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WITCHER [35]
3 years ago
10

Amanda wants to pay the same amount each month on her car loan. The number of the payment various inversely with the monthly pay

ment amount. If each monthly payment is $175, she makes 48 payments. how much would her monthly payment be if she wanted to pay off the loan in 30 Months?
Mathematics
1 answer:
SashulF [63]3 years ago
3 0

Answer:

$280

Step-by-step explanation:

175 x 48 = 8400

8400 / 30 = 280

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liberstina [14]

2=6-3 is the direct variation

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kicyunya [14]

Answer:

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Step-by-step explanation:

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7 0
2 years ago
Six more than 1/5 a number is 12. What is the number
suter [353]

Answer:

Step-by-step explanation:

let x be the number

6+1/5x=12

1/5x=12-6

1/5x=6

x=5*6

x=30

check: 6+(1/5*30)=6+6=12

8 0
2 years ago
6 pounds of apples and 3 pounds of oranges cost 24 dollars.5 pounds of apples and 4 pounds of oranges cost 23 dollars. Set up a
vitfil [10]

Answer:cost of each pound of apple= $3

And cost of each pound of orange =$2

Step-by-step explanation:

Step 1

Let cost of apples = x

And cost of Oranges =y

Let 6 pounds of apples and 3 pounds of oranges cost 24 dollars be represented as

6 x + 3y= 24----- equation 1

Also, Let 5 pounds of apples and 4 pounds of oranges cost 23 dollars be represented as

5x+ 4y= 23----- equation 2

Step 2

6 x + 3y= 24----- equation 1

5x+ 4y= 23----- equation 2

Using substitution method to solve the equation

6 x + 3y= 24

24-6x=3y

y= 24-6x/3 = 8-2x

Y= 8-2x

Substituting the value of y= 8-2x  into equation 2

5x+ 4( 8-2x)= 23

5x+ 32 -8x= 23

32-23= 8x-5x

9=3x

x=9/3

x=3

Putting the value of x= 3 in equation 1 and solving to find y

6 x + 3y= 24

6(3) +3y= 24

18+3y=24

3y= 24-18

3y=6

y=6/3= 2

Therefore the cost of each pound of apple= $3

And cost of each pound of orange =$2

8 0
3 years ago
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and 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% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, 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.

3) Calculate the statistic  

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

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

4) 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 right tailed test the p value would be:  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and 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% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0
3 years ago
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