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3 years ago

7400 dollars is placed in an account with an annual interest rate of 7%. To the nearest

Mathematics

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

23.4 years.

Step-by-step explanation:

7400 dollars is placed in an account with an annual interest rate of 7%.

Now, let us assume that the interest is compounded annually and it will take x years for account value to reach 36000 dollars.

Hence, from the formula of compound interest, we can write that

$36000 = 7400(1 + \frac{7}{100})^{x}$

⇒ $(1.07)^{x} = 4.865$

Taking ln both sides we get,

x ln (1.07) = ln (4.865)

⇒ x = 23.4 years.

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3 years ago

How do you write 9.14 X 10^-5 in standard form?

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3 years ago

(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

Linear equation topic. (Lisa is 5 years younger than his elder brother,Daniel.The product of their ages is 2/3 of their grandmot

alisha [4.7K]

Let Daniel's age be 'x' years , hence, Lisa's age is (x - 5) years. Let Grandma's age be 'y' = 54 years.
Hence,
x(x - 5) = 2/3 × y
x² - 5x = 2 × 54 years / 3
x² - 5x = 36 years
x² - 5x - 36 = 0
∆ = (-5)² - 4(1)(-36) = 25 + 144 = 169
√∆ = √169 = 13
$x\: = \: \frac{ - b \: \frac{ + }{ } \: \sqrt{ {b}^{2} \: - \: 4ac} }{2a}$
x1 = (5+13)/2, x2 = (5-13)/2
x1 = 9, x2 = -4
Since, age can't be negative,
Daniel's age = x = 9 years.

7 0

4 years ago