Mr. Floyd, the school librarian, is ordering new keyboards and mice for all of the school's computer labs. Each keyboard costs $13.50, and each mouse costs $6.50. There are 3 computer labs in the school, and each computer lab has s computer stations.Pick all the expressions that represent how much Mr. Floyd will spend on new keyboards and mice.

Answers

Answer 1
Answer:

Answer:

13.50 + 6.50 x ( 3 x s) =

Step-by-step explanation:

I don't know what the options are, however that is how I would write it.


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The table below shows the linear function f.x: -6,-5,-4,-3,-2f(x): 22,19,16,13,10Determine the average rate of change of the given function over an interval [-5, -2].A. -1/3b.1/3c.-3D.3

The distance between -6 and 15 on a number line is 9?

Answers

The distance between -6 and 15 on a number line is 21.

What is number line?

A straight line with numbers placed at equal intervals or segments along its length. Number line is a straight line with a "zero" point in the middle, with positive and negative numbers listed on either side of zero and going on indefinitely.

Given:

The distance between -6 and 15 on a number line is 9 is true or false.

We have to find the the distance between -6 and 15 on a number line.

According to given question we have

15 - (- 6)

= 15 + 6

= 21

The distance between -6 and 15 on a number line is 9 is false.

Therefore, the  distance between -6 and 15 on a number line is 21.

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Answer:

21 units. your answer is false.

Step-by-step explanation:

-6 to 0 = 6

0 to 15 = 15

you are counting Units not physical numbers. don't forget that!

correct distance is 21

A farmer owns 30 acres of land on which he wishes to grow corn and barely. The cost per acre for seedcorn is $30, and the cost per acre for barely seed is $20. The farmer can invest a maximum of $600 in seed for the two crops. During the cultivation process, the farmer needs to spray fertilizers and insecticides specific to each crop. This costs $10 per acre for corn and $15 per acre for barely. He can invest only $400 in this process.A) Write the two inequalities that are deciding factors for the number of acres of each crop the farmer will plant, based on the amount of money the farmer will spend on planting and cultivating the two crops.

B) replace the inequality signs in the two any qualities with equal signs. For a graft representing the two equations that influence the farmers choice of how much of each crop to grow.

C) should the lines be dilated or solid? Give reasons for both lines. What area should be shaded?


Help please

Answers

ok hola bro graicas por los punto                            qui    :

Determine the value of x for which a || b if m∠1 = 60 - 2x and m∠5 = 70 - 4x.50
5
4
10

Answers

x is equal  to 10 cuz u would sub 70-60


Sales of Volkswagen's popular Beetle have grown steadily at auto dealerships in Nevada during the past 5 years. The sales manager has predicted in 2004 and 2005 sales would be 410 VW's. Using exponential smoothing with a alpha = 0.30 develop a forecast for 2006 through 2010. What is the forecast value for 2010?Please answer to one decimal place. (Example: 466.1)Formula: Ft = Ft-1 + alpha(At-1 - Ft-1)Actual sales data:2005 = 4502006 = 4952007 = 5182008 = 5632009 = 584

Answers

Answer:

Step-by-step explanation:

Given the data:

Year___Actual sale (At) ___forecast(Ft)

2005__450_____________410

2006__495____________ 422

2007__518____________ 443.9

2008_ 563____________ 466.1

2009_584____________ 495.2

2010__

Using the formula :

Ft = Ft-1 + alpha(At-1 - Ft-1)

Ft-1 = previous year forecast

At-1 = previous year actual

Alpha = 0.3

Forecast:

2006:

410 + 0.3(450 - 410) = 422.0

2007:

422 + 0.3(495 - 422) = 443.9

2008:

443.9 + 0.3(518 - 443.9) = 466.1

2009:

466.1 + 0.3(563 - 466.1) = 495.2

2010:

495.2 + 0.3(584 - 495.2) = 521.8

Forecasted value for 2010 = 521.8

Final answer:

By utilizing the method of exponential smoothing with alpha=0.3 and applying the forecasting formula iteratively, the forecasted sales of Volkswagen Beetles for 2010 in Nevada is 525.7 units.

Explanation:

The exponential smoothing method is commonly used in business and economics for forecasting future data in situations where historical data is available. The forecast value for 2010 using exponential smoothing with alpha = 0.30 can be obtained by implementing the provided formula in an iterative manner. This formula takes into account the actual data point from the previous year (At-1) and the forecasted data point for that year (Ft-1).

Let us use the forecast value for 2004 and 2005, which was predicted as 410 for each year, as our initial forecast value (F1).

By applying the formula Ft = Ft-1 + alpha * (At-1 - Ft-1) iteratively for each year from 2006 to 2009, we get:

  • Forecast for 2006 (F_2006) would be 410 + 0.30*(450-410) = 422.
  • Then, the forecast for 2007 (F_2007) is 422 + 0.30*(495-422) = 443.9.
  • The forecast for 2008 (F_2008) is 443.9 + 0.30*(518-443.9) = 466.1.
  • For 2009 (F_2009), the forecast is 466.1 + 0.30*(563-466.1) = 496.37.
  •  

 

Using these forecasted rates, we then forecast for the year 2010. The forecast for 2010 (F_2010) is calculated by 496.37 + 0.30*(584-496.37) giving 525.67 (rounded to one decimal place). Hence the forecast for the year 2010 using exponential smoothing with the alpha as 0.30 is estimated to be 525.7 VolksWagen Beetles.

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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 139 millimeters, and a variance of 49. If a random sample of 34 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 1.8 millimeters

Answers

The probability that the sample mean would differ from the population mean by more than 1.8 millimeters is approximately 0.0668.

What is the standard deviation?

A standard deviation (σ) is a measure of the distribution of the data in reference to the mean.

The standard deviation of the population is $√(49) = 7$ millimeters. The standard error of the sample mean is then

$(7)/(√(34)) = (7)/(5.874) \approx 1.2$millimeters.

The probability that the sample mean would differ from the population mean by more than 1.8 millimeters is the probability that it falls outside of the interval $(139 - 1.8, 139 + 1.8)$. We can use the standard normal distribution to approximate this probability.

First, we need to convert the difference between the sample mean and the population means to standard units.

The difference of 1.8 millimeters is :

$(1.8)/(1.2) = 1.5$ standard units.

Then, we can use the standard normal distribution to find the probability that the sample mean falls outside of this interval.

This probability is equal to $1 - 2\Phi(-1.5) \approx 0.0668$,

where $\Phi$ is the standard normal cumulative distribution function.

Therefore, the required probability is approximately 0.0668.

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23.5 in expanded form164.38 in expanded form
209.106 in expanded form

can you help me

Answers

Answer:

 20  

+ 3  

+   0.5

 100  

+ 60  

+ 4  

+   0.3

+   0.08

200  

+ 0  

+ 9  

+   0.1

+   0.00

+   0.006

Step-by-step explanation:

Hope this helps :D

Answer:

23.5 in expanded form is: 20 + 3 + 0.5

164.38 in expanded form is: 100 + 60 + 4 + 0.3 + 0.08

209.106 in expanded form is: 200 + 0 + 9 + 0.1 + 0.00 + 0.006

Step-by-step explanation:

Show the value of each digit when writing in expanded form. Use plus signs (+) between the values. Knowing the definition of expanded notation can help. Expanded notation means:

"Writing a number to show the value of each digit. It is shown as a sum of each digit multiplied by its matching place value (ones, tens, hundreds, etc.)."

Other Questions
In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 241.4 sec. and a standard deviation of 57.59 sec. Use a 0.05 significance level and the accompanying minitab display to test the claim that the sample is from a population of songs with a mean great thatn 210 sec. What do these results suggest about the advice given in the manual.The mini tab displays the following:One-sample TTest of mu=210 vs.>210N Mean St. Dev SE Mean 95% lower bound T p40 241.40 57.59 9.11 226.06 3.45 0.001A H0 u>210 sec. H1 u < 210secB H0 u=210 sec. H1 u < 210secC H0 u<210 sec. H1 u> 210secD H0 u=210 sec. H1 u> 210secIdentify the test statistic:T =Identify the P-ValueP-value=Stat the final conclusion that addresses the original claim. Choose from below:A. Reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.B. Fail to reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length great thatn 210 sec.C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.D. Fail to reject H0. There is sufficient evidence to support the claim tha tthe sample is from a population of songs with a mean lenght greater than 210 sec.What do the results suggest about the advice given in the manual?A. The results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.B The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound adviceC. The results suggest that 241.4 seconds is the best song lenght.D. The results are inconclusive because the average length of a hit song is constantly changing.