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Oksana_A [137]
3 years ago
13

1a-f please help i would appreciate it.

Mathematics
1 answer:
monitta3 years ago
4 0

Answer:

c is the answer

Step-by-step explanation:

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Leviafan [203]
It is biased because they surveyed people who were at a park, therefore they enjoy parks and would want more in the area
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A ball has a circumference of 9.42 cm. what is the diameter
Keith_Richards [23]
2.998 correct me if wrong
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3 years ago
The sample consisted of 50 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 25 day students,
bulgar [2K]

Answer: The test statistic is t= -0.90.

Step-by-step explanation:

Since we have given that

n₁ = 50

n₂ = 25

\bar{x_1}=3.02\\\\\bar{x_2}=3.04\\\\\sigma_1=0.08\\\\\sigma_2=0.06

So, s would be

s=\sqrt{\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2-2}}\\\\s=\sqrt{\dfrac{50\times 0.08^2+25\times 0.06^2}{50+25-2}}\\\\s=0.075

So, the value of test statistic would be

t=\dfrac{\bar{x_1}-\bar{x_2}}{s(\dfrac{1}{n_1}+\dfrac{1}{n_2})}\\t=\dfrac{3.02-3.04}{0.074(\dfrac{1}{50}+\dfrac{1}{25})}\\\\t=\dfrac{-0.04}{0.074(0.02+0.04)}\\\\t=\dfrac{-0.04}{0.044}\\\\t=-0.90

Hence, the test statistic is t= -0.90.

7 0
3 years ago
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

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3 years ago
What is the greatest common factor of 63135
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