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Veseljchak [2.6K]
2 years ago
11

Jim bought 2 packs of batteries from a store. The price of each pack was the same. After he bought the batteries, his account ba

lance showed a change of −$23.94. What would have been the change to Jim's account balance had he only bought 1 pack of batteries from the store?
Mathematics
1 answer:
a_sh-v [17]2 years ago
6 0

Answer:

Divide $23.94 by 2. 23.97/2=11.97. The change would have been -$11.97 :) hope this helps

Step-by-step explanation:

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HELP PLEASE BRAIN FOR CORRRECT
N76 [4]

Answer:

1/6

Step-by-step explanation:

you divide y by x

y ÷ x

each column when you you divide y by x the answer is 1/6

8 0
2 years ago
Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters
s344n2d4d5 [400]

The various answers to the question are:

  • To answer 90% of calls instantly, the organization needs four extension lines.
  • The average number of extension lines that will be busy is Four
  • For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$

Generally, the equation for is  mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

Read more about signal

brainly.com/question/14699772

#SPJ1

3 0
1 year ago
#23-6: The pool concession stand made $5,800 in June and $6,300 in July. What is the percent of increase in sales? Round the ans
katrin [286]

Answer:

8.6%

Step-by-step explanation:

To find the percent change, you will need to compute the positive difference and then divide the difference by the original (the older amount).

So the positive difference will be obtain by doing larger minus smaller:

 6300

- 5800

-----------

   500

The older amount was 5800.

So 500/5800 is the answer as a un-reduced fraction.

I'm going to reduce it by dividing top and bottom by 100:

500/5800  = 5/58

5/58 is the answer as a reduced fraction.

5 divided by 58 gives=0.086206897  in the calculator .

Approximately 0.0862 is the answer as a decimal.

To convert this to a percentage, multiply it by a 100:

8.62%

Rounded to the nearest tenths is 8.6%

-------------

So 5800+5800(.0862) should be pretty close to 6300 (not exactly though since we rounded).

5800+5800(.0862)=6299.96 using the calculator.

8 0
3 years ago
Given g(x) = x^2-x , find g(2/3).<br><br> show work please
castortr0y [4]
I hope this helps you



x=2/3



g (2/3)=(2/3)^2-2/3



g (2/3)= 4/9-2.3/3.3



g (2/3)=4/9-6/9



g (2/3)= -2/9
7 0
2 years ago
How do I find QS here?
juin [17]
Answer: 12.
To solve this problem, let's first solve for x. Thi is easiest done by figuring what QR is in terms of x using two equations, both from different lines. 
In the first line: QR = 15 - 4x. 
In the third line:QR = 13x - 1 - x = 12x - 1. 
Now, we have to set these equations equal to each other. 15 - 4x = 12x - 115 = 16x - 116x = 16x = 1
Next, we take line three, 13x - 1, and substitute x as 1. The answer is 12.
4 0
3 years ago
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