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UkoKoshka [18]
2 years ago
11

The formula to find the perimeter of a rectangle is P = 2l + 2w. Solve for w.

Mathematics
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

W =<u> P - 2L </u>

         2

Step-by-step explanation:

perimeter P = 2L + 2W

P - 2L = 2W

W =<u> P - 2L </u>

         2

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I need help please answer asap​
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48 sq

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Your question is not well understood

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Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
1. The data given below includes data from 42 ​candies, and 7 of them are red. The company that makes the candy claims that 33​%
jeyben [28]

Answer:

The 95% confidence interval for true proportion of red candies is (5%, 27%).

The true proportion of red candies made by the company is different from 33%.

Step-by-step explanation:

The hypothesis to determine whether the claim made by the candy company is correct or not is:

<em>H</em>₀: The true proportion of red candies made by the company is 33%, i.e. <em>p</em> = 0.33.

<em>Hₐ</em>: The true proportion of red candies made by the company is different from 33%, i.e. <em>p</em> ≠ 0.33.

A (1 - <em>α</em>)% confidence interval can be constructed to check this claim.

The decision rule is:

If the confidence interval consists the null value then the null hypothesis will not be rejected. Otherwise it will be rejected.

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The information provided is:

<em>n</em> = 42

<em>X</em> = number of red candies = 7

Compute the sample proportion of candies that are red as follows:

\hat p=\frac{X}{n}=\frac{7}{42}=0.167

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table for the critical value.

Compute the 95% confidence interval for true proportion of red candies as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.167\pm 1.96\times \sqrt{\frac{0.167(1-0.167)}{42}}\\=0.16\pm0.1137\\=(0.0463, 0.2737)\\\approx(0.05, 0.27)

The 95% confidence interval for true proportion of red candies is (5%, 27%).

The confidence interval does not contains the null value.

Thus, the null hypothesis will be rejected at 5% level of significance.

Conclusion:

The true proportion of red candies made by the company is different from 33%.

7 0
3 years ago
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