Final answer:
By using the principle of inclusion-exclusion, we can find that the number of families that had only a parakeet is 5, making option C the correct answer.
Explanation:
To solve this problem, we can use the principle of inclusion-exclusion for sets. According to the survey:
100 families were surveyed.35 had a dog.28 had a cat.10 had both a dog and a cat.42 had neither a cat nor a dog nor a parakeet.0 had a cat, a dog, and a parakeet.First, we need to determine the number of families that had either a cat or a dog or both, which is given by the formula for the union of two sets:
Families with a cat or dog = Families with a dog + Families with a cat - Families with both a dog and a cat
Families with a cat or dog = 35 + 28 - 10 = 53 families
Now, we subtract this number from the total number of families to find out how many families had a pet that was not a cat or a dog:
Families with other pets = Total families - Families with a cat or dog - Families with neither pet
Families with other pets = 100 - 53 - 42 = 5 families
Since none of the families had all three pets and we are assuming that 'other pets' only includes parakeets, those 5 families must have had only a parakeet. Therefore, the answer is 5 families had a parakeet only.
Please explain.
A quiz has 3 multiple-choice questions with 4 possible answer choices each. For each question, there is only 1 correct answer.
A student chooses each answer at random. What is the probability that the student will answer all three questions correctly?
Answer:
25% probability
Step-by-step explanation:
If there are three questions with four possible choices, there are twelve total answer choices. if you only get one answer as the correct one in each question then there's three out of the twelve answer choices that are correct in total. Basically, 3/12 = .25 = 25%. that's for total.
Also, with each question having 4 answers with 1 correct, 1/4 = .25 = 25% as well.
I hope this helped!
Answer: 25% is the possibility.
Step-by-step explanation:
So 1 multiple choice will have 4 possible answer.
2 multiple choice will have 4 possible answer.
3 multiple choice will have 4 possible chances.
So it’s out of a 100. You do 100 divided by 4.
Since there are 4 possible answer and it gives us 25%.
To check if our answer is correct then you should do 25 divided by 4 which gives us 100%.
The average time required to complete an accounting test has been determined to be 55 minutes. Assuming that times required to take tests are exponentially distributed, how many students from a class of 30 should be able to complete the test in between 45 and 60 minutes?
Answer: 3
Step-by-step explanation:
Given : The average time required to complete an accounting test : [tex]\lambda = 55 \text{ minutes}=0.9167\text{ hour}[/tex]
Interval = (45, 60) minutes
In hour : Interval = (0.75, 1)
The cumulative distribution function for exponential function is given by :-
[tex]F(x)=1- e^{-\lambda x}[/tex]
For [tex]\lambda =0.9167\text{ hour}[/tex]
[tex]P(X\leq1)=1- e^{-(0.9167) (1)}=0.6002[/tex]
[tex]P(X\leq0.75)=1- e^{-(0.9167)(0.75)}=0.4972[/tex]
Then ,
[tex]P(0.75<x<1)=P(X\leq1)-P(X\leq0.75)\\\\=0.6002-0.4972=0.103[/tex]
Now, the number of students from a class of 30 should be able to complete the test in between 45 and 60 minutes =
[tex]0.103\times30=3.09\approx3[/tex]
Hence, the number of students should be able to complete the test in between 45 and 60 minutes =3
Professor Halen has 184 students in his college mathematics lecture class. The scores on the midterm exam are normally distributed with a mean of 72.3 and a standard deviation of 8.9. How many students in the class can be expected to receive a score between 82 and 90?
Answer: 21
Step-by-step explanation:
Given : The scores on the midterm exam are normally distributed with
[tex]\mu=72.3\\\\\sigma=8.9[/tex]
Let X be random variable that represents the score of the students.
z-score: [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=82
[tex]z=\dfrac{82-72.3}{8.9}\approx1.09[/tex]
For x=90
[tex]z=\dfrac{90-72.3}{8.9}\approx1.99[/tex]
Now, the probability of the students in the class receive a score between 82 and 90 ( by using standard normal distribution table ) :-
[tex]P(82<X<90)=P(1.09<z<1.99)\\\\=P(z<1.99)-P(z<1.09)\\\\=0.9767-0.8621=0.1146[/tex]
Now ,the number of students expected to receive a score between 82 and 90 are :-
[tex]184\times0.1146=21.0864\approx21[/tex]
Hence, 21 students are expected to receive a score between 82 and 90 .
The number of students who can be expected to receive a score between 82 and 90 is approximately 21.
The number of students who can be expected to receive a score between 82 and 90 is calculated by finding the area under the normal distribution curve between these two scores. This requires standardizing the scores and using the standard normal distribution table or a calculator with normal distribution capabilities.
First, we need to find the z-scores for both 82 and 90 using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where ( X ) is the score in question, [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
For a score of 82:
[tex]\[ z_{82} = \frac{82 - 72.3}{8.9} \approx \frac{9.7}{8.9} \approx 1.09 \][/tex]
For a score of 90:
[tex]\[ z_{90} = \frac{90 - 72.3}{8.9} \approx \frac{17.7}{8.9} \approx 1.98 \][/tex]
Next, we look up these z-scores in the standard normal distribution table to find the corresponding area under the curve to the left of each z-score.
For [tex]\( z_{82} \approx 1.09 \)[/tex], the area to the left is approximately 0.8621.
For [tex]\( z_{90} \approx 1.98 \)[/tex], the area to the left is approximately 0.9762.
The area between the two z-scores is the difference between the two areas:
[tex]\[ P(82 < X < 90) = P(X < 90) - P(X < 82) \][/tex]
[tex]\[ P(82 < X < 90) \approx 0.9762 - 0.8621 \approx 0.1141 \][/tex]
To find the expected number of students, we multiply this probability by the total number of students:
[tex]\[ \text{Number of students} = 184 \times 0.1141 \approx 21.0 \][/tex]
Since we cannot have a fraction of a student, we would round to the nearest whole number.
Therefore, the number of students who can be expected to receive a score between 82 and 90 is approximately 21.
Suppose you send out your newest "tweet" to your 5000 Twitter followers. You suspect that the change in the number of followers that have seen your tweet is proportional to the ratio of the number of followers that have seen the tweet and the number of followers that have not seen the tweet. If 10 followers have seen the tweet 5. after 1 minute, write a differential equation that models the number of followers that have seen the tweet, including any initial condition. [Do not solve the differential equation.]
Answer: Suppose we send out our newest "tweet" to our 5000 Twitter followers.
If 10 followers have seen the tweet after 1 minute, then the differential equation can be written as ;
Let us first assume that at time "t" , "n" followers have seen this tweet.
So, no. of follower who have not seen this tweet are given as : 5000 - n
ratio = [tex]\frac{n}{5000 - n}[/tex]
∴ we get ,
[tex]\frac{\delta x}{\delta t}[/tex] ∝ [tex]\frac{n}{5000 - n}[/tex]
[tex]\frac{\delta x}{\delta t}[/tex] = k×[tex]\frac{n}{5000 - n}[/tex] ------ (1)
where k is the proportionality constant
At t = 0 , one follower has seen the tweet.
So n(0) = 0 ------ (2)
So n(1) = 10 ------ (3)
∴ equation (1), (2) and (3) together model the no. of followers that have seen the tweet.
Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel to the line x = −1 + 4t, y = 6 − 4t, z = 3 + 6t
Answer:
The vector equation of the line is [tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex] and parametric equations for the line are [tex]x=4t[/tex], [tex]y=11-4t[/tex], [tex]z=-8+6t[/tex].
Step-by-step explanation:
It is given that the line passes through the point (0,11,-8) and parallel to the line
[tex]x=-1+4t[/tex]
[tex]y=6-4t[/tex]
[tex]z=3+6t[/tex]
The parametric equation are defined as
[tex]x=x_1+at,y=y_1+bt,z=z_1+ct[/tex]
Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.
From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.
The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are
[tex]x=4t[/tex]
[tex]y=11-4t[/tex]
[tex]z=-8+6t[/tex]
Vector equation of a line is
[tex]\overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}[/tex]
where, [tex]\overrightarrow {r_0}[/tex] is a position vector and [tex]\overrightarrow {v}[/tex] is cosine of parallel vector.
[tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex]
Therefore the vector equation of the line is [tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex] and parametric equations for the line are [tex]x=4t[/tex], [tex]y=11-4t[/tex], [tex]z=-8+6t[/tex].
The vector equation and parametric equations for the line through the point (0, 11, −8) and parallel to the given line are established by using the point as the initial point and the direction ratios from the parallel line.
Explanation:The task is to find a vector equation and parametric equations for a line passing through the point (0, 11, −8) and parallel to given line equations x = −1 + 4t, y = 6 − 4t, z = 3 + 6t. To find these equations, we utilize the given point as the initial point and extract the direction ratios from the coefficients of t in the given parametric equations of the parallel line, which are (4, −4, 6).
The vector equation of the line can be written as ℓ = ℓ0 + t·d, where ℓ0 is the position vector of the initial point, and d is the direction vector. Since the given point is (0, 11, −8) and the parallel line's direction vector is (4, −4, 6), the vector equation is ℓ = (0, 11, −8) + t(4, −4, 6).
The parametric equations are derived directly from the vector equation. These are:
These equations represent the trajectory of the line through space, governed by the parameter t.
Every day a student randomly chooses a sandwich for lunch from a pile of wrapped sandwiches. If there are six kinds of sandwiches, how many different ways are there for the student to choose sandwiches for the seven days of a week if the order in which the sandwiches are chosen matters?
If the order matters and there are six kinds of sandwiches a student can choose for each of the seven days, there are 6^7, or 279,936, combinations possible. The calculation is based on the permutation rule of counting principle.
Explanation:The student can select sandwiches in different ways following the rules of counting principle or more specifically permutation. Since the student can choose from six sandwiches each day, and this choice is made seven times (for seven days), the choice each day is an independent event because the choice of sandwich one day does not affect what he or she can choose the subsequent day.
The total number of ways the student can select sandwiches is given by raising the total number of choices (6) by the total number of days (7). So, there are 6^7 or 279,936 possible combinations of sandwiches for the week.
Learn more about Counting Principle, Permutation here:https://brainly.com/question/14632628
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samantha wants to sort her greeting cards into boxes of 24 cards each. she has 312 greeting cards. How many boxes will she need?
312 greeting cards in 24-card boxes require 13 boxes.
312/24=13
Samantha will need 13 boxes to sort her 312 greeting cards.
To determine the number of boxes needed, we divide the total number of greeting cards by the number of cards that can fit into one box. Samantha has 312 greeting cards, and each box can hold 24 cards.
First, we perform the division:
[tex]\[ \frac{312}{24} = 13 \][/tex]
Since we are dealing with whole boxes, we do not need to consider any remainder because Samantha cannot use a fraction of a box. Therefore, Samantha will need exactly 13 boxes to accommodate all 312 greeting cards.