What is the circumference of a circle if the diameter is 25 feet. Show answer as a decimal.

Estimate. then record the product. 640×3 =1800+12+3=1815 so how do I estimate?

pentagon [3]

Round each number and then multiply

6 0

3 years ago

naomi has earned $54 mowing lawns the past two days. she worked 2 1/2 hours yesterday and 4 1/4 hours today. if naomi is pains t

Vaselesa [24]

Answer:

Naomi charges $8 an hour to mow lawns

Step-by-step explanation:

first day // 2 1/2= 2.5 hrs

2nd day // 4 1/4= 4.25 hrs

2.5hrs + 4.25hrs = 54hrs // 6.75hrs = 54hrs

/6.75 /6.75

= 8hrs

Really hope this helps!

8 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

4 years ago

This market is a duopoly and the demand schedule as well as the total revenue is provided below. The number of customers you see

-Dominant- [34]

Answer:

??

Step-by-step explanation:

8 0

2 years ago

Which of the following questions could be used to represent the equation 4 - X= 28?

rosijanka [135]

Answer:

$-24$

Step-by-step explanation:

$4 - -24= 28\\x=-24$

8 0

3 years ago