This table shows how many sophomores and juniors attended two school events. A student is selected randomly from this group.

Mathematics

1 answer:

Gnoma [55]2 years ago

8 0

Answer: B. 0.45

Step-by-step explanation:

From the given table, the total number of students = 137

The number of students are sophomores =35+42=77

Let A be the event that students are sophomores.

Then probability that students are sophomores is given by :

$\text{P(A)}=\dfrac{77}{137}$

The number of sophomores who attended the jazz concert = 35

Let B be the event that students attended the jazz concert .

The probability that students attended the jazz concert and are sophomores is given by :

$\text{P(A and B)}=\dfrac{35}{137}$

Now, the probability of that the student attended the jazz concert, given that the students is sophomore is given by :-

$P(B|A)=\dfrac{\text{P(A and B)}}{\text{P(A)}}\\\\=\dfrac{\dfrac{35}{137}}{\dfrac{77}{137}}\\\\\\=\dfrac{35}{77}=0.454545454545\approx0.45$

You might be interested in

What value of x is in the solution set of 3(x – 4) ≥ 5x 2? A) -10 B) -5 C) 5 D) 10

Flura [38]

First, use the distributive property to expand the first equation.

3(x-4)=3*x+3*-4
=3x-12
Then we solve this as if it were an equation

3x-12≥5x+2
Minus 3x is
-12≥2x+2
Minus 2 is
-14≥2x

Which gets us that $x \leq 7$ (We had to flip the sign because we divided by a negative number).

You'll notice that this matches none of the choices, however, we are looking for the solution set, not the solution. In this case, it means any number that is less than 7. The correct answer is -10.
<em />:)

8 0

2 years ago

Read 2 more answers

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are

kondor19780726 [428]

Answer: The number of different combinations of 2 vegetables are possible = 15 .

Step-by-step explanation:

In Mathematics , the number of combinations of selecting r values out of n values = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Given : Number of available vegetables = 6

Then, the number of different combinations of 2 vegetables are possible will be :

$^6C_2=\dfrac{6!}{2!(6-2)!}=\dfrac{6\times5\times4!}{2\times4!}=15$

Hence , the number of different combinations of 2 vegetables are possible = 15 .

5 0

2 years ago

A Gallup Poll in July 2015 found that 26% of the 675 coffee drinkers in the sample said they were addicted to coffee. Gallup ann

emmasim [6.3K]

Answer:

The 95% confidence interval for the percent of all coffee drinkers who would say they are addicted to coffee is between 21% and 31%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.