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3 years ago

I love you who held me thank you ❤️

Mathematics

1 answer:

suter [353]3 years ago

7 0

Answer:

1) Since AP/AB=PQ/BC, 6x/(3x+5)=x/1. Cross multiply the denominator and you will get 6x=(3x+5)*x. Simplifying it more and you are left with 3x^2-x=0

2) 3x^2-x=0. x(3x-1)=0. x=0 or x=1/3 but x=0 can't be a solution since it would make the figure absurd. x=1/3 cm

3) According to the figure PB=AB-AP=3x+5-6x=-3x+5.

PB=-3*(1/3)+5=4 cm

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A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10

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Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$

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A cylindrical container has a radius of 0.3 meter and a height of 0.75 meter the container is filled with kerosene. the density

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The volume of the cylinder is:

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Answer:

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Step-by-step explanation:

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A = allowance
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