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2 years ago

I NEED HELP(10 POINTS) ON THIS PROBLEM

Mathematics

1 answer:

agasfer [191]2 years ago

8 0

(1250-140)/2

1110/2

answer= 550

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Ms. Bell's mathematics class consists of 12 sophomores, 8 juniors, and 13 seniors.

ycow [4]

Answer:715 ways

Step-by-step explanation:

Given

There are 12 sophomores

8 Juniors

and 13 seniors

Ms. bell create a 4- member committee of seniors

No of ways of selecting 4 seniors out of 13

$=^{13}C_4$

$=\frac{13!}{9!4!}$

$=\frac{13\times 12\times 11\times 10}{4\times 3\times 2\times 1}$

$=715$

So, there are 715 ways of forming a 4 member senior committee.

3 0

2 years ago

A cheetah can run about 68miles per hour. how fast can it run in meters per minute? use 1mi= 1609m

Olin [163]

Answer:

Approximately 26,82 meters per minute

Step-by-step explanation:

If 1 mile = 1609 meters, then the cheetah runs 1609 in 60 minutes, and we want to find the x meters it runs in 1 minute.

We do x = (1 * 1609) / 60 = 26,82 meters per minute ( approximately )

8 0

2 years ago

Read 2 more answers

Please help! Will mark brainliest

kobusy [5.1K]

Answer:

B. 15"

Step-by-step explanation:

$a^{2}+b^2=c^2\\9^2+12^2=c^2\\81+144=c^2\\225=c^2\\\sqrt{225} = 15$

7 0

2 years ago

CAN YOU HELP ME PLEASE

dalvyx [7]

Answer:

just doing this for the gram

Step-by-step explanation:

say hi

7 0

2 years ago

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

2 years ago