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Fynjy0 [20]
3 years ago
14

Shaquil needs to borrow $400. The loan company charges him $80 in fees for the loan. Shaquil calculates that the annual percenta

ge rate (APR) on his loan is 250%. Approximately what is the term of Shaquil’s loan? a. 7 days b. 29 days c. 46 days d. 73 days
Mathematics
2 answers:
lesya [120]3 years ago
6 0

Answer:

(B)  

Step-by-step explanation:

PRINCIPAL AMOUNT(p) = $400

RATE (r)= 250% = 2.5

INTEREST = $80

t = time (in days) = t/365

By using the formula,

r = \frac{i}{pt}

2.5 = \frac{80*365}{400*t}

t = \frac{364*80}{400*2.5}

t = 29.2

t = 29 days (approx)

Hence option (B) is correct.


mina [271]3 years ago
6 0
The answer is B.) 29 Days
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We have to choose the function from options, the graph of which has an axis of symmetry at x = 3.

The function in the fourth option will have an axis of symmetry at x = 3.

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3 years ago
Jim gets a weekly allowance he keeps $8 for spending money saves $5 and gives $2 to charity how much is Jim's weekly allowance
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Step-by-step explanation:

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A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

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

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha 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 true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

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

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

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

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
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