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2 years ago

Someone please help me

Mathematics

1 answer:

natka813 [3]2 years ago

3 0

Answer:

I would say ray but im not 100% sure.

Step-by-step explanation:

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a store is giving away $10 coupons to every 7th person to enter the store and a $25 coupon to every 18th person too emter the st

almond37 [142]

7,14,21,28,35,42,49,56,63,70,77,84,91,98,105,112,119,126
18,36,54,72,90,108,126

So the first person to get both of them is the 126th person

3 0

3 years ago

Please help me answer.

katen-ka-za [31]

Answer:

Slope-intercept: y=3/4x+1

Sorry idk how to do general form

Step-by-step explanation:

Slope-intercept form is y=mx+b, m being the slope, b being the y-intercept.

The slope of this graph is 3/4, and the y-intercept is 1.

4 0

3 years ago

Read 2 more answers

The sum of 32,476 and a number P is divided by 381, the quotient is 100 and remainder is zero. Find the number P.

Misha Larkins [42]

Answer:

p = 5,624

Step-by-step explanation:

(32,476 + p) / 381 = 100

multiply both sides by 381

32,476 + p = 38,100

Subtract 32476 from both sides

p = 5,624

6 0

3 years ago

3x + 5 and x=2, im having trouble understanding how im suppose to do this

velikii [3]

3x + 5 (first plug in 2 for x )

3(2) + 5 =
6 + 5 = 11

Hope this helps . Give brainliest

6 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