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3 years ago

5

Mark studied a group of 30 whales. If each will wait approximately 3.8×10 to the fifth power, pounds find the total weight of al

l 30 whales express your answer in scientific notation

1 answer:

natta225 [31]3 years ago

7 0

Answer:

11.4 x 10 ^ 6

Step-by-step explanation:

First, you have to convert 3.8 x 10^5 out of scientific notation. To do that, multiply 3.8 * 10^5, you'll get 380,000. This is the weight of each whale.
Now multiply the weight of each whale times 30. 380,000 * 30 = 11,400,000
Convert back to scientific notation. 11.4 x 10^6

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Answer:

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Using bootstrap, the standard error is: 0.050

Comparison:

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Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

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Solving (a): Standard error using formula

First, calculate the proportion of A

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

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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
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

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So, we have:

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

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

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

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

$SE_{p_A-p_B} = 0.050$

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