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anygoal [31]
3 years ago
11

Find the number of the different ways, for 3 students to sit on 7 seats in one row.

Mathematics
2 answers:
qwelly [4]3 years ago
3 0

Answer:

210

Step-by-step explanation:

Here comes the problem from Combination.

We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.

Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7

Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6

Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5

Hence the total number of ways for three students to be seated will be

7 x 6 x 5

=210

natulia [17]3 years ago
3 0
The answer to the question is 210
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Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
What inequality matches the statement : A number is at most 40
Misha Larkins [42]

Answer:

N ( a number) \leq 40

Step-by-step explanation:

Or just say

N \leq 40

8 0
3 years ago
Read 2 more answers
Sam wants to surround his garden with fencing. He is using a drawing with a scale factor of 4 feet for every 1 inch.
solmaris [256]
24! 2 plus 4 is 6 and 6 times 4 for the scale factor
5 0
2 years ago
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The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer se
tankabanditka [31]

Answer:

A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

<u><em>Difference Scores Calculations</em></u>

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

<u><em>T-value Calculation</em></u>

<u><em>Pooled Variance= Sp²</em></u>

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.

7 0
2 years ago
Try to simplify 72/6+85/6
Reil [10]
72/6+ 85/6
= (72+ 85)/6
= 157/6
= (156+1)/ 6
= 156/6+ 1/6
= 26+ 1/6
= 26 1/6

The final answer is 26 1/6~
4 0
3 years ago
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