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a_sh-v [17]
3 years ago
6

Find two numbers whose sum is -2 and whose product is -63

Mathematics
1 answer:
djverab [1.8K]3 years ago
3 0
Answer: -9 and 7

explanation:
-9+7=-2
-9x7=-63
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5(-2 -2p) simplify expression
I am Lyosha [343]

Answer:

-10 -10p

Step-by-step explanation:

5(-2 -2p)

Distribute the 5

5*-2 +5(-2p)

-10 -10p

6 0
3 years ago
Read 2 more answers
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
Use the expression below:
skelet666 [1.2K]

Answer:

Part A: 8b -12c -16

Part B: 4(2b - 3c -4)

Part C: 36

Step-by-step explanation:

Part A:

5b + 3c - 20 + 3b - 15c + 4

= 8b -12c -16

Part B:

Factorising 8b -12c -16

= 4(2b - 3c -4)

Part C:

= 4[2(2)-3(-3)-4)]

= 4[4+9-4]

= 4[9]

= 36

7 0
3 years ago
How many cubes are needed to build this structure?
Elena-2011 [213]
Hello!

I believe that 8 cubes are needed to build the given figure

I hope this helps!
6 0
3 years ago
Read 2 more answers
4.5/y = 12.5/4 PLEASE HELP!!! SOS
garik1379 [7]

Answer:

y = 1.44

Step-by-step explanation:

What are you aiming to do here?  Please share all instructions with each problem.

If you want to solve 4.5/y = 12.5/4 for y:  Multiply both sides by 4y:

18 = 12.5y.  Then y = 1.44

7 0
3 years ago
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