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3 years ago

24/60 in simplest form explain

2 answers:

const2013 [10]3 years ago

8 0

2/5
In order to find this question quickly, find the greatest common variable between the two numbers. In this case, the number is 12. Now, divide both numbers by their greatest common variable. 24/12= 2 and 60/12= 5.

nekit [7.7K]3 years ago

5 0

24/60 Frist you have to find the greatest common factor which is 12 then you divide 24 and 60 by 12 and you get 2/5

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A flower show has an inventory of 60 tulips and 100 roses. Which equation is correct?

Zinaida [17]

Answer:

The correct equitation is 60=\frac{3}{5}\times10060=

×100 OR 100=\frac{5}{3}\times60100=

×60

Step-by-step explanation:

To determine:

A flower show has an inventory of 60 tulips and 100 roses?

Information Fetching and Solution Steps:

A flower show has an inventory of

60 tulips

100 roses

So,

60=\frac{3}{5}\times10060=

×100

100=\frac{5}{3}\times60100=

×60

Other options can not be true as they do not satisfy the equation. For example,

60 = 3\times 10060=3×100 can not be true as the L.H.S ≠ R.H.S

Therefore, the correct equitation is 60=\frac{3}{5}\times10060=

×100 OR 100=\frac{5}{3}\times60100=

×60

5 0

3 years ago

A math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probabili

vovangra [49]

Answer:

(a) The probability that a male is selected, then two females is 0.4352.

(b) The probability that a female is selected, then two males is 0.3348.

(c) The probability that two females are selected, then one male is 0.4352.

(d) The probability that three males are selected is 0.0717.

(e) The probability that three females are selected is 0.1583.

Step-by-step explanation:

We are given that a math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment.

(a) The probability that a male is selected, then two females is given by;

Number of ways of selecting a male from a total of 11 male = $^{11}C_1$

Number of ways of selecting two female from a total of 14 female = $^{14}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{11}C_1 \times ^{14}C_2}{^{25}C_3}$

= $\frac{\frac{11!}{1! \times 10!} \times \frac{14!}{2! \times 12!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{1001}{2300}$ = 0.4352

(b) The probability that a female is selected, then two males is given by;

Number of ways of selecting a female from a total of 14 female = $^{14}C_1$

Number of ways of selecting two males from a total of 11 male = $^{11}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_1 \times ^{11}C_2}{^{25}C_3}$

= $\frac{\frac{14!}{1! \times 13!} \times \frac{11!}{2! \times 9!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{770}{2300}$ = 0.3348

(c) The probability that two females is selected, then one male is given by;

Number of ways of selecting two females from a total of 14 female = $^{14}C_2$

Number of ways of selecting one male from a total of 11 male = $^{11}C_1$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_2 \times ^{11}C_1}{^{25}C_3}$

= $\frac{\frac{14!}{2! \times 12!} \times \frac{11!}{1! \times 10!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }