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3 years ago

6

• 1. If 16 ounces of bulk rice costs $2.50, how much would 24 ounces cost? Solve the problem and then verify your response using

another method.

2 answers:

jasenka [17]3 years ago

6 0

$60 hope this helps you!:)

Maru [420]3 years ago

6 0

$9.6 is the awnser i think

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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 ---- <em>see attachment for the generated data</em>

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

2 years ago

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0

3 years ago

Help with both questions offering 10 points

Alexandra [31]

Answer:

$\:\mathrm{First\: problem:\:}-2,\\\:\mathrm{Second\: problem:\: }-3$

Step-by-step explanation:

The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$

Plugging in any two points in the table we have:

<u>First problem:</u>

<u /> $m=\frac{1-5}{2-0}=\frac{-4}{2}=\fbox{$-2$}$ <u />

<u>Second problem:</u>

<u /> $m=\frac{-1-5}{2-0}=\frac{-6}{2}=\fbox{$-3$}$ <u />

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3 years ago

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4xy + y = w<br><br> solve for y<br><br> pls show work

Soloha48 [4]

Answer:

y = w/4x+1

Step-by-step explanation:

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2 years ago

What kind of change occurs when a figure is translated?

vagabundo [1.1K]

It depends on what way the figure is translated, but it's possible it could be a dilations, reflection, rotation, etc.

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3 years ago

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