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Mashutka [201]
3 years ago
14

James works in a sporting goods store and earns $324 a week and 5% of his sales. One week James earned $432. What were his sales

that week?
Mathematics
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

His sales that week were $2,160.

Step-by-step explanation:

First, you have to subtract $324 from the amount he earned that week, to find the 5% he got from sales:

$432-$324=$108

Now, you know that he received $108 that represent 5% of his sales and you can use a rule of three to find the amount that represents 100% which would be his sales that week:

 5%   →    108

100% →      x

x=(100*108)/5=2160

According to this, the answer is that his sales that week were $2,160.

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One car rental charges $30 per day plus &0.25 per mile driven. A second company charges $40 per day plus &0.10 per mile
Alexeev081 [22]
I think it’s $70.35 because $30+40=70 then 0.35+0.10=0.35 so if you add $70+0.35 then you will get $70.35
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3 years ago
-8(a - 4) + 3a = 2(4a + 9) + 1
ankoles [38]

Answer: a=1

Step-by-step explanation:

-8(a-4)+3a=2(4a+9)+1          given

-8a+32+3a=8a+18+1            distribute

+8a             +8a

32+3a=16a+18+1                  subtraction property of equality

    -3a  -3a

32=13a+18+1                         addition property of equality

32=13a+19                             add like terms

-19        -19

13=13a                                    addition property of equality

÷13 ÷13

1=a                                          multiplication property of equality

a=1 is the answer. Hope this helps!

4 0
3 years ago
A random sample of a specific brand of snack bar is tested for calorie count, with the following results: Assume the population
LenKa [72]

Answer:

The 95% confidence interval for the population mean (calorie count of the snacks bars) is (130.32, 161.68).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"A random sample of a specific brand of snack bar is tested for calorie count, with the following results: </em><em>149, 145,140,160,149,153,131,134,153</em><em>. Assume the population standard deviation is </em><em>σ=24</em><em> and that the population is approximately normal. Construct a 95% confidence interval for the calorie count of the snack bars."</em>

We start by calculating the mean of the sample:

M=\dfrac{1}{9}\sum_{i=1}^{9}(149+145+140+160+149+153+131+134+153)\\\\\\ M=\dfrac{1314}{9}=146

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is know and is σ=24.

The sample mean is M=146.

The sample size is N=9.

As σ is known, the standard error of the mean (σM) is calculated as: \sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{24}{\sqrt{9}}=\dfrac{24}{3}=8

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=1.96 \cdot 8=15.68

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 146-15.68=130.32\\\\UL=M+t \cdot s_M = 146+15.68=161.68

The 95% confidence interval for the population mean is (130.32, 161.68).

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