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3 years ago

A standard deck of 52 cards contains 4 aces. What is the probability of randomly drawing a card that is not an ace?

Mathematics

1 answer:

Stells [14]3 years ago

8 0

I There would be 52-4 = 48 cards that are not aces.

The probability of picking a non ace card would be 48 out of 52, which is written as 48/52, which reduces to 12/13

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70 x 10 ^ 3 word form

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Seven multiplied by ten to the third power

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...................Simplify 43.45

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Answer:

9.31st 91 35 5789 400th

Step-by-step explanation:

43.45+567=

9.3

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Find the value of the length <br> x<br> rounded to 1 DP.

bagirrra123 [75]

Length of x is 98.2 m

<u>Step-by-step explanation:</u>

Step 1:

Use the trigonometric ratio tan 27° to find the common side of both the right angled triangles.

tan 27° = opposite side/adjacent side = opposite side/9

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A recycling box has a square base with sides measuring 18 inches. The box is 16 inches tall. How much recycling can the inside o

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The volume of a cube = length x width x height. Since they told us this box has a square base, the length and width are both 18 inches.

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2 years ago

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

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Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

2 years ago