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crimeas [40]
1 year ago
9

Myrtle has a credit card that uses the average daily balance method. For the first 21 days of one of her billing cycles, her bal

ance was $2030, and for the last 9 days of the billing cycle, her balance was $1450. If her credit card's APR is 23%, which of these expressions could be used to calculate the amount Myrtle was charged in interest for the billing cycle?
(a)- (0.23/365 x30)÷(21x $2030+ 9x$1450/30)
(b)- (0.23/365 x30)÷(9x $2030+ 21x$1450/30)
(c)- (0.23/365 x31)÷(9x $2030+ 21x$1450/31)
(d)- (0.23/365 x31)÷(21x $2030+ 9x$1450/31)

Mathematics
2 answers:
Musya8 [376]1 year ago
8 0

Answer:

Step-by-step explanation:

levacccp [35]1 year ago
5 0

Answer:

A

I had this question yesterday

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Last Month the electric meter reading was 7985 kWh. This month the reading was 8328 kWh. If electricity costs P3.40 per kWh,how
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D2-9k+3 for d=10 and k=9
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22

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2 years ago
what is the geometric mean between 2 and 18? also if you answer this question is there anyway you could include the way you solv
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Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
xxTIMURxx [149]

Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0
2 years ago
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