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2 years ago

Please help me out! :)

Mathematics

1 answer:

skad [1K]2 years ago

8 0

3x-11=x+23
-x -x
2x-11=23
+11 +11
2x=34
x=17

3(17)-11 17+23
=40 =40

explanation ;
you would equal them because they are congruent . & you substitute x for 17 in each equation to find out the degrees of both of the angles :)

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Prof. Lazy decides to assign final grades in CSE 312 by ignoring all the work the students have done and instead using the follo

Leya [2.2K]

Answer:

According to Section 1031 of Internal Revenue Code, an exchange of like-kind property is referred to as like-kind exchange, if the exchange meets the following criteria:

The property should be exchanged only for the 'like-kind' (same class) property.
The 'like-kind' property that is in exchange should be either used in operations of business or held for investment purposes.
The exchange should be done under specific timing restrictions or requirements (for indirect exchanges through third parties).

Step-by-step explanation:

A) Explanation: Since both the properties in exchange are real properties which are used in business, this is a like-kind exchange

Description Amount ($)

Fair market value of bowling alley 120,000

Add: Mortgage value L is relieved off 40,000

Amount realized 160,000

Less: Adjusted basis (175:000)

Realized loss by L (15,000)

C) Explanation: Since the loss realized is due to exchange, the loss is not recognized.

D) Explanation: Since the loss is not recognized, the character of gain or loss could not be

E) Explanation: The total of realized loss, $15,000 (From Requirement (b)) is deferred.

F) Determine the basis of property acquired in exchange, bowling alley.

Description Amount ($)

Fair market value of bowling alley 120,000

Add: Deferred loss 15,000

Basis of bowling alley 135,000

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2 years ago

A truck takes 6 hours yo travel 623km. Calculate for the average speed

Allushta [10]

The formula of average speed is distance/time. So 623/6= 103.833333 repeatings let just put it around 103.84km/s^2

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1 year ago

What is the square root of 138

melamori03 [73]

The answer is 11.7473

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2 years ago

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Can you explain what went wrong and the actual answers these 3

vlada-n [284]

For the first one, they got rid of the 6x when they shouldn’t have, and the correct answer is x = 21.5

For the second one, 47 would be subtracted by 180, not 90, and the correct answer is 133

For the third one, 5x - 10 doesn’t equal 115, 5x - 10 equals 180 - 115, and the correct answer is x = 17

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2 years ago

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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.

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2 years ago