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3 years ago

4(x−3)=1/2(x/2+5) how many solutions

Mathematics

1 answer:

charle [14.2K]3 years ago

7 0

I think it might be six, sorry if its wrong, but I hope its right!!

Hope this helps!!!!

love: Deku <3

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Maria bought a bag of marbles. There were twice as many purple ones as red ones. There were 93 marbles in all. How many of each

Leto [7]

Answer:62 purple

31 red

Step-by-step explanation:

93/3= 31 to find how many of each color.

31+31=62 purple

31 red

3 0

3 years ago

Read 2 more answers

Please answer question for me ASAP!!!!!

antiseptic1488 [7]

Answer:

Step-by-step explanation:

Any number and variable combined is a term.

So 8 (#1 - Constant), + 19y (#2 -Coefficient + Variable), - 4z (#3 -Coefficient + Variable)

3 terms - B

5 0

3 years ago

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

3 years ago

A jar contains 133. A bigger jar contains 1 2/7 times as many pennies.what is the value of the pennies in the bigger jar

adoni [48]

A jar contains 133 pennies....a bigger jar contains 1 2/7 times as much

so the bigger jar contains :
1 2/7 * 133
9/7 * 133 =
1197/7 =
171 pennies <===

4 0

3 years ago

What is the domain of f(x)=9-x²<br> A(.fx)≥9<br> B.All real numbers<br> C.-3≤x≤3<br> D. x≤9

Ber [7]

Answer:

B. all real numbers

Step-by-step explanation:

hope that helps

3 0

4 years ago

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