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sergey [27]
3 years ago
15

Olivia has $240 in her bank account. Each month, her bank deducts a $12.50 fee for maintaining a balance below $250. If Olivia m

akes no other deposits or withdrawals, how much money will be in her account at the end of five months.
Mathematics
2 answers:
Vikentia [17]3 years ago
8 0
Olivia will have $177.50 in the end of five months after all fees.
lukranit [14]3 years ago
5 0
Let's find the equatio first to represent this problem.
240-12.50x=y
X is the number of months, and y is the remaining amount of money in her account. We're are looking for 5 months, so let's plug in 5 for x.
240-12.50(5)=y
240-62.5=y
177.50=y
Olivia will have $177.50 at the end of five months.
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Answer:

z=\frac{0.614 -0.64}{\sqrt{\frac{0.64(1-0.64)}{145}}}=-0.652  

The p value would be given by:

p_v =P(z  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance

Step-by-step explanation:

Information given

n=145 represent the random sample taken

X=89 represent the number  of people who believed the supermarket brand was as good as the national brand

\hat p=\frac{89}{145}=0.614 estimated proportion of people who believed the supermarket brand was as good as the national brand

p_o=0.64 is the value to test

\alpha=0.05 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true proportion is less than 0.64, the system of hypothesis are.:  

Null hypothesis:p \geq 0.64  

Alternative hypothesis:p < 0.64  

The statistic is given by:

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

Replacing the info we got:

z=\frac{0.614 -0.64}{\sqrt{\frac{0.64(1-0.64)}{145}}}=-0.652  

The p value would be given by:

p_v =P(z  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance

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