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Vera_Pavlovna [14]
2 years ago
15

The following table shows last year's revenue for each Herman's Hoagies location.

Mathematics
2 answers:
SOVA2 [1]2 years ago
8 0

Answer:

The revenue for Granton location is 175 thousand dollars

Step-by-step explanation:

Given

Cedarton 121

Rimber 189

Linton 147

Mean = 158

Required

Revenue for Granton location.

To calculate the revenue for Granton location, we make use of mean formula.

Mean is calculated by Summation of Observation divided by number of observations.

Since Granton location is unknown; Let it be represented by letter G.

So, the summation of observation becomes 121 + 189 + 147 + G

Summation = 457 + G

The number of observations = 4

Recall that Mean = Summation ÷ Number

By substituting 158 for mean, 457 + G for summation and 4 for number, we have

158 = (457 + G) ÷ 4

158 = ¼(457 + G)

Multiply both sides by 4

4 * 158 = = 4 * ¼(457 + G)

632 = 457 + G

Make G the subject of formula

G = 632 - 457

G = 175

Hence, the revenue for Granton location is 175 thousand dollars

Leona [35]2 years ago
5 0

Answer:

175

Step-by-step explanation:

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ch4aika [34]

The area of the bare spot the gardener needs to replant is 201 square feet

Step-by-step explanation:

We have to find the area of the circular bare spot

Given

Radius of bare spot = r = 8 inches

The formula for area of a circle is given by:

Area=A = \pi r^2

putting the value

A = 3.14 * (8)^2\\= 3.14 * 64\\=200.96

Rounding off to nearest foot

201 ft^2

Hence,

The area of the bare spot the gardener needs to replant is 201 square feet

Keywords: Area, radius

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2 years ago
Alicia sews costumesfor a school play.She takes an average of 86 minutes to sew each costume.How long would she take to sew 16 o
Goryan [66]

Answer:

1376 minutes

Step-by-step explanation:

Alicia sews costumes for schools

She takes 86 minutes to sew one costume

Therefore the time taken to sew 16 costumes can be calculated as follows

86 mins= 1 costume

x= 16 costume

Cross multiply

x= 86×16

= 1376 minutes

3 0
2 years ago
Roger claims that the two statistics most likely to change greatly when an outlier is added to a small data set are the mean and
Leni [432]

Answer:

No, the Roger’s claim is not correct.

Step-by-step explanation:

We are given that Roger claims that the two statistics most likely to change greatly when an outlier is added to a small data set are the mean and the median.

This statement by Roger is incorrect because the median is unaffected by the outlier value and only the mean value gets affected by the outlier value.

As the median represents the middlemost value of our dataset, so any value which is an outlier will be either at the start or at the end will not the median value. So, the median will not likely change when an outlier is added to a small data set.

Now, the mean is the average of all the data set values, that is the sum of all the observations divided by the number of observations. The mean will get affected by the outlier value because it take into account each and every value of the data set.

Hence, the mean will likely to change greatly when an outlier is added to a small data set.

7 0
2 years ago
Find the dimensions of the rectangle with area 289 square inches that has minimum perimeter, and then find the minimum perimeter
nalin [4]
 <span>To minimize the perimeter you should always have a square. 
sqrt(289) = 17 
The dimensions should be 17 X 17 

To see , try starting at length 1, and gradually increase the length. 
The height decreases at a faster rate than the length increases, up until you reach a square. 

Or if you want to use algebra, Say the width is 17-x 
Then the length is 289/(17-x) 

Now, this is bigger than 17+x, as shown here: 
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which is true. 
so the perimeter would be bigger than 2 * (17- x + 17 + x) = 2 * (2 * 17) = 4 * 17 

Again, the dimensions should be a square. 17 X 17.</span>
4 0
2 years ago
The "hang time" of a football is the amount of time the football stays in the air after being kicked. Zara kicks a football, and
FrozenT [24]

Answer:

Step-by-step explanation:

The graph is a nonlinear graph.

The hang time is 3 seconds.

The h- coordinate of the top of the bump is about 11 m, so the maximum height is 11 m after rounding to the nearest meter.

For t between 0 and 1.5, the height is increasing.

I hope this helps.  I found  it on Khan Academy.

3 0
2 years ago
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