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sukhopar [10]
3 years ago
14

Ahmed and Gavin are playing both chess and checkers. The probability of Ahmed winning the chess games is 45%. The probability of

Ahmen winning the checkers games is 0.36
Which of these events is more likely

A: Ahmed wins the chess game

B: Ahmed wins the checkers game

C: Neither. Both events are equally likely
Mathematics
2 answers:
natta225 [31]3 years ago
7 0

Answer:

Option A. Ahmed wins the chess game

Step-by-step explanation:

we know that

The probability of Ahmed winning the chess games is 45%

45%=045/100=0.45

The probability of Ahmen winning the checkers games is 0.36

Compare the probability

0.45> 0.36

so

Is more likely that  Ahmed wins the chess game

Evgesh-ka [11]3 years ago
6 0

Answer:

A

Step-by-step explanation:

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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
crimeas [40]

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
3 years ago
Question<br> 9<br> What is the circumference of a circle whose diameter is ?<br> 2
ZanzabumX [31]

6.28 because pi x diameter = circumference.

( diameter = half of a circle)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Write the equation of the line PARALLEL to the line y = 4x - 9 that passes through the
german

Answer:

y = 4x + 12

Step-by-step explanation:

The slope of the line y = 4x - 9 is 4.

The slope of the line we are looking for is also 4.

y - y1 = m(x - x1)

y - 4 = 4[x - (-2)]

y - 4 = 4(x + 2)

y - 4 = 4x + 8

y = 4x + 12

Answer: y = 4x + 12

6 0
3 years ago
Read 2 more answers
Please help me that’s would be great and I promise to mark brainlest!!
defon

Option A is correct.

If we evaluate the function at x=0, i.e. at the beginning, we have

f(0)=75 \cdot 2^0 = 75 \cdot 1 = 75

with each minute, you multiply the current number of cell by 2, so they double every minute.

Option B is incorrect, because if the number of cells increases by 2 every minute, the function would be

f(x) = 75 + 2k

Option C is incorrect, because after one minute (i.e. when x=1) we have

f(1) = 75\cdot 2^1 = 75 \cdot 2 = 150

cells, so 75 is not the number of cells after one minute

Option D is incorrect in both the initial value and the behaviour of the function

6 0
3 years ago
Evaluate the expression. 5^2 ( The two goes to the top) - 4 · 6 + 11
Thepotemich [5.8K]

HI Sydney!

Question:

Solve 5^2 - 4 x 6 + 11

Solution:

Remember to follow PEMDAS,

= 5^2 - 4 x 6 + 11

= 25 - 4 x 6 + 11

= 25 - 24 + 11

= 1 + 11

= 12

Answer: 12

4 0
4 years ago
Read 2 more answers
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