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Anika [276]
3 years ago
8

Please help!! Giving brainliest and extra points!! (Click on the photo)

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
8 0

Answer:

B) 24

Step-by-step explanation:

Hw = 20% = 0.2

120(0.2) = 24

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The $299.99 of a cell phone Brittany purchesd was on sail for 10% off. What amount did Brittany save?
Oksi-84 [34.3K]

Answer:

10* 299.99=2999.9

2999.9/100=29.999

this would round up to 30, so Brittany saved $30. : )

6 0
3 years ago
3. Find the value of a and b when<br>5+√6/5-√6 = a+b√6​
mr Goodwill [35]

The answer for A is,

a = 5 +  \frac{ \sqrt{6} }{ \sqrt{5} }  -  \sqrt{6}  -  \sqrt{6b}

And the answer for B is,

b = - 1 +  \frac{5 +  \frac{ \sqrt{6} }{ \sqrt{5} } - a  }{  \sqrt{6} }

hope it helps.

sorry of its wrong.

4 0
3 years ago
Michela says that the modes of the two data are the same so the median and mean must also be the same. what is michela’s error?
aev [14]

Answer: median and mean are the same

Step-by-step explanation:

The error committed by Michela is by saying

The median and mean are the same.

Because in statistics, we can't use mode to predict the median and the mean.

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=6%5Cfrac%7B1%7D%7B5%7D%20%2B8%5Cfrac%7B4%7D%7B5%7D" id="TexFormula1" title="6\frac{1}{5} +8\fr
Allushta [10]
6 + 8 = 14

1/5 + 4/5 = 5/5 or 1

14 + 1 = 15
4 0
3 years ago
Read 2 more answers
A dataset shows that, on average, smokers do not live as long as nonsmokers and heavy smokers do not live as long as light smoke
leva [86]
<span>Given that A dataset shows that, on average, smokers do not live as long as nonsmokers and heavy smokers do not live as long as light smokers.

If you compute the least squares line for y = age at time of death and x = number of packets per day typically smoked, you will notice that as the number of packets per day a person smokes increases, the age at time of death of the person decreases.

Thus, the graph wil have a negative slope.

Therefore, the slope of your regression line will be less than 0.</span>
8 0
3 years ago
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