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2 years ago

3/4 + 5/4 with explanation and steps....

Mathematics

2 answers:

zhenek [66]2 years ago

7 0

Answer:

step 1

3/4+5/4

step 2

dinominator is same so we have to put 4

step 3

3+5/4

step 4

7/4

answer is 7/4

Ray Of Light [21]2 years ago

3 0

Answer:

Step-by-step explanation:

Given

$\frac{3}{4}$ + $\frac{5}{4}$

Since the denominators are common, both 4 then add the numerators leaving the denominator

= $\frac{3+5}{4}$

= $\frac{8}{4}$

= 2

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Two groups of students were asked how far they lived from their school. The table below shows the distances in miles:

pychu [463]

Answer:

C. The value for Group A is double the value for Group B

Step-by-step explanation:

First, you add all the numbers from each group.

A: The distance in total is 39.06

B: The distance in total is 19.53

To find the mean, which is another way of saying the average, you divide the sum with how many numbers there are. (hope this makes sense, I suck at wording)

A: There are 9 numbers in total, hence, 39.06/9 = 4.34

B: There are 9 numbers in total, hence, 19.53/9 = 2.17

Those are the mean distance for the two groups.

2.17 * 2 = 4.34

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3 years ago

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A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

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3/4 + 5/4 with explanation and steps.... ​

3/4 + 5/4 with explanation and steps....