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love history [14]
2 years ago
10

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

selected (without replacement) to participate in a survey. Consider the following events: A = Exactly 1 of the three selected is a freshman B = Exactly 2 of the three selected are juniors Find the following probability. If needed, round to FOUR decimal places. Pr(A∩B) = ___________
Mathematics
1 answer:
ycow [4]2 years ago
3 0

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

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REY [17]
0.96 or 19/20

0.96 as fraction = 96/100 = 19.2 / 20      5 into 96 is 19.2 and 5 into 100 is 20

So 0.96 is the same as 19.2/20

Since 19.2 / 20  and 19 / 20 have the same denominator,  therefore the 19.2 / 20 is bigger than 19 / 20.

So the 0.96 is bigger than the 19 / 20.
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2 years ago
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
2 years ago
Lucia is making a 21.6 centimeter beaded string to hang in the window. She decides to put a green bead every 0.4 centimeters and
olasank [31]
21.6 ÷ 0.4 = 54 times for green
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Together, 90 beads. Separately, 54 green and 36 purple.
3 0
2 years ago
Read 2 more answers
6x+7y=4x+4y6x+7y=4x+4y6, x, plus, 7, y, equals, 4, x, plus, 4, y Complete the missing value in the solution to the equation. ((l
tia_tia [17]

Answer:

<h3>The missing value in the solution to the equation is <u>6</u>.</h3><h3>So, the complete solution is (6, -4).</h3>

Step-by-step explanation:

Given:

6x+7y=4x+4y.

(  ,-4)

Now, to complete the missing value in the solution to the equation.

<em>As, the equation is:</em>

<em />6x+7y=4x+4y<em />

<em>And, the solution is:</em>

(x,y)=(\ \ ,-4)

Now, solving the equation by putting the value of y=-4 :

6x+7(-4)=4x+4(-4)\\\\6x+(-28)=4x+(-16)\\\\6x-28=4x-16\\\\

<em>Adding both sides by 28 we get:</em>

<em />6x=4x+12<em />

<em>Subtracting both sides by </em>4x<em> we get:</em>

<em />2x=12<em />

<em>Dividing both sides by 2 we get:</em>

x=6.

<u><em>Thus, the solution to the equation is (6, -4).</em></u>

Therefore, the missing value in the solution to the equation is <u>6</u>.

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2 years ago
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AfilCa [17]
It is called a scale factor :)
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