All categories

3 years ago

How do u turn .875 into a fraction

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

6 0

You put .875 over the number 1 to make it a fraction so $\frac{.875}{1}$

You might be interested in

Fill in the blank for Remainder Therom<br><br><br> x 3 + x 2 - 21x - 45 = (x + 3)(x - 5)(x+?)

icang [17]

Not sure about remainder theorem but I'm sure that the last terms should all multiply tho the last term

see the expanded form is -45

so the last terms of each binomial should multiply to -45

3 times -5 times ?=-45
-15 times ?=-45
divide by -15
?=3

the question mark is 3

5 0

3 years ago

Read 2 more answers

Nesreen wants to find out how often people in her town visit the cinema.

cricket20 [7]

Tbh I don’t know and I hope someone gives you the answer

3 0

3 years ago

PLEASE ITS MY LAST QUESTION IM ALMOST DONE HELPPP

arlik [135]

They took $10 from 10 people that saw and wanted to try the ad.

6 0

3 years ago

Read 2 more answers

Answer 10,11,12,13,14

QveST [7]

10
13.50 per set
+.45 per day value
she got 20 sets
5 days after would be
13.50 ×20=270
value of 20 sets at 13.50 a set on day one was 270.00
.45×20=9.00
the 20 sets gain 9 dollars a day
9×5=45
9 dollars a day times 5 days is 45 dollars
45+270=315

11

4 0

3 years ago

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each stu

V125BC [204]

Answer:

$\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1\times ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^{12}C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Which is the required expression.

3 0

3 years ago

Read 2 more answers