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4 years ago

What is 4/7 + 1/3 into a fraction?

1 answer:

7 0

To find 4/7 + 1/3, I would first want to find the least common denominator. The least common denominator for theses is 21. So, To get from 7 to 21, I multiply three. What I multiply on top I must do on the bottom, so your new fraction is 12/21. For the other fraction, I multiplied 3 and 7 together to get a common denominator of 21. So the other new fraction is 7/21.

Now all you have to do is add these together. 7/21 + 12/21 = 19/21.

Thus, 4/7 + 1/3 = 19/21.

Hope this helps :)

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What is the product of 7/16, 4/3, and 1/2?

Contact [7]

7/16 * 4/3 * 1/2 = 28/96 = 0.29

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4 years ago

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A diagonal of a rectangle is 8cm and forms an angel measuring 30 with one side.Find the area of the rectangle.

LenKa [72]

Answer:

16 * √3 cm² or approximately 27.71 cm²

Step-by-step explanation:

Given the 8 cm length of the diagonal and the 30° (π/6 radians) angle with one of the sides, we can calculate the width and height of the rectangle using:

8 * cos(π/6) = 8 * ((√3)/2) = 4 * √3

8 * sin(π/6) = 8 * (1/2) = 4

Multiply those two values together to get the area and you get

16 * √3 cm²

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3 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

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3 years ago

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3 years ago

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Alja [10]

12 4/11

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3 years ago

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