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3 years ago

Sean bought 68 3/4 inches of rope . How many 15-inches ropes can he make from it

2 answers:

6 0

Length of rope bought by Sean = 68 3/4 inches
   = 275/4 inches
Then
The number of 15 inch rope that can be made by Sean = (275/4)/15
   = 275/(4 * 15)
   = 275/60
   = 4.58
So we can see from the above deduction that Sean can make 4 complete 15 inches ropes and there will be 0.58 inches of rope excess. <span>I hope the procedure and the method of solving is absolutely clear to you. This is the easiest way to get to the solution for such problems.</span>

lutik1710 [3]3 years ago

6 0

I first divided 3/4 tk her 0.75
I then got 68.75 as the inches of rope.

If you then divide 68.75/15
you get 4.58333.....

Sean can make 4.58 - 15inch ropes

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The attendence at a water park varies directly as the temperature and invesely as the number of days of rain during the month. I

Margarita [4]

Let A be attendance, t the temperature and n the number of days. Then,
A $\alpha$ $\frac{t}{n}$
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Where k is a constant of proportionality;
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Rewriting the equation and replacing the value of k,
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3 years ago

Which technique is most appropriate to solve this equation?

Free_Kalibri [48]

quadratic formula

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x(x-9)+4(x-9)

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3 years ago

In our class, there are 14 female and 11 male students. a. Pick one student randomly, what is the probability you get a female s

Reika [66]

Answer:

a) 56% probability you get a female student.

b) 30.33% probability that both are female

c) 43.52% probability that you get 2 female and one male

d) 18.26% probability that you get 1 female and 3 male

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in whih the students are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

a. Pick one student randomly, what is the probability you get a female student?

Desired outcomes:

1 female student from a set of 14. So

$D = C_{14,1} = \frac{14!}{1!13!} = 14$

Total outcomes:

1 student from a set of 25. So

$T = C_{25,1} = \frac{25!}{1!24!} = 25$

Probability:

$P = \frac{D}{T} = \frac{14}{25} = 0.56$

56% probability you get a female student.

b. Pick two students randomly, what is the probability that both are female?

Desired outcomes:

2 female students, from a set of 14. So

$D = C_{14,2} = \frac{14!}{2!12!} = 91$

Total outcomes:

2 students from a set of 25. So

$T = C_{25,2} = \frac{25!}{2!23!} = 300$

Probability:

$P = \frac{D}{T} = \frac{91}{300} = 0.3033$

30.33% probability that both are female

c. Pick three students randomly, what is the probability that you get 2 female and one male?

Desired outcomes:

2 female students, from a set of 14, and one male student, from a set of 11. So

$D = C_{14,2}*C_{11,1} = \frac{14!}{2!12!}*\frac{11!}{1!10!} = 1001$

Total outcomes:

3 students from a set of 25. So

$T = C_{25,3} = \frac{25!}{3!22!} = 2300$

Probability:

$P = \frac{D}{T} = \frac{1001}{2300} = 0.4352$

43.52% probability that you get 2 female and one male

d. Pick four students randomly, what is the probability that you get 1 female and 3 male?

Desired outcomes:

1 female student, from a set of 14, and 3 male students, from a set of 11.

$D = C_{14,1}*C_{11,3} = \frac{14!}{1!13!}*\frac{11!}{8!3!} = 2310$

Total outcomes:

4 students from a set of 25. So

$T = C_{25,4} = \frac{25!}{4!21!} = 12650$

Probability:

$P = \frac{D}{T} = \frac{2310}{12650} = 0.1826$

18.26% probability that you get 1 female and 3 male

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3 years ago

The random variable KK has a geometric distribution with mean 16. Which of the following is closest to the standard deviation of

asambeis [7]

Answer:

E 240

Step-by-step explanation:

Geometric Distribution mean = (1- p) / p

16 = (1 - p) / p

16p = 1 - p ; 17p = 1 ; p = 1/17 = 0.058

Standard Deviation = (1 - p) / p^2

= (1 - 0.058) / (0.058)^2

0.942 / 0.003364

= 280 , closest to 240

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3 years ago

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Answer:

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3 years ago