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2 years ago

Simon can see two lights, light A and light B.

Mathematics

1 answer:

xz_007 [3.2K]2 years ago

3 0

Answer:

2 Times

Step-by-step explanation:

Simon can see two lights A and B

Light A flashes every 15 seconds.
Light B flashes every 18 seconds.

First, we determine the next time both lights will flash together.

This is done by finding the Least Common Multiple of 15 and 18.

15=3X5

18= $2X3^2$

LCM of 15 and 18= $2X3^2X5=90$

This means that lights A and B will first flash together after 90 seconds.

Now, 240/90= $2.\overline{6}$ seconds

Therefore, the number of times more both light will flash together in the next 4 Minutes= 2 Times

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I believe that the answer to the question would be 4.5

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Solve for y 4/7t+2:7=6

Art [367]

Answer:

1/7

Step-by-step explanation:

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1 year ago

A triangular road sign has a height of 3 feet and a base of 2.5 feet. How much larger in area is this sign than one with a heigh

jeka57 [31]

Answer:

Not larger or smaller; the areas are the same

Step-by-step explanation:

Recall that the formula for the area of a triangle of base b and height h is

A = (1/2)(b)(h).

In the first case, h = 3 ft and b = 2.5 ft. This results in an area of 3.75 ft^2.

In the second case, h = 2.5 ft and b = 3 ft. The area is the same: 3.75 ft^2.

This is because A = (1/2)(b)(h) has the same value regardless of the order in which b and h are used as multipliers.

There is no difference in the areas of the two signs. The areas are the same.

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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 R 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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2 years ago

The range of 16,23,55,64,49

miskamm [114]

Answer: 48

Step-by-step explanation: To find the range of the data set shown here, remember that the range is the difference between the greatest number in the data set and the least number in the data set.

Greatest number → 64

Least number → 16

Now, we need to subtract 16 from 64.

64 - 16 = 48

Therefore, the range of the data set is 48.

7 0

3 years ago