All categories

3 years ago

5% of 35? P L E A S E H E L P M E T H A N K Y O U

Mathematics

2 answers:

Crazy boy [7]3 years ago

7 0

It would be 1.75 thats it your welcome

Tanzania [10]3 years ago

3 0

The answer would be 1.75

You might be interested in

How do you Find the acute Angle A when sinA=0.616?

kirill [66]

Answer:

arcsin0.616

Step-by-step explanation:

arcsino.616

4 0

3 years ago

How you you factor this expression?

ioda

You have to use factoring by grouping.. multiply by 8 and 15 and what factors add to -14.

The factors are 6 and -20

(2x-5)(4x+3)

4 0

3 years ago

emma will roll two number cubes labeled 1 through 6. She will record the sum of the two numbers after each roll. She will roll t

SashulF [63]

Answer:

75 times

Step-by-step explanation:

First of all let's ook at the 36 possible outcomes and their sums.

2=1,1

3=1,2 2,1

4=1,3 ; 2,2 ; 3,1

5=1,4 ; 2,3 ; 3,2 ; 4,1

6=1,5 ; 2,4 ; 3,3 ; 4,2 ; 5,1

7=1,6 ; 2,5 ; 3,4 ; 4,3 ; 5,2 ; 6,1

8=2,6 ; 3,5 ; 4,4 ; 5,3 ; 6,2

9=3,6 ; 4,5 ; 5,4 ; 6,3

10=4,6 ; 5,5 ; 6,4

11=5,6 ; 6,5

12=6,6

Looking at the above, there are 4 sums equal to 5 out of a total of 36 sums.

Thus, probability of getting a sum equal to 5 is;

P(5) = 5/36

Now,since the cubes are rolled 540 times, number of times a sum equal to 5 will appear will be;

(5/36) x 540 = 75 times

4 0

3 years ago

What percent is 608 of 845 million?

astraxan [27]

To find the percent of 608 of 835 million, we do:
608 ÷ 845, 000, 000 = 0.0000007.19

7 0

3 years ago

If a computer network has 60 switching nodes, in how many ways can 2 or 3 nodes fail?___________

notka56 [123]

Answer:

There are 35990 ways in which 2 or 3 nodes fail

Step-by-step explanation:

Given : A computer network has 60 switching nodes.

To Find : In how many ways can 2 or 3 nodes fail?

Solution:

We are supposed to find no. of ways can 2 or 3 nodes fail.

So, we will use combination here .

No. of ways can 2 or 3 nodes fail= $^{60}{C_2 +^{60}C_3$

Formula : $^nC_r =\frac{n!}{r!(n-r)!}$

No. of ways can 2 or 3 nodes fail= $\frac{60!}{2!(60-2)!}+\frac{60!}{3!(60-3)!}$

= $35990$

Hence there are 35990 ways in which 2 or 3 nodes fail

5 0

3 years ago