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alexira [117]
4 years ago
9

Jackson purchased some power tools totaling $1,543 using a six month deferred payment plan with an interest rate of 23.76%. He d

id not make any payments during the deferment period. What will the total cost of the power tools set be if he must pay off the power tools within two years after the deferment period?
$1,543.00
$1,735.63
$2,197.44
$2,746.80
Mathematics
1 answer:
skad [1K]4 years ago
4 0
Suppose he makes the payment with two equal annual instalments, the present value of the amount he is owing is $1,543 , the interest rate is 23.76% = 0.2376.

The amount of payment he makes in two of the periodic payments is given by:

PV = P\left( \frac{1-(1+r)^{-n}}{r} \right) \\  \\ \Rightarrow1,543=P\left( \frac{1-(1+0.2376)^{-2}}{0.2376} \right) \\  \\ =P\left( \frac{1-(1.2376)^{-2}}{0.2376} \right)=P\left( \frac{1-0.6529}{0.2376} \right) \\  \\ =P\left( \frac{0.3471}{0.2376} \right)=1.4609P \\  \\ \Rightarrow P= \frac{1,543}{1.4609} =1,056.20

Therefore, in 2 years, the amount he has paid for the tools is 2(1,056.20) = 2,112.40
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3 years ago
Just curious, I want to see who can get this. tooooootttalllllly know it already
Aneli [31]

Answer:

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3 years ago
The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed
algol [13]

Answer:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

p_v = P(\chi^2_{5} >5.860)=0.32

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number:      1,    2 ,   3 ,  4 , 5    ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The observed values are given:

O_{1}=27   O_{2}=31

O_{3}=42  O_{4}=40

O_{5}=28  O_{6}=32

The expected values are given by:

E_{1} =\frac{1}{6}*200=33.33   E_{2} =\frac{1}{6}*200=33.33

E_{3} =\frac{1}{6}*200=33.33   E_{4} =\frac{1}{6}*200=33.33

E_{5} =\frac{1}{6}*200=33.33   E_{6} =\frac{1}{6}*200=33.33

And now we can calculate the statistic:

\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860

Now we can calculate the degrees of freedom for the statistic given by:

df=Categories-1=6-1=5

And we can calculate the p value given by:

p_v = P(\chi^2_{5} >5.860)=0.32

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

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3 years ago
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Serggg [28]

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5 0
2 years ago
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Rudik [331]
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5 0
3 years ago
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