Please help I’m really need the help (NO LINKS ) (NO LYING ) PART 2 again

Mathematics

1 answer:

ohaa [14]2 years ago

6 0

Answer:

A. 15

B. 25.6

C. 41.6

Step-by-step explanation:

You need to divide

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What fraction is equal to the decimal 0.32? Enter your answer in the box in simplest form.

AVprozaik [17]

The answer is 8/25 I believe

3 0

2 years ago

A classroom has ten students. Three students are freshman, two are sophomores, and five are juniors. Three students are randomly

ycow [4]

Answer:

0.25

Step-by-step explanation:

We have a total of ten student, and three students are randomly selected (without replacement) to participate in a survey. So, the total number of subsets of size 3 is given by 10C3=120.

On the other hand A=Exactly 1 of the three selected is a freshman. We have that three students are freshman in the classroom, we can form 3C1 different subsets of size 1 with the three freshman; besides B=Exactly 2 of the three selected are juniors, and five are juniors in the classroom. We can form 5C2 different subsets of size 2 with the five juniors. By the multiplication rule the number of different subsets of size 3 with exactly 1 freshman and 2 juniors is given by

(3C1)(5C2)=(3)(10)=30 and

Pr(A∩B)=30/120=0.25

3 0

3 years ago

A bag contains 42 red marbles, 6 white marbles, and 8 gray marbles. You randomly pick out a marble, record its color, and put it

Elanso [62]

For any single draw,

$\mathbb P(\text{white})=\dfrac6{42+6+8}=\dfrac6{56}$
$\mathbb P(\text{gray})=\dfrac8{42+6+8}=\dfrac8{56}$

Drawing a white marble or a gray marble are disjoint events; only one of them can happen. So

$\mathbb P(\text{white or gray})=\mathbb P(\text{white})+\mathbb P(\text{gray})-\underbrace{\mathbb P(\text{white and gray})}_0$
$\mathbb P(\text{white or gray})=\dfrac6{56}+\dfrac8{56}=\dfrac{14}{56}$

Out of 224 draws, you should expect $\dfrac{14}{56}\times224=56$ of the marbles to be either white or gray.