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3 years ago

-20/35 * 15/7 equal to in fraction form

Mathematics

1 answer:

pychu [463]3 years ago

4 0

Answer:-60/49

Step-by-step explanation:

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A car manufacturer sent out survey cards to owners who had purchased new cars. The survey card only had boxes to check for repli

pantera1 [17]

Answer: I think the answer is B. I don't fully understand the question but it seems like that would be the answer. You might double check though.

Step-by-step explanation:

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3 years ago

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Solve the system of equation Y=4x-11 Y=x+13 X= Y =

Phoenix [80]

Answer:

(8,21)

Step-by-step explanation:

Since both equations equal y, set both equaitions equal to each other

and solve for x

4x-11=x+13 , subtract x from both sides

3x-11=13, add eleven to both sides

3x=24, divide both sides by 3 to get x alone

x=8

Substitute x=8 into either equation to solve for y

y=4(8)-11

y=32-11

y=21

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3 years ago

PLEASE HELP!! 50 points and brainliest!!!

Kitty [74]

EQUIVALENT WOULD BE THE ANSWER

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3 years ago

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Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago

I need help with these.

sattari [20]

Answer:

See below.

Step-by-step explanation:

5^2 + 12^2 = x^2

25 + 144 = x^2

169 = x^2

x = 13

3^2 + x^2 = [sqrt(10)]^2

9 + x^2 = 10

x^2 = 1

x = 1

1^2 + x^2 = 4^2

1 + x^2 = 16

x^2 = 15

x = sqrt(15)

[sqrt(27)]^2 + x^2 = 6^2

27 + x^2 = 36

x^2 = 9

x = 3

15/5 = c/sqrt(29)

5c = 15 * sqrt(29)

c = 3sqrt(29)

26/13 = x/12

2 = x/12

x = 24

7 0

2 years ago