term (a1) to the nth term (an), then multiply by n/2

-----------------

As a check, here are the 19 terms listed out and added up. We get 912 like expected.

3+8+13+18 +23+28+33+38 +43+48+53+58 +63+68+73+78 +83+88+93 = 912

There are 19 values being added up in that equation above. I used spaces to help group the values (groups of four; except the last group which is 3 values) so it's a bit more readable.

5 0

3 years ago

Read 2 more answers

Which of the following is equal to tan(A)?

Scrat [10]

Answer: cot B

Step-by-step explanation: if i am right mark me as brainliest

8 0

3 years ago

Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0

3 years ago

How many tens places are in 49?

kow [346]

49 = 4 tens + 9 ones

4 0

3 years ago

Read 2 more answers

Working out as well please. Thanks

dimulka [17.4K]

6.1/5 is bigger because 100÷5=20 7.1% of 40=0.4 × 20= 8 40+8 = 48 8=?? Hope this helps

7 0

3 years ago