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frutty [35]
2 years ago
10

Sales tax is a function of the price of an item. The amount of sales tax is 0.08 times the price of the item. Use h to represent

the function.
a.



Please select the best answer from the choices provided

A
B
C
D

It’s C

Mathematics
1 answer:
Margaret [11]2 years ago
5 0

Option C:

h(x) = 0.08x

Solution:

Given data:

Sales tax is a function of the price of an item.

Let h be the function.

Take x be the price of an item.

Amount of sales tax = 0.08 times price of the item

                                  = 0.08 × x

Amount of sales tax = 0.08x

⇒ h(x) = 0.08x

Hence the function is h(x) = 0.08x.

Option C is the correct answer.

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Jack bought 4 dozen eggs at $10 per dozen. 6 were broken. what percent of his money goes to waste
Korvikt [17]

So, if he has 4 dozen eggs, ten dollars per dozen, he has spent $40 on 48 eggs. <em>This means the cost of one egg = 48/40. </em>This can simplify to <em>6/5 </em>which is equal to<em> $1.20. </em>This means the value of one egg is $1.20, including the broken ones! So if six were broken, we multiply 1.20 x 6 which equals <em>$7.20!</em>

Use this "formula" to help find percentages

<em>Part/Total = %( Percentage )/ 100</em>

Now that we know how much money has gone to waste, we can plug in the known values into this "formula."

<em>7.2/48 = x/100</em>

Solve accordingly; cross multiply, 720 = 48x; divide both sides of the equation by 48 to isolate the variable, 720/48 = 48x/48; now you have your final answer which is:

15 = x; going back to the "formula" this means 15% of his money has gone to waste. I hope this helped! :)

5 0
2 years ago
Students who party before an exam are twice as likely to fail as those who don't party (and presumably study). If 20% of the stu
True [87]

Answer:

The fraction of the students who failed to went partying = \frac{1}{10}

Step-by-step explanation:

Let total number of students = 100

No. of students partied are twice the no. of students who not partied.

⇒ No. of students partied = 2 × the no. of students who are not partied

No. of students partied before the exam = 20 % of total students

⇒ No. of students partied before the exam = \frac{20}{100} × 100

⇒ No. of students partied before the exam =  20

No. of students who not partied before the exam = \frac{20}{2} = 10

Thus the fraction of the students who failed to went partying = \frac{10}{100} = \frac{1}{10}

8 0
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Answer:

74

Step-by-step explanation:

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Consider the graph of Miriam's bike ride to answer the
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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:  

p_v =P(Z  

c) If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

8 0
2 years ago
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