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Zolol [24]
2 years ago
8

Miss Jones attended a a southern living home party she purchased a vase which cost $34.99The shipping charge was three dollars a

nd 8.25% sales tax was paid on the subtotal which included the shipping charge how much did you pay for the vase
Mathematics
1 answer:
jenyasd209 [6]2 years ago
8 0
The answer is $41.12
(Work is below)

First add the total of the vase and shipping-
Vase:                    $34.99
Shipping:              $   3.00
                         +__________
                             $37.99

Then calculate sales tax by multiplying $37.99 x 0.0825 -
Above Total:         $37.99
Sales Tax Rate:    $   0.0825
                         x__________
                              $  3.1341
You round down because ^ less than five (five or above you would round up)

Sales Tax:             $3.13

Then you add the sales tax to the above total

Vase:                    $37.99
Shipping:              $  3.13
                         +__________
                             $41.12
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Compute the cost of driving for two different job options.4 $Driving costs 0.50 dollars per mile. Driving occurs 250 days per ye
ohaa [14]

Answer:

Compute the cost of driving for two different job options.

Driving costs 0.50 dollars per mile. Driving occurs 250 days per year.

Job 1

Miles each way 5

What is the cost of driving?

Job 2

Miles each way 50

What is the cost of driving?

4 0
2 years ago
Cynthia rounds a number, x,to one decimal place. the result is 6.3.
gladu [14]

The answer is: 6.25\leq6.3

The explanation is shown below:

1. Cynthia rounds the number, which is identified as x, to one decimal place and the result is 6.3.

2. Based on this, we know that x could have been between 6.25 and 6.35. Therefore, the error interval for x is given by:

 6.25\leq6.3

Where \leq indicates that the value 6.25 is included and indicates that the value 6.35 is not included (Because if x had been exactly 6.35, Cynthia would round up to 6.4).

7 0
2 years ago
Suppose you were exploring the hypothesis that there is a relationship between parents’ and children’s party identification. Wou
alukav5142 [94]

Answer:

No

It could be purely due to chance.

Step-by-step explanation:

A population is defined as the whole group which has the same characteristics. For example a population of the college belongs to the same college . But a sample may be an element of a population.

So it is not necessary for a population to have the same characteristics as the sample.

But it is essential for the sample to have at least one same characteristics as the population.

So we would not be correct in inferring that such a relationship also exists in the population.

It is a hypothesis which can be true or false due to certain conditions or limitations as the case maybe.

For example in a population of smokers some may be in the habit of taking cocaine. But a sample of cocaine users does not mean the whole population uses it.

It could be purely due to chance if we find out that there is a relationship between parents’ and children’s party identification in the population.

7 0
2 years ago
Function ggg can be thought of as a scaled version of f(x)=x^2f(x)=x 2 f, left parenthesis, x, right parenthesis, equals, x, squ
emmainna [20.7K]

Answer:

g(x)=15/20 x^2

Step-by-step explanation:

write 15/20 as a fraction then x^2 comes after it

I took the khan test and aced it

5 0
2 years ago
Read 2 more answers
In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0
2 years ago
Read 2 more answers
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