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Norma-Jean [14]
1 year ago
12

Ann's car can travel 228 miles on 6 gallons of gas. a. Write an equation to represent the distance, y, in miles Ann's car can tr

avel on x gallons of gas. b. Ann's car used 7 gallons of gas during a trip. How far did Ann drive?
Please show your work sow I know how to do it don't just tell me the answer explain Thank you
Mathematics
1 answer:
Murrr4er [49]1 year ago
7 0

Answer:

288 gallons ÷ 6 miles = 48 miles per gallon  

48 miles per gallon • 7 miles = 366 miles  

Step-by-step explanation:

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A product is composed of four parts. In order for the product to function properly in a given situation, each of the parts must
svet-max [94.6K]

Answer:

P(working product) = .99*.99*.96*.96 = .0.903

Step-by-step explanation:

For the product to work, all four probabilities must come to pass, so that

P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

where

P(Part-1) = 0.96

P(Part-2) = 0.96

P(Part-3) = 0.99

P(Part-4) = 0.99

As all parts are independent, so the formula is P(A∩B) = P(A)*P(B)

P (Working Product) =  P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

P (Working Product) = 0.96*0.96*0.96*0.99*0.99

P(Working Product) = 0.903

3 0
1 year ago
Harry's soccer team plays 2 nonconference games for every 3 games that they play against conference opponents. If y represents t
Sergeeva-Olga [200]
The correct answer is y=2/3x
5 0
1 year ago
Read 2 more answers
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
1 year ago
Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.
Ghella [55]

Answer:  The required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

Step-by-step explanation:  Given that Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Also, Heloise has written \frac{1}{4} as many articles as Mustafa has and Gia has written \frac{3}{2} as many articles as Mustafa has.

We are to write an inequality to determine the number of articles, m, Mustafa could have written for the school newspaper.  Also, to solve the inequality.

Since m denotes the number of articles that Mustafa could have written. Then, according to the given information, we have

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.

And the solution of the above inequality is as follows :

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22\\\\\\\Rightarrow \dfrac{4x+x+6x}{4}>22\\\\\\\Rightarrow 11x>88\\\\\Rightarrow x>\dfrac{88}{11}\\\\\Rightarrow x>8.

Thus, the required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

4 0
2 years ago
Read 2 more answers
One week, Claire earned $272.00 at her job when she worked for 17 hours. If she is paid the same hourly wage, how many hours wou
DochEvi [55]

Answer: 448 hours

Step-by-step explanation:

272 divided by 17 = hourly wage

hourly wage= 16$

448 divided by 16 = 28 weeks

28 x 16= 448 hours

4 0
1 year ago
Read 2 more answers
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