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Ray Of Light [21]
3 years ago
15

Doug read three books that averaged 156 pages each. He also read 12 short stories that were each six pages long. How many pages

did Doug read altogether?
Mathematics
2 answers:
Maru [420]3 years ago
5 0

Answer:

72

Step-by-step explanation:

you just have to multiply 12 by 6 and then you get 72. so its 72 pages

Naily [24]3 years ago
3 0

Answer:

540

Step-by-step explanation:

Total pages in the 3 books= 156x3=468

Total pages in the 12 short stories= 6x12=72

468+72=540 pages

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Natalija [7]

Answer:

<em><u>Choice C :</u></em>  6.3*10^{6}

<em><u>Step-by-step explanation:</u></em>

Choice A won't be it because it's only 3 million.

Choice B won't work because it's 83 million.

Choice C <u>WILL WORK </u>because it's 98 million.

Choice D won't work because it's only 11 million.

7 0
4 years ago
Tiya flipped a coin 40 times. The coin landed heads up 16 times and tails up 24 times.
beks73 [17]

Answer:

16:40, 0.4

Step-by-step explanation:

experimental probability would be a ratio consisting of the number of times the coin lands on heads to the total number of times the coin is flipped so it would be 16 to 40 or simply 16:40

theoretical probability would be a way to show the likeliness of the coin landing on heads and can be found by dividing the number of flips that is expected to land heads up by the total number of flips so you divide 16 by 40 which equals 0.4

4 0
3 years ago
Someone help ASAP please
notsponge [240]

for number 10a is 4,10,16,22

10c is 3,6,12,24

8 0
4 years ago
Jack is car salesman who works strictly on commision. For every car he sells, he earns 2% commision. Cars sell for $35,000 each.
Dafna11 [192]
He earns 2% commission for every car...and each car sells for 35,000....
so for every car, he earns (0.02(35000) = $ 700

700c = 5000
c = 5000/700
c = 7.14....so he would basically have to sell 8 cars because 7 cars wouldnt quite be enough
5 0
3 years ago
Solve this system of linear inequalities ( will give brainiest)
lakkis [162]
From the graphs you can see that the graph of the line y= \frac{1}{3} x-3 lies under the graph of the line y= \frac{1}{3} x-1. Then all solutions of unequality y\le \frac{1}{3} x-3 are solutions of unequality y\le \frac{1}{3} x-1, but not all solutions of unequality y\le \frac{1}{3} x-1 are solutions of unequality y\le \frac{1}{3} x-3. 
For example, if x=1, y=-2
-2\le  \frac{1}{3} \cdot 1-1 =- \frac{2}{3} \\ -2\ \textgreater \  \frac{1}{3} \cdot 1-3=- \frac{8}{3}.
Answer: Correct choice is B.



3 0
4 years ago
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