All categories

2 years ago

A piece of fabric is for 1/6 yards long if 2 1/4 yards are cut off The Yards remain I'll see you

Mathematics

1 answer:

ehidna [41]2 years ago

7 0

Answer:

2 1/12 yds remain

Step-by-step explanation:

1. You first change the denominators, so then the fractions will be alike. 2 1/4-1/6

2. Find the LCD or Least Common Denominator. 12 can go into and it is alike, so 2 3/12-2/12

3. Now we can find the answer!

So the answer is 2 1/12

Please mark me brainliest!

You might be interested in

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

2 years ago

Which graph solves the following system? x+2y=4 5x−2y=8

12345 [234]

Answer:

elimination method

x+2y=4 1

5x-2y=8 2

1+2

6x=12

x=2

plug into x+2y=4

2+2y=4

2y=4-2

2y=2

y=1

(2,1)

so graph 1

6 0

3 years ago

Emil is itemizing deductions on his federal income tax return and had $1900 in non-reimbursed work expenses last year. If his AG

a_sh-v [17]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices that can be found from other source:

A.$340
B.$1560 
C.$38 
D.$1862

Below is the solution:

78000 x .0.02=1560
1900 - 1560=340

5 0

3 years ago

Read 2 more answers

25000 rounded to the nearest ten thousand

kozerog [31]

Answer:

30000

Step-by-step explanation:

You round up because it is 5 or < making the 5 a 0 and the 2 a 3. Bringing in the extra 0's makes it 30,000

3 0

2 years ago

Read 2 more answers