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3 years ago

Convert 4 1/8 ÷ 2 3/4 to improper fractions

Mathematics

2 answers:

Dmitry_Shevchenko [17]3 years ago

8 0

Answer:

3/2

Step-by-step explanation:

4 1/8 / 2 3/4

= 33/8 / 11/4

= 3/2

Delicious77 [7]3 years ago

7 0

Answer:

(33/8)/(15/4)=1 1/10

Step-by-step explanation:

To convert to improper fractions, you have to multiply the denominator by the number to the left and add the numerator when converting a mixed number. So,

8*4=32 ,,,,, 32+1=33 But we still have to divide it by the denominator so it's 33/8

Do the same thing for the other mixed number. To solve with the improper fractions, you must multiply by the reciprocal of the 2nd number in your expression.

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Find the slope of a line that passes through each pair of points M(-2,3),N(7,-4)

yan [13]

-9/7

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3 years ago

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

8 0

3 years ago

X+y = -16<br> —3х – у = 30

lys-0071 [83]

Answer:

x = -7, y = -9

Step-by-step explanation:

The coefficient (the number before) the y is the same but the signs are opposite so add the equations together:

x-3x+y-y=-16+30

-2x=14

x=-7

y=-16-x=-16--7=-9

8 0

3 years ago

Convert 4 1/8 ÷ 2 3/4 to improper fractions​

Convert 4 1/8 ÷ 2 3/4 to improper fractions