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faltersainse [42]
3 years ago
13

How do i use permutation on 5P3?

Mathematics
1 answer:
Vika [28.1K]3 years ago
5 0

You use permutation formula

n!/(n-r)!

n = 5

r = 3

5!/(5-3)!

= 120/2=60


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Five hundred yards of fence is to be used to enclose a rectangular area next to a straight river. The river bank acts as one sid
fenix001 [56]

Answer:

the requirements are missing:

  1. express the height of the rectangle in terms of w
  2. express the area of the rectangle in terms of w

1) the height of the rectangle is:

w + h + h + w = 500 + w

2w + 2h = 500 + w

2h = 500 + w - 2w = 500 - w

h = (500 - w) / 2 = 250 - 0.5w

2) the are of the rectangle is:

w x h = w x (250 - 0.5w) = 250w - 0.5w²

3 0
2 years ago
How can i find the volume of this figure help please
valentinak56 [21]

Answer:

The answer is 463 inch cubed

Step-by-step explanation:

you do 8 x 8 x 7 which will give you 448 volume for just the cube.

then you do 5 x 3 x 1 = 15

you add 448 with 15 you will get 463

6 0
3 years ago
(7y - 2n) - (5y + 3a) =<br><br> Pls help
Sati [7]

Answer:

2y-2n-3a

Step-by-step explanation:

(7y-2n) - (5y+3a)

7y-2n-5y-3a

2y-2n-3a

4 0
2 years ago
Read 2 more answers
Luis has a $10 and three $5 bills. He spends $12.75 on entrance fee to an amusement park and $8.50 on snacks. How much does he h
Ulleksa [173]
3x5=$15
15+10=$25 total to spend

He spends 12.75+8.5= $21.25

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6 0
2 years ago
A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take
ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0
2 years ago
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