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tigry1 [53]
3 years ago
5

5.3 times 10.4 is??? (Show work)

Mathematics
2 answers:
lora16 [44]3 years ago
8 0

Answer:

= 55.12

Solution:

There's 2 total decimal places in both numbers.

Ignore the decimal places and complete the multiplication as if operating on two integers.

Rewrite the product with 2 total decimal places.

Answer = 55.12

Therefore:

5.3 × 10.4 = 55.12

ololo11 [35]3 years ago
7 0

Answer:

do the work on your calculator to double check yourself :)

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It was estimated that 800 people would attend the play, but 752 actually attended. what is the percent error?
natima [27]

Answer:

94 percent

Step-by-step explanation:

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(1,5), (9,85), (2,10),(6,38),(4,3),(12,107),(7,64), (12,86) ,(7,47), (9,64), (4,27) The line is in the form y=mx+b.y=mx+b. What
alexandr1967 [171]

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should be the right answer since I used a calculator :p

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-4(4x-9)=2x<br> I need help
Stolb23 [73]

Step-by-step explanation:

-4(4x-9)=2x

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3 years ago
Read 2 more answers
Find the 50th term of the sequence 5,-2,-9,-16 ,,,
Mademuasel [1]
Common difference is -2-5 = -7
first term is 5

a_50 = 5 -7(50-1) = -338

2) d = 3 - (-7) = 10
 a = -7

a_110 = -7 + 10(110 - 1) = 1083

d = -27 - (-22) = -5

second -7 - 5=  -12
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7 0
3 years ago
There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit
Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a \frac{3}{11} chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is \frac{8}{11}, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is\frac{2}{11}, and the probability of picking the second red ball is \frac{1}{10} and the probability of picking the green ball is\frac{1}{9}. We want to multiply thm again, as per the multiplication rule like the last problem.

\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}

5 0
3 years ago
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