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oksano4ka [1.4K]
1 year ago
15

Maxine and her daughter are going grocery showing after she gets off work. Maxine sees that apples are $0.25, while bananas are

$0.79. She has $2.50 to spend on apples, x, and bananas. y. Write the equation, in standard form that represents her situation.
HELP ME PLEASE
Mathematics
1 answer:
Butoxors [25]1 year ago
3 0

Her situation can be modeled by the linear equation (in standard form):

x*$0.25 + y*$0.79 = $2.50

<h3>How to write the equation that represents her situation?</h3>

The variables that we will use here are:

x =  number of apples that she can buy.

y = number of bananas that she can buy.

We know that she has $2.50 to spend, that each apple costs $0.25 and each banana costs $0.79, then the cost of the x apples and y bananas is:

x*$0.25 + y*$0.79

and that must be equal to the amount she has to spend, then we can write the linear equation:

x*$0.25 + y*$0.79 = $2.50

That is the linear equation in standard form that represents her situation.

Learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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Answer:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Step-by-step explanation:

Information given

\bar X=8.19 cm^3 represent the sample mean

s=0.8 cm^3 represent the sample deviation

n=17 sample size  

\mu_o =9.02 represent the value to verify

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 9.02  

Alternative hypothesis:\mu < 9.02  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

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Step-by-step explanation:

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