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3 years ago

Tickets to the museum cost $7.50 each. Students receive a 15% discount on admission

Mathematics

1 answer:

Ugo [173]3 years ago

5 0

Answer:

$ 6.375

Step-by-step explanation:

Cost of ticket = $7.50

Discount = 15%

Discount price = 15/100 x 7.50 = 1.125

Cost of student ticket = $7.50 - 1.125 = $ 6.375

I hope im right!!

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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

Sarah bought a lawnmower for $320. She signed up for the buy now pay later plan at the store with the following conditions: $100

kolbaska11 [484]

Answer:

85%

Step-by-step explanation:

8 0

3 years ago

Which of these is a simplified form of the equation 7p + 4 = – p + 9 + 2p + 3p? (1 point) 13p = 13 3p = 5 7 = 4 11 = 15

Rudiy27

Combine the "like terms" in <span>7p + 4 = – p + 9 + 2p + 3p:

7p + 4 = 4p + 9 Again, combine like terms, obtaining 3p = 5.

Solving for p, p = 5/3</span>

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3 years ago

for each one fifth, use a horizontal line or lines to show fractions equivalent to 1/5 and write the equivalent fractions

Zina [86]

Answer:

2/10 3/15

Step-by-step explanation:

Simplifying these answers equals 1/5

3 0

3 years ago

Let an integer be chosen at random from the integers 1 to 30 inclusive. Find the probability that the integer chosen is divisibl

MAVERICK [17]

There are a total of 30 integers and 10 of these are divisible by 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, and 30). So the probability of getting an integer divisible by 3 is $\frac{10}{30}$ = 1/3. The answer is letter C.

4 0

3 years ago