All categories

2 years ago

In a class of 25 students, 15 of them have a cat,

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

7 0

Answer: 9/25

Step-by-step explanation:

Since 3 of them have neither a cat or a dog, that means 22 (25-3) of them have a dog, a cat, or both.

16 + 15 = 31, which is 9 more than 22. That means 9 of them have both a dog and a cat.

Out of the 25 people, 9 have both a cat and dog, which means the probablity a student chosen at random has both is 9/25.

You might be interested in

Chapter 12 F

marshall27 [118]

Answer:

(m+n) -a = r

Step-by-step explanation:

first, you have to find what m+n is because that is their total budget, then you have to remove how much their apartment costs from their budget, because r is the amount of free cash they have. once A is subtracted from m+n, you have r

8 0

2 years ago

where would an imaginary line need to be drawn to reflect across an axis of symmetry so that a regular pentagon can carry onto i

irina [24]

Answer:

An imaginary line would need to be drawn at any angle to the center of the side opposite the angle to reflect across an axis of symmetry so that a regular pentagon can carry onto itself. Hope this answer helps.

5 0

2 years ago

FRÉD AND ETHEL WACHED TV FOR SEVERAL HOURS ONE AFTERNOON. THEY BEGAN AT 1:30 P.M AND CONTINUED WACHING TV FOR 3 HOURS AND 45 MIN

QveST [7]

$30 + 45 =75$
$3 + 1 = 4$
There are 60 minutes in one hour, so 75 minutes is equal to 1 hour and 15 minutes.

$4 + 1 = 5$
Their mom turned the TV off at 5:15.

7 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

2 years ago

The dimensions of a conical funnel are shown below: A conical funnel is shown with the height of the cone as 7 inches and the ra

Nataly [62]

The first thing we should know in this case is that by definition the flow of liquid is given by:
Q = V / t
Where,
V: volume
t: time
The volume of the cone is given by:
V = (pi * r ^ 2 * h) / (3)
Where,
r: radio
h: height
Substituting the values we have:
V = (3.14 * ((3) ^ 2) * 7) / (3)
V = 65.94 in ^ 3
We now turn off the time of the flow equation:
t = V / Q
Substituting values:
t = (65.94) / (14)
t = 4.71 minutes
Answer:
It will take for all the liquid in the funnel to pass through the nozzle about:
t = 4.71 minutes

5 0

3 years ago

Read 2 more answers