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4 months ago

10) Chris invested $5,500. The investment had a simple interest rate of 3.5%. If he plans to leave the

Mathematics

1 answer:

Oksi-84 [34.3K]4 months ago

5 0

Chris will have $7810 at the end of 12 years in his account if the simple interest rate is 3.5%.

According to the question,

We have the following information:

Chris invested $5,500. The investment had a simple interest rate of 3.5%. And he plans to leave the investment alone for 12 years.

Now, we know that the following formula is used to find the simple interest on an investment:

Interest = principal*rate*time/100

Principal = $5500

Rate = 3.5

Time = 12 years

Interest = (5500*3.5*12)/100

Interest = $2310

Now, the total amount in his account will be:

5500+2310

$7810

Hence, Chris will have $7810 at the end of 12 years in his account if the simple interest rate is 3.5%.

To know more about simple interest rate here

brainly.com/question/7279636

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Answer:

The area of the triangle is of 8 units of area.

Step-by-step explanation:

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The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (1,0) (5,0) (3,4). So

$A = \pm 0.5 \left|\begin{array}{ccc}1&0&1\\5&0&1\\3&4&1\end{array}\right|$

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2 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

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Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

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Step-by-step explanation:

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Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

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$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

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$p_B = 0.20$

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$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

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Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
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Click on Generate 1000 samples ---- see attachment for the generated data

From the randomization sample, we have:

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$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

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2 years ago

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Sergio [31]

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2 years ago