3 years ago

Please answer please

Mathematics

2 answers:

Mice21 [21]3 years ago

7 0

Answer:

Step-by-step explanation:

28 × 3 = 84 + 12 = 96 I think, I'm not sure

riadik2000 [5.3K]3 years ago

5 0

Answer:

Step-by-step explanation:

1 sticker pack has 28 stickers

3(28) = 84

84+12

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Make x the subject of h = 4(x + 3y) + 2

ratelena [41]

H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x

6 0

3 years ago

As part of a survey, 300 girls were asked to name their favorite sport. The results showed that 12 of the girls named bowling as

Delvig [45]

You do (12/300)x100 and it is going to give you the a answer

4 0

3 years ago

If thirty students come to school and 10% go home sick, how many students went home sick? Must show your work.

Lunna [17]

3

if 30 = 100% and 10% went home sick, so 30*0.10
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7 0

3 years ago

Part 2

cluponka [151]

Answer:

$\displaystyle f(t) = -16t^2 + 192t; Maximum$

Step-by-step explanation:

Maximum Value → Parabola opens downward [−A]

Minimum Value → Parabola opens upward [A]

See graph above

I am joyous to assist you anytime.

6 0

3 years ago

Read 2 more answers

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