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Q5: The probability that event A occurs is 5/7 and the probability that event B occurs is 2/3 . If A and B are independent event

bekas [8.4K]

Answer:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

Step-by-step explanation:

We define two events for this case A and B. And we know the probability for each individual event given by the problem:

$p(A) = \frac{5}{7}$

$p(B) = \frac{2}{3}$

And we want to find the probability that A and B both occurs if A and B are independent events, who menas the following conditions:

$P(A|B) = P(A)$

$P(B|A) = P(B)$

And for this special case we want to find this probability:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

7 0

3 years ago

Consider the reaction. Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3

umka2103 [35]

The equilibrium constant of the reaction is 0.0030.

The equation of the reaction is given as;

N2(g) + 3H2(g) ⇆ 2NH3 (g)

We have the following information at equilibrium;

[NH3] = 0.105 M

[N2] = 1.1 M

[H2] = 1.50 M

Hence, we can calculate the equilibrium constant using the relation;

K = [NH3]^2/ [N2] [H2]^3

Substituting values;

K = [0.105]^2/[ 1.1] [1.50]^3

K =0.011025 /3.7125

K= 0.0030

Learn more: brainly.com/question/17960050

3 0

2 years ago

for a trip, one high school rented and filled 2 vans and 8 buses with 176 students.another high school instead fit its 164 stude

vodka [1.7K]

Answer:

A bus carries 20 students while the van carries 8 students

Step-by-step explanation:

Let's assume that the number of students that the bus can carry is X And

The number of students that the van can very be y

From first statement,

8x + 2y = 176______ first equation

From Second statement,

7x + 3y = 164______ second equation

The above is a simultaneous equation and we will use the elimination method to solve for both x and y

But if we take a look at both equations,we will find out that we can't make use of the elimination method since the coefficients of both x and y aren't same in both equations.

What we have to do now is to multiply equation 1 by 3 and multiply equation 2 by 2 so as to get the coefficients of y in both equation to be the same.

Initially, both equation are

8x + 2y = 176____ first equation

7x + 3y = 164____ second equation

They will now be :

24x + 6y = 528(was multiplied by 3)

14x + 6y = 328(was multiplied by 2)

Now subtract equation 2 from 1 and we have

10x = 200

X = 20

Now substitute x = 20 in equation 1 above

8x + 2y = 176

(8 × 20) + 2y = 176

2y = 176 - 160

Y = 8

Remember that we assumed that the number of students that each bus carried was x and van was y,now we know the values of c and y to be 20 and 8 respectively.

Therefore, each bus carried 20 students and each van had 8 students on it

8 0

4 years ago

Read 2 more answers

Please helpppp tyyy so much

SIZIF [17.4K]

Answer:

`d

Step-by-step explanation:

you would have to add the dimes and quarters

5 0

3 years ago