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Nastasia [14]
3 years ago
8

Cos and sinc=_____ yards rounded to nearest hundredth place

Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Answer: c = 8 yards

Work Shown:

cos(angle) = adjacent/hypotenuse

cos(45) = c/(8*sqrt(2))

c = 8*sqrt(2)*cos(45)

c = 8*sqrt(2)*(1/sqrt(2))

c = 8

----------------

Optionally, you can use the 45-45-90 triangle template which says that the length of the leg of this triangle is x and the hypotenuse is x*sqrt(2)

Since 8*sqrt(2) is the given hypotenuse, this must mean x*sqrt(2) = 8*sqrt(2) so x = 8

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Answer:

15 percent

Step-by-step explanation:

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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

c)

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

4 0
3 years ago
D. How many dozen doughnuts are there in 46.2 doughnuts?
eimsori [14]

Answer:

3.85

Step-by-step explanation:

46.2/12=3.85

5 0
4 years ago
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Simon used 3 pears and 9 apples to make a fruit salad. What was the ratio of the number of pears to the number of apples in the
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Answer:

Step-by-step explanation:

Ratio is an expression that shows the relationship between two items in terms of quantity. It shows how much of one item can be found in another.

Simon used 3 pears and 9 apples to make a fruit salad.

This is expressed in terms of quantity.

The ratio of the number of pears which simon used to the number of apples which simon used will be

Number of pears which simon used ÷ the number of apples which simon used.

The ratio is 3/9.

Simplifying further to its lowest term,

It becomes 1/3

This is done by dividing the numerator and denominator by 3

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Answer:

7.   30%

8.   70%

Step-by-step explanation:

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