A birthday cake was cut into equal pieces, and four pieces were eaten. The fraction is 3/7 how many are left

Mathematics

1 answer:

GaryK [48]3 years ago

7 0

3 are left.
3/7 means there were 7 slices, but it's missing 4 slices. (Eaten)

i hope this helps :)

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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.

6 0

3 years ago

What is the value of x?

frez [133]

$\left\{\begin{array}{ccc}x+2y=4&\text{multiply both sides by 5}\\4x-5y=42&\text{multiply both sides by 2}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}5x+10y=20\\8x-10y=84\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad13x=104\qquad\text{divide both sides by 13}\\\\\boxed{x=8}\\\\Answer:\ \boxed{\boxed{x=8}}$