All categories

3 years ago

What is 56:21 in simplest form

Mathematics

2 answers:

postnew [5]3 years ago

5 0

You divide both sides by 7, which is the highest common factor, and it become:
8:3

Hope this helps :)

lana [24]3 years ago

3 0

We can divide each digit by 7. So 56/7=8 and 21/7=3 so.... 8:3

You might be interested in

the basement has an area of 864 square feet. the width of the basement is two thirds its length. What is the length of the basem

Leno4ka [110]

A = 864
w = 2/3l

A = lw

Plug in what we know:

864 = l(2/3l)

Multiply:

864 = 2/3l^2

Divide 2/3 to both sides or multiply by its reciprocal, 3/2:

864 * 3/2 = l^2

2592/2 = l^2

1296 = l^2

Find the square root of both sides:

l = 36

So the length of the basement is 36 feet.

7 0

3 years ago

Read 2 more answers

A dozen eggs cost $1.32 at the grocery store how much profit would a restaurant make if they sold 1 scrambled egg for 99 cents

PIT_PIT [208]

They would make $11.77 :)

5 0

3 years ago

PLEASE HELP IM DESPERATE I DID THIS KHAN AT LEAST 7 TIMES IM SO DEPRESSED I JUST WANT TO GET THIS DONE WITH!

crimeas [40]

Answer:

y = (-3/2) x + 3

Step-by-step explanation:

The line passes through (0, 3) and (2, 0)

Slope = (0 - 3)/(2 - 0) = -3/2

Let y = (-3/2) x + b

Use (0, 3) to get b:

3 = (-3/2)*0 + b, so b = 3

Then y = (-3/2) x + 3

4 0

2 years ago

The sum of two numbers is 39.Three times the smaller number exceeds the larger number by 81. Find the numbers

Mice21 [21]

Let smaller no be x and bigger be y

x+y=39--(1)
3x=y+81
y=3x-81
y=3(x-27)--(2)

Put it in eq(1)

x+3(x-27)=39
x+3x-81=39
4x=81+39
4x=120
x=30

Now

x+y=39
y=39-30
y=9

The numbers are 9 and 30

7 0

2 years ago

A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that a

madam [21]

Answer:

Option D. 938/969

Step-by-step explanation:

At most 2 defective means 2 or less than 2 bulbs are defective

So, we have 3 cases:

a) No defective bulb. b) 1 defective bulb. c) 2 defective bulbs

Case a) No defective bulb

Total number of bulbs = 20

Number of defective bulbs = 5

Number of non-defective bulbs = 15

Total number of ways to select 0 defective bulb = 15C4 = 1365

Case b) 1 defective bulb

Total number of ways to select 1 defective bulb = 15C3 x 5C1 = 2275

Case c) 2 defective bulbs

Total number of ways to select 2 defective bulbs = 15C2 x 5C2 = 1050

Therefore, total number of ways to select at most 2 defective bulbs = 1365 + 2275 + 1050 = 4690

Total number of ways to select 4 bulbs from 20 = 20C4 = 4845

Therefore, probability of selecting at most 2 defective bulbs = $\frac{4690}{4845}=\frac{938}{969}$

Therefore, option D gives the correct answer.

6 0

3 years ago