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4 years ago

9

She invited 150 friends. But she only had 5 activities for her friends and her party only lasted 4 hours. How much time will eac

h friend get on each activity

1 answer:

DanielleElmas [232]4 years ago

4 0

Answer:

19.20 seconds.

Step-by-step explanation:

Given:

She invited 150 friends.

But she only had 5 activities for her friends and her party only lasted 4 hours.

Question asked:

How much time will each friend get on each activity ?

Solution:

First of all convert 4 hours into seconds:-

1 hour = 3600 seconds

4 hours = 3600 $\times$ 4 = 14400 seconds

<u>By unitary method:</u>

150 friends get time in 5 activities = 14400

1 friend get time in 5 activities = 14400 $\div$ 150 = 96 seconds

That means <u>each friend get 96 seconds in all 5 activities.</u>

Each friend get time on each activity = ?

Each friend get time on 5 activities = 96

Each friend get time on 1 activity = 96 $\div$ 5 = 19.20 seconds

Thus, each friend will get 19.20 seconds on each activity.

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Improper fraction for 3 1/14

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Answer: 43/14

Step-by-step explanation: You take you 3 and times it by 14 and + 1 because the equation is saying 14/14 3 times plus the 1/14,

So 3 times 14=42

42+1=43 and then you take the 43 and put it over the 14 so now you have your answer: 43/14

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3 years ago

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The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distribu

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Answer:

a) Standard error = 0.5

b) 99% Confidence interval: (19.2125,21.7875)

Step-by-step explanation:

We are given the following in the question:

Population mean = 20.50 mpg

Sample standard deviation = 3.00 mpg

Sample size , n = 36

Standard Error =

$=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5$

99% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.575$

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3 years ago

For each statement, write what would be assumed and what would be proven in a proof by contrapositive of the statement. Then wri

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Answer:

a)

Given Statement - If x and y are a pair of consecutive integers, then x and y have opposite parity.

Proof by Contrapositive:

Assumed statement: Suppose that integers x and y do not have opposite parity.

Proven Statement: x and y are not a pair of consecutive integers.

Proof -

x = 2u₁ , y = 2u₂

Then

(x, x+1) = (2u₁ , 2u₁ + 1) = (Even, odd)

If y = 2u₁ + 1

Not possible

⇒x and y are not a pair of consecutive integers.

Hence proved.

Proof by Contradiction:

Assumed statement: Suppose x and y are not a pair of consecutive integers.

Proven Statement: Suppose x and y do not have opposite parity.

Proof -

If x and y are not a pair of consecutive integers.

⇒ either x and y are odd or even

If x and y are odd

⇒x and y have same parity

Contradiction

If x and y are even

⇒x and y have same parity

Contradiction

(b)

Proof by Contrapositive:

Assumed statement: Let n be an integer such that n is not odd (i.e. n is an even integer)

Proven Statement: n² is not odd (i.e n² is even)

Proof -

Let n is even

⇒n = 2m

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Hence proved.

Proof by Contradiction:

Assumed statement: Let n be an integer such that n² be odd.

Proven Statement: suppose that n is not odd (i.e n is even)

Proof -

Let n² is odd

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3 years ago

If X and Y are independent continuous positive random

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$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

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$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

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$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

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and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

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