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Butoxors [25]
2 years ago
7

plssssssssssssssssssssss help Six friends, four boys and two girls, went to a movie theater. They wanted to sit in a way so no g

irl sits on either first or last chair. How many such arrangements are possible?
Mathematics
2 answers:
Airida [17]2 years ago
7 0

Answer:

672 ways

Step-by-step explanation:

The total number of ways the friends were supposed to sit without restrictions would have been = 6!

= 720 possible ways.

If the girs should sit on the first or last sit

The possible arrangement= 2!4!

= 2*24

= 48

But now the two girls should be arranged in such a way that the don't sit on either the first or the last chair.

So the possible way now is= way without restriction - way if the girls are to sit at either first or last sit

= 720-48

= 672

tatyana61 [14]2 years ago
5 0

Answer:

672.

Step-by-step explanation:

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Paul bought 9 total shirts for a total of $72. Tee shirts cost $10 and long sleeves shirts cost
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2 years ago
*31. Priya and Ravish planned to participate in a cycle race to be organised for National integration. They decided
Julli [10]

Answer:

36 minutes

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Step-by-step explanation:

The answer is the LCM (least common multiple) of 12 and 18.

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18 = 3^2 x 2

=>LCM of 12 and 18 = 2^2 x 3^2 = 4 x 9 = 36

=> After 36 minutes they meet again at the starting point

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7 0
2 years ago
the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of
soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = \sigma = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

z=\frac{x-u}{\sigma}

Using the values, we get:

z=\frac{25-22}{4}=0.75

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20

Thus, a test score of 20 represent the 31st percentile of the distribution.

3 0
2 years ago
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