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frosja888 [35]
3 years ago
14

The table shows the weights of a sample of packages carried by an air freight company last month. Find the mean deviation for th

e data in the quantity column.
Select one:

8.09

1.52

2.84

6.18

Mathematics
1 answer:
podryga [215]3 years ago
3 0

Answer:

1.52

Step-by-step explanation:

\frac{5 + 9 + 5 + 6 + 8 }{5} = \frac{33}{5} = 6.6

distance from 6.6

5,  1.6

9,  2.4

5,  1.6

6,   .6

8,  1.4

\frac{1.6 + 2.4 + 1.6 + .6 + 1.4}{5} = \frac{7.6}{5} = 1.52

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Find the discount if a $69.99 item is on sale for $39.99.
inn [45]
The easiest way to do it (for me) is to find 1% of the regular price. I think we can round the prices to $70,00 and $40,00 to show nice percentage.

70 / 100 = 0,7

So 1% is 0,7. Now we can divide $40,00 by 0,7 to calculate how many percent it is:

40 / 0,7 = 57 (after rounding it to the nearest whole number)

So the discount was 100 - 57 = 43 percent.

Doublecheck (optional):

70 * 57% =
= 70 * 57/100 =
= 7 * 57/10 =
= 7 * 5 7,10 =
= 7 * 5,7 =
= 39,9 (the price after the discount)

70 * 43% =
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= 7 * 43/10 =
= 7 * 4 3/10 =
= 7 * 4,3 =
= 30,1 (the value of the discount)

The difference of 0,1 is because of the rounding, but it's correct.

8 0
3 years ago
Is x + 1 a factor of the polynomial function p(x) = x^3 – 7x^2 - 10x + 16​
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Answer:

slove it by yourself dear

Step-by-step explanation:

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Differences in linear and quadratic equations
BabaBlast [244]

Answer:

linear equation has two variables doesn't involve any power

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Step-by-step explanation:

A <u>linear equation</u> in two variables doesn't involve any power higher than one for either variable. It has the general form Ax + By + C = 0, where A, B and C are constants. ... A <u>quadratic equation</u>, on the other hand, involves one of the variables raised to the second power.

7 0
3 years ago
Please I'm in urgent need of help. what must be added to make (1/4)x^2-(2/3)xy a perfect square?.​
Ipatiy [6.2K]

Answer:

\bigg(\frac{2}{3}  {y} \bigg)^{2}

STEP BY STEP EXPLANATION

\frac{1}{4}  {x}^{2}  -  \bigg( \frac{2}{3}  \bigg)xy \\  \\  =   \bigg(\frac{1}{2}  {x} \bigg)^{2}  - 2. \bigg(\frac{1}{2}  {x} \bigg)\bigg( \frac{2}{3} y \bigg ) +  \bigg(\frac{2}{3}  {y} \bigg)^{2} \\  \\  = \bigg(\frac{1}{2}  {x}  -  \frac{2}{3} y\bigg)^{2}  \\

To make \red{\bold{\frac{1}{4}  {x}^{2}  -  \bigg( \frac{2}{3}  \bigg)xy}} a perfect square we should add \purple{\bold{\bigg(\frac{2}{3}  {y} \bigg)^{2}}}

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