Ask question

Ask question

All categories

3 years ago

11

If you are trying to compare 2 loans, one that pays 3.15% compounded mounthly for 4 years and another that pays 2.95% compounded

weekly for 4.5 years, what is the APY of the cheaper loan?

1 answer:

Afina-wow [57]3 years ago

3 0

.......................

You might be interested in

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

3 years ago

What is 28 divided by 1.0000

Leto [7]

28 divided by 1.0000 is 20

4 0

3 years ago

Read 2 more answers

PLEASEEEE HELP MEEE I"LL GIVE U BRAINLEST!!

DedPeter [7]

Answer:

m=13

Step-by-step explanation:

156 divided by 12mph = 13 miles

6 0

3 years ago

Write as a decimal. If necessary, use a bar to indicate which digit or group of digits repeats

Gnoma [55]

2.83333333(repeating)

8 0

3 years ago

What is the value of Left-bracket (two-thirds) Superscript 0 Baseline Right-bracket Superscript negative 3?

Vinil7 [7]

Answer:

The value of the expression is 1.

Step-by-step explanation:

The expression is:

$[(\frac{2}{3})^{0}]^{-3}$

Exponent rules:

$a^{0}=1\\\\(a^{m})^{n}=a^{m\times n}\\\\a^{-m}=\frac{1}{a^{m}}$

Simplify the expressions as follows:

$[(\frac{2}{3})^{0}]^{-3}=[1]^{-3}$

$=\frac{1}{1^{3}}\\\\=\frac{1}{1}\\\\=1$

Thus, the value of the expression is 1.

3 0

3 years ago

Read 2 more answers