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xeze [42]
3 years ago
12

When a student so than Xbox on eBay for $157, she was charged a two-part final value fee: 5.05% of the first $35 of the selling

price plus 3.05% of the remainder of the selling price over $35. Find the fee to sell the Xbox on EBay.
Mathematics
1 answer:
oee [108]3 years ago
7 0
0.0505(35) = 1.7675...rounds to 1.77

157 - 35 = 122

0.0305(122) = 3.721...rounds to 3.72

the fee is : 1.77 + 3.72 = $5.49
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sattari [20]
Five fifths are in one whole. 
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6 0
3 years ago
Read 2 more answers
A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation o
olasank [31]

Answer:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

Step-by-step explanation:

Information given

\bar X=9.41 represent the sample mean

s=0.24 represent the sample standard deviation

n=81 sample size  

\mu_o =9.33 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:  

Null hypothesis:\mu \leq 9.33  

Alternative hypothesis:\mu > 9.33  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

4 0
4 years ago
Solve by simplifying the problem. There are 18 students in Mr. Alvarez’s art class. Throughout the year, every student must pair
Alla [95]
We are given with an 18 student class that is working a project worked by pairs. Each pair works on a project. In this case, there are 9 teams of 2 individuals working on a single project. Hence for a year or school year, there are 9 projects accomplished each time.
5 0
3 years ago
Use the raw data below to create a table that can be used to create a histogram with 5
schepotkina [342]

Answer:

Five classes

Step-by-step explanation:

1. Sort the data

You get

1.1, 1.3, 1.6, 1.7, 2, 2.6, 2.7, 2.9, 3.2, 3.2, 3.5, 3.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9, 4.9, 5.3, 5.7, 6.4, 6.5, 7.1, 7.5, 7.6, 8.1, 8.2, 9.2, 9.4

2. Calculate the range

Range = Max - Min = 9.4 - 1.1 = 8.3

3. Calculate the class width

Divide the range by the number of classes

8.3/5 = 1.7

Round this up to 2.

4. Decide where to start the histogram

You could use classes: 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 -11

However, even numbers are easier to read.

It would be preferable to use classes: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10

5. Prepare a frequency distribution table

Note: Each class does not include its largest possible value. Thus, a value of 2 goes into Class 2 - 4, (not Class 0 - 2).

\begin{array}{cc}\textbf{Miles run} & \textbf{No. of students} \\0 - 2 & 4 \\2 - 4 & 9 \\4 - 6 & 8 \\6 - 8 & 5 \\8 - 10 & 4 \\\end{array}

4 0
3 years ago
I need help with exercises 3,4, and 6 please show all work and don’t answer number 5 just tell me if it is correct
joja [24]
*If you need me to do #5, DM me!

3. 
The area of a triangle can be given by (just plug and chug as always):
A_t = \frac{1}{2} bh = \frac{1}{2}(3)(4) = 6ft^2
The area of the triangle is 6ft².

4. I will divide into a triangle and a rectangle (because the actual equation for the area of a pentagon requires it to be a perfect pentagon). Let's do the triangle first (height is 3 because you subtract 12 from 15):
A_t = \frac{1}{2} bh = \frac{1}{2}(8)(3) = 12m^2
A_r = bh = 12(8) = 96m^2
Now we just add them:
96+12 = 108m^2

So, the area of that pentagon is 108m².

5. You are actually wrong on this one because the area of a triangle is:
A_t = \frac{1}{2}bh
So, just halve your answer and it will be correct.

6. We can just split it into 4 triangles of equal area and then multiply the area of 1 triangle by 4 to get the total area. Let's do just that:
A_t = \frac{1}{2}bh = \frac{1}{2}(2.5)(5) = 6.25cm^2
Multiply by 4 to get total area:
6.25*4 = 25cm^2

So, the area of the given rhombus is 25cm².
8 0
3 years ago
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