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omar tariq has a credit card that uses the unpaid balance method to compute the finance charge. His monthly periodic rate is 2.4

Korvikt [17]

Answer:

(a) unpaid balance = $667.95

(b) Previous balance = $761.67

(c) his new balance = $683.98

Step-by-step explanation:

In the question, we are given the following:

Monthly periodic rate = 2.4%, or 0.024

Curren charges = $256.28

Total payment made = $350

Finance charge = $18.28

Therefore, we now proceed as follows:

(a) unpaid balance

This can be calculated as follows:

Unpaid balance = Previous balance + Current charges - Total payment made ................. (1)

Note that from part (b), Previous balance is equal to $761.67.

Substituting this and other values from the question into equation (1), we have:

Unpaid balance = $761.67 + $256.28 - $350 = $667.95

(b) Previous balance

This can be calculated as follows:

Previous balance = Finance charges / Monthly periodic rate = $18.28 / 2.4% = $18.28 / 0.024 = $761.67

(c) his new balance

This can be calculated as follows:

New balance = Unpaid balance * (1 + Monthly periodic rate) = $667.95 * (1 + 2.4%) = $667.95 * (1 + 0.024) = $667.95 * 1.024 = $683.98

7 0

3 years ago

A cylinder has a height of 1.2 cm and the following base. 5 mm

koban [17]

Step-by-step explanation:

this picture will help

you

3 0

3 years ago

Advertisers want to create "awareness," which is whether a consumer even knows about a business. If your business starts with 3%

zepelin [54]

Necesito los puntos suerte de mi the same and I have no idea why I was in a relationship lol

4 0

2 years ago

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

PLZ HELP ME BY THE END OF THE DAY

nlexa [21]

$3x^{2} +1=28$