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Emery bought 3 cans of beans that had a total weight of 2.4 pounds. If each can of beans weighed the same amount, which model co

Alexxx [7]

Answer:

y=0.8x

A 2-column table with 3 rows. Column 1 is labeled number of cans with entries 5, 15, 20. Column 2 is labeled total weight (in pounds) with entries 4, 12, 16.
On a coordinate plane, the x-axis is labeled number of cans and the y-axis is labeled total weight (in pounds. A line goes through points (5, 4) and (15, 12).

Step-by-step explanation:

Statement 1

If 3 cans of beans weigh 2.4 pounds

Then 1 Can will weigh (2.4 ÷ 3)=0.8 Pounds

If y is the total weight of x number of cans, then: y=0.8x

Statement 2

If x=5, then y=0.8(5)=4

If x=15, then y=0.8(15)=12

If x=20, then y=0.8(20)=16

Therefore the below statement applies:

A 2-column table with 3 rows. Column 1 is labeled number of cans with entries 5, 15, 20. Column 2 is labeled total weight (in pounds) with entries 4, 12, 16.

Statement 3

From the pair of points above, we have (5,4) and (15,12). Therefore if on a coordinate plane, the x-axis is labeled number of cans and the y-axis is labeled total weight (in pounds.) A line goes through points (5, 4) and (15, 12).

3 0

4 years ago

Read 2 more answers

Calculus question?

Ann [662]

Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)

Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.

Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.

I suppose this is possible.
y' = 3/x^(1/2) + 6x

y' = $\frac{3 + 6x^{3/2}}{x^{1/2}}$

Frankly I like the first answer better, but you have a choice of both.

5 0

3 years ago

What is the area of this parallelogram?

Hunter-Best [27]

Answer:

28 in²

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Which expression is equivalent to...

torisob [31]

Answer:

$x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$ i.e answer A.

Step-by-step explanation:

This question involves knowing the following power/exponent rule:

$\sqrt[n]{x^m} = x^\frac{m}{n} \\\\so \sqrt[7]{x^2} = x^\frac{2}{7} \\\\and \\\\ \sqrt[5]{y^3} = y^\frac{3}{5} \\$

Next, when a power is on the bottom of a fraction, if we want to move it to the top, this makes the power become negative.

so the y-term, when moved to the top of the fraction, becomes:

$y^{-\frac{3}{5} } \\$

So the answer is: $x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$

7 0

3 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

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

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

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

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

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

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

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

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

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

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

$SE_{p_A-p_B} = 0.052$

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 ---- see attachment for the generated data

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

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

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

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

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

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

$SE_{p_A-p_B} = 0.050$

5 0

3 years ago