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2 years ago

Help me please this is hard

Mathematics

2 answers:

soldi70 [24.7K]2 years ago

4 0

$> \: is \: number \: 3 \: you \: hve \: to \: go \: in \: order \: and \: see \: what \: numbers \: are \: bigger$

Ghella [55]2 years ago

3 0

The answer might be < but if you want do >

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What is the median of the data set?

gizmo_the_mogwai [7]

7. because you would put the number set into order from least to greatest first, ti would look like this: 2,3,4,5,9,10,12,12. then go to the middle, crossing one from each side off at one time. Then you get 5 and 9 in the middle. Then the middle between 5 and 9 is seven, you can check this by: 5+9=14, then 14/2= 7.
Your median is 7.

6 0

2 years ago

PLEASE HELP THIS IS MY LAST QUSTION OF THE DAY!!! :)<br> thanks in advanced

notsponge [240]

So using standard form you convert the equation to 5x - 2y = -10 then use slope m of a line of the form Ax + By = C equals negative a over b. It would be a = 5 b = -2 m = -5/-2 m = 5/2. Have a good day the answer is 5/2

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3 years ago

Part A

Liono4ka [1.6K]

Because i did the math and i had this question before

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2 years ago

An operator in the plant has a monthly salary of $2575. Thirty-three percent (33%) of her salary is deducted for income tax.

katovenus [111]

2575X.33=849.75

2575-849.75=1,725.25

6 0

3 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$

$S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\$

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73$

5 0

3 years ago