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xz_007 [3.2K]
3 years ago
14

Simplify the ratio 18 : 15

Mathematics
2 answers:
Olenka [21]3 years ago
4 0
I believe the answer would be 6:5 because you divide both 18 and 15 by 3.
kirill115 [55]3 years ago
4 0
18/15 = (3*6)/(3*5) = 6/5

hope this is understandably sure right easy 
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I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b
Bezzdna [24]
<h3>Answer:   15</h3>

========================================================

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\

Unfortunately we don't reach 365 or larger.

Let's try n = 11

n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for <u>n = 15</u> instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

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2 years ago
The seventh-grade class is building target areas for a PE activity. The bases for the game will be circular in shape. The diamet
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3 years ago
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It is possible to score higher than 1600 on the combined mathematics and reading portions of the SAT, but scores 1600 and above
Citrus2011 [14]

Answer:

The proportion of scores reported as 1600 is 0.0032

Step-by-step explanation:

Let X be the score for 1 random person in SAT combining maths and reading. X has distribution approximately N(μ = 1011,σ = 216).

In order to make computations, we standarize X to obtain a random variable W with distribution approximately N(0,1)

W = \frac{X-\mu}{\sigma} = \frac{X-1011}{216} \simeq N(0,1)

The values of the cummulative distribution function of the standard Normal random variable, lets denote it \phi are tabulated, you can find those values in the attached file. Now, we are ready to compute the probability of X being bigger than 1600

P(1600< X) = P(\frac{1600-1011}{216} < \frac{X-1011}{216}) = P(2.7269 < W) = 1- \phi(2.7269)\\= 1- 0.9968 = 0.0032

Hence, the proportion of scores reported as 1600 is 0.0032.

Download pdf
5 0
4 years ago
For a singing contest in which 42,000 votes were cast, the winner received 3/5 of the votes. How many votes did the winner not r
zaharov [31]
The winner did not receive 2/5 of the votes, so in numerical values, that would be 16, 800 votes
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3 years ago
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A cube has edge length 5in use a exponent to express volume
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5 cubed ( or an exponent of three )
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