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Setler79 [48]
3 years ago
7

If Alonzo got 4 bull-eyes out of 22 shots, what is his percentage of perfect shots ?

Mathematics
1 answer:
allsm [11]3 years ago
3 0
4 percent of 22 equals 0.88 
hope that helps ya 
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Emily estimated that there were 315 students at a soccer game the actual number of students at the game with 350 what was the pe
Readme [11.4K]

Answer:

About 11.10 Percent off. The change from 315 to 350 was an increase


Step-by-step explanation:


4 0
3 years ago
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

              b. P(1000<x<2000) = 0.2325

              c. P(x<1000) = 0.6321

              d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

P(0<x<1000) = e^{-1}-1

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0
3 years ago
Please help me out i will reward brainliest
Nadusha1986 [10]
I can not see the picture so i caN NOT HELP SORRY

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%2816%20%2B%204%29%20%5Ctimes%209%20%3D%20%2816%20%5Ctimes%20x%29%20%2B%20%284%20%5Ctimes%20y%
RSB [31]
Isolate the variable by dividing each side by factors that don't contain the variable.

x= 13/4 - y/4



If you are solving for x (only)
X = 45/4 - y/4

If you are solving for y (only)
Y= 45- 4x
6 0
3 years ago
Simplify -6 1/4 - (-2 1/8)
OLEGan [10]
- 6   \frac{1}{4}  +  2  \frac{1}{8}

- 6   \frac{1}{4}  -  (- 2  \frac{1}{8}  )

(-6 + 2 ) + ( -\frac{1}{4}  +   \frac{1}{8} ) )

(-4) + ( - \frac{1}{8})

- 4  \frac{1}{8}
6 0
3 years ago
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