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2 years ago

Cai says you can divide both quantities in a ratio by the same nonzero number to

Mathematics

1 answer:

alexandr1967 [171]2 years ago

6 0

Answer:

correct

Step-by-step explanation:

dividing both quantities in a ratio by the same non zero number does indeed find an equivalent ratio.

for example

60 : 36 ( divide both parts by 6 )

= 10 : 6 ← equivalent ratio ( divide both parts by 2 )

= 5 : 3 ← equivalent ratio in simplest form

in fact dividing the original by the HCF of the 2 quantities gives the ratio in simplest form immediately , that is

60 : 36 ( divide both parts by 12 )

= 5 : 3

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Jordan drove for 6 hours at 55 miles per hour, while Matt drove for 3 hours at 60 miles per hour. If Isaac drove 82 miles longer

mihalych1998 [28]

If you look at the equation, we have to multiply the hour and the mph(miles per hour) together to find how many miles they drove. For Jordan, we would have to multiply 6 times 55 which is 330. Jordan drove 330 miles. For Matt, we have to multiply 3 times 60, which gets us 180. Together, they would have drove 510 miles. If Isaac drove 82 miles longer than Jordan and Matt drove combined, we would get 582 miles. Isaac drove 592 miles.

5 0

4 years ago

The area of a trapezoid is 243cm2. The height is 18cm and the length of one of the parallel sides is 10cm. Find the length of th

Irina18 [472]

Answer:

$Side_2 = 17\ cm$

Step-by-step explanation:

Given

Shape: Trapezoid

$Area = 243cm^2$

$Height = 18cm$

$Side_1 = 10cm$

Required

Determine the length of the second parallel side

The area of a trapezoid is:

$Area = \frac{1}{2}(Side_1 + Side_2) * Height$

Substitute values for Area, Height and Side1

$243 = \frac{1}{2}(10 + Side_2) * 18$

Multiply both sides by 2

$2 * 243 = 2 * \frac{1}{2}(10 + Side_2) * 18$

$486 = (10 + Side_2) * 18$

Divide both sides by 18

$27 = 10 + Side_2$

$Side_2 = 27 - 10$

$Side_2 = 17\ cm$

Hence;

<em>The length of the second parallel side is 17cm</em>

8 0

3 years ago

A while back, either james borrowed $12 from his friend rita or she borrowed $12 from him, but he can’t quite remember which. ei

LenaWriter [7]

To answer the question let x be the amount of money James will have after he sees Rita. If it was him who owed Rita $12 and he intent to pay, he will have $42.80 - $12 after seeing her. However, if it was Rita who owed him the amount, the money he will have after seeing her will be $42.80 + $12. Thus, the equations are,
x = 42.80 - 12 and x = 42.80 + 12

8 0

4 years ago

Read 2 more answers

Ms Davis has 6 red pens, 4 blue pens and 8 black pens. All of her pens are the same size She puts all of them in a bag What is t

solmaris [256]

Answer:

Step-by-step explanation:

Add all of them =18 and there is 4 blue ones

4/18

4 0

4 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

4 years ago