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3 years ago

an oil spreads 25 square meters every 10 minutes. how much area will the oil spill cover after 1 hour?

1 answer:

3 0

You know that every 10 minutes the oil moves 25 meters so and in 1 hour there are 60 minutes the equation is 60/10(25) and then you solve

The answer is 150 meters squared

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What is the value of cos theta subscript one

erik [133]

Ok so I really don’t know and I’m just doing this cause I have to do it for some reason so uhhhhhhhh yeah bye

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2 years ago

What is the slope of the line that passes through the points listed in the table?

brilliants [131]

Answer:

B. 2

Step-by-step explanation:

(4, 3) (7, 9) M = Slope

M = $\frac{y^2-y^1}{x^2-x^1}$ Slope formula

$M=\frac{9-3}{7-4}$ Using the slope formula, plot both x and y intercepts

$M=\frac{6}{3}$ Slope needs to be simplified into a whole number

$M=2$

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2 years ago

the digit at the tens place of a two—digit number is 4 times the digit at ones place if the sum of this number and the number fo

zysi [14]

Answer:

Step-by-step explanation:

The number 1 is in the ones place. 1x4=4, so 4 is in the tens place. If you reverse 41, you get 14. 41+14=55.

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2 years ago

The sum of two numbers is 19. Their difference is 65. What are the numbers?

Ludmilka [50]

X = 42 and y = - 23 are the two numbers

3 0

2 years ago

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

2 years ago