The probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].
Further explanation:
Given:
The probability [tex]p[/tex] that an appliance is currently repaired is [tex]0.5[/tex].
The number of complex [tex]n[/tex] are [tex]100[/tex].
Calculation:
The [tex]\bar{X}[/tex] follow the Binomial distribution can be expressed as,
[tex]\bar{X}\sim \text{Binomial}(n,p)[/tex]
Use the normal approximation for [tex]\bar{X}[/tex] as
[tex]\bar{X}\sim \text{Normal}(np,np(1-p))[/tex]
The mean [tex]\mu[/tex] is [tex]\fbox{np}[/tex]
The standard deviation [tex]\sigma\text{ } \text{is} \text{ } \fbox{\begin{minispace}\\ \sqrt{np(1-p)}\end{minispace}}[/tex]
The value of [tex]\mu[/tex] can be calculated as,
[tex]\mu=np\\ \mu= 100 \times0.5\\ \mu=50[/tex]
The value of [tex]\sigma[/tex] can be calculated as,
[tex]\sigma=\sqrt{100\times0.5\times(1-0.5)} \\\sigma=\sqrt{100\times0.5\times0.5}\\\sigma=\sqrt{25}\\\sigma={5}[/tex]
By Normal approximation \bar{X} also follow Normal distribution as,
[tex]\bar{X}\sim \text{Normal}(\mu,\sigma^{2} )[/tex]
Substitute [tex]50[/tex] for [tex]\mu[/tex] and [tex]25[/tex] for [tex]\sigma^{2}[/tex]
[tex]\bar{X}\sim\text {Normal}(50,25)[/tex]
The probability that at least [tex]60[/tex] are currently being repaired can be calculated as,
[tex]\text{Probability}=P\left(\bar{X}>60)\right}\\\text{Probability}=P\left(\frac{{\bar{X}-\mu}}{\sigma}>\frac{{(60-0.5)-50}}{\sqrt{25} }\right)\\\text{Probability}=P\left(Z}>\frac{{59.5-50}}{5}}\right)\\\text{Probability}=P\left(Z}>\frac{9.5}{5}}\right)\\\text{Probability}=P(Z}>1.9})[/tex]
The Normal distribution is symmetric.
Therefore, the probability of greater than [tex]1.9[/tex] is equal to the probability of less than [tex]1.9 [/tex].
[tex]P(Z>1.9})=1-P(Z<1.9)\\P(Z>1.9})=1-0.9722\\P(Z>1.9})=0.0278[/tex]
Hence, the probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].
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Answer Details:
Grade: College Statistics
Subject: Mathematics
Chapter: Probability and Statistics
Keywords:
Probability, Statistics, Appliance, Apartment complex, Binomial distribution, Normal distribution, Normal approximation, Central Limit Theorem, Z-table, Mean, Standard deviation, Symmetric.
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.
ONE HALF OF AN ANNUALLY SALARY OF $35,700. IS APPROXIMATELY
Answer: Approximately $18,000
Step-by-step explanation:
Since you are approximating, the 7 indicates the need to round the original number up to 36,000. Then divide by 2 since you want half, thus getting $18,000.
Find the intersection and union of the sets A and B given below:
A = { a, m, w, u, 7, y } , B = { w, y, g, 7}
Answer:
Intersection: {w, 7, y}
Union: {a, m, w, u, 7, y, g}
Step-by-step explanation:
The intersection of two or more sets are the elements they have in common, that means the elements that are in all the sets at the same time.
For example, "a" is in A but not in B, that's why is not in the intersection. On the other hand, "w" is in both sets, so it's in the intersection.
The union are all the elements that are in one set or the other, but we don't add the element twice if it's in both sets.
For example, "a" is in A so we add it to the union. "w" is in A so we add it to the union but it's also in B, we don't add it again because it is already in the union.
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.
Two friends visited Washington D.C. for the weekend. Person 1 rode the subway one stop, 5 times at the peak fare price and four times at the off peak fare price for a total cost of $17.40. Person 2 rode with Person 1, 2 times at the peak fare price and three times at the off peak fair price for a total cost of $9.20. How much was the peak fare price?
Answer:
The peak fare price is $2,2
Step-by-step explanation:
Considering
peak fare price= xoff peak fare price = yPerson 1
5x + 4y= 17,40
Person 2
2x + 3y= 9,20
Isolating the x
2x= 9,20-3y
x=(9,20-3y)/2=4,6 -1,5y
x=4,6 -1,5y
Replacing in the other equation
5 (4,6 -1,5y) + 4y=17,40
23-7,5y+4y=17,40
23-17,40=7,5y-4y
5,6=3,5y
y=1,60 (this is the off peak fare)
x= 4,60-1,5y
x= 4,60-1,5*1,60
x=2,2 (this is the peak fare)
Final answer:
To determine the peak fare price, we formed a system of linear equations based on the given data. By using the elimination method, we solved the system to find that the peak fare price is $2.20.
Explanation:
We have a system of linear equations representing the total costs of peak and off-peak fares for two friends using public transportation in Washington D.C. Let's denote the peak fare as P and the off-peak fare as Op. According to the problem, we are given the following two equations based on the trips taken and the total cost paid by each person:
5P + 4Op = $17.40 (Person 1)
2P + 3Op = $9.20 (Person 2)
To find the peak fare price, we need to solve this system of equations. We can use a variety of methods, including substitution or the elimination method.
Step 1 - Use the Elimination Method:
Multiply the second equation by 2 to align the Op terms:
4P + 6Op = $18.40 (after multiplying by 2)
Step 2 - Subtract the resulting equation from the first person's total cost:
5P + 4Op - (4P + 6Op) = $17.40 - $18.40
P - 2Op = -$1.00
Step 3 - Solve for P:
Isolate P by adding 2Op to both sides:
P = 2Op - $1.00
Now, substitute P into the second equation from Person 2's total cost:
2(2Op - $1.00) + 3Op = $9.20
4Op - $2.00 + 3Op = $9.20
7Op - $2.00 = $9.20
Add $2.00 to both sides to get 7Op alone:
7Op = $11.20
Divide by 7 to find the off-peak fare:
Op = $11.20 / 7
Op = $1.60
Use the value of Op to find P:
P = 2($1.60) - $1.0
P = $3.20 - $1.00
P = $2.20
The peak fare price is $2.20.
g A standard deck of playing cards is shuffled and three people each choose a card. Find the probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement, and if the cards are chosen without replacement. (a) The cards are chosen with replacement. (b) The cards are chosen without replacement.
Answer:
The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 .
The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058
Step-by-step explanation:
Total no. of cards = 52
Spade cards = 13 (Black)
Club cards = 13 (Black)
Heart cards = 13 (Red)
Diamond cards = 13(Red)
Total red cards = 26
With replacement case:
Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]
Now card is replaced
Probability of getting spade on second draw = [tex]\frac{13}{52}[/tex]
Now card is replaced
Probability of getting red card on third draw = [tex]\frac{26}{52}[/tex]
So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement = [tex]\frac{13}{52} \times \frac{13}{52} \times \frac{26}{52}[/tex]
=[tex]\frac{1}{32}[/tex]
= [tex]0.03125[/tex]
Without replacement case:
Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]
Remaining cards = 51
Remaining spade cards = 12
Probability of getting spade on second draw = [tex]\frac{12}{51}[/tex]
Remaining cards = 50
Probability of getting red card on third draw = [tex]\frac{26}{50}[/tex]
So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement = [tex]\frac{13}{52} \times \frac{12}{51} \times \frac{26}{50}[/tex]
= [tex]\frac{13}{425}[/tex]
= [tex]0.03058[/tex]
Hence The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 . The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058
When the cards are chosen with replacement, the probability is 169/2704. When the cards are chosen without replacement, the probability is 39/884.
Explanation:When the cards are chosen with replacement, the probability that the first two cards chosen are spades and the third card is red can be found by multiplying the probabilities of each event happening. There are 13 spades in a deck of 52 cards, so the probability of choosing a spade on the first draw is 13/52. Since the cards are chosen with replacement, the probability of choosing a spade on the second draw is also 13/52. The probability of choosing a red card on the third draw is 26/52, since there are 26 red cards in the deck. Therefore, the overall probability is (13/52) * (13/52) * (26/52) = 169/2704.
When the cards are chosen without replacement, the probability changes. The probability of choosing a spade on the first draw is 13/52. Since the first card is not replaced, there are now 51 cards left and 12 spades remaining. So the probability of choosing a spade on the second draw is 12/51. Finally, there are 25 red cards left out of 50 total cards, so the probability of choosing a red card on the third draw is 25/50. Therefore, the overall probability is (13/52) * (12/51) * (25/50) = 39/884.
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Although not so common now, consider a phone plan that does not have unlimited texting. Your base plan costs you $30 a month and each text sent or received is only 12 cents. Write an equation to describe the total monthly cost of your bill with x text messages and y cost.
In one month 217,000 messages were sent between two brothers in Philadelphia. What was their approximate bill? (This is a true story!)
Answer:
The bill will be $26070.
Step-by-step explanation:
Your base plan costs you $30 a month
And each text sent or received is only 12 cents or $0.12.
Let the messages sent be = x
So, cost for messages become = [tex]0.12x[/tex]
Let total bill cost be = y
Now, total bill will comprise of monthly plan charges plus message charges.
So, equation forms:
[tex]y=30+0.12x[/tex]
Now, given is that In one month 217,000 messages were sent between two brothers in Philadelphia.
So, their approximate bill will be :
[tex]y=30+0.12(217000)[/tex]
=>[tex]y=30+26040[/tex]
=> y = 26070
Therefore, bill will be $26070.
The answer explains how to write an equation for the total monthly cost of a phone plan based on text message usage and calculates the bill for 217,000 messages sent between two brothers in Philadelphia.
Explanation:To write an equation for the total monthly cost of a phone plan with a base cost and a cost per text message:
Let x be the number of text messages sent or received.Let y be the total monthly cost.Using the given information, the equation is: y = $30 + 0.12xFor the specific case of 217,000 messages sent between two brothers:
Substitute x = 217,000 into the equation: y = $30 + 0.12(217,000)Calculate to find their approximate bill.A multiple choice test contains 10 questions, each with 3 possible answers (of which only one is correct). If a student answers each question by rolling a die and choosing the first answer if the die shows 1 or 2, the second answer if the die shows 3 or 4, or the third answer if the die shows 5 or 6, what is the probability that the student will get exactly 6 correct answers? more than 6 correct answers?
Answer:
1. The probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].
2. The probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].
Step-by-step explanation:
From the given information it is clear that
The total number of equations (n) = 10
The probability of selecting the correct answer (p)= [tex]\frac{1}{3}[/tex]
The probability of selecting the incorrect answer (q)= [tex]1-p=1-\frac{1}{3}=\frac{2}{3}[/tex]
According to the binomial distribution, the probability of selecting r items from n items is
[tex]P=^nC_rp^rq^{n-r}[/tex]
where, p is probability of success and q is the probability of failure.
The probability that the student will get exactly 6 correct answers is
[tex]P(r=6)=^{10}C_6(\frac{1}{3})^6(\frac{2}{3})^{10-6}[/tex]
[tex]P(r=6)=210(\frac{1}{3})^6(\frac{2}{3})^{4}=\frac{1120}{19683}[/tex]
Therefore the probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].
The probability that the student will get more than 6 correct answers is
[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{10-7}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{10-8}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{10-9}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{10-10}[/tex]
[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{3}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{2}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{1}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{0}[/tex]
[tex]P(r>6)=120\times \frac{8}{59049}+45\times \frac{4}{59049}+10\times \frac{2}{59049}+1\times \frac{1}{59049}=\frac{43}{2187}[/tex]
Therefore the probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].
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.
Assume the random variable x is normally distributed with mean muμequals=8787 and standard deviation sigmaσequals=55. Find the indicated probability. P(x less than<7979)
Answer: 0.4404
Step-by-step explanation:
Let the random variable x is normally distributed .
Given : Mean : [tex]\mu=\ 87[/tex]
Standard deviation : [tex]\sigma= 55[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 79
[tex]z=\dfrac{79-87}{55}\approx-0.15[/tex]
The p-value = [tex]P(x<79)=P(z<-0.15)[/tex]
[tex]=0.4403823\approx0.4404[/tex]
Hence, the required probability : [tex]P(x<79)=0.4404[/tex]
solve l 3x-7 l = 2
and solve l 2x-5 l< or equal to 8
the lines (l) signifies absolute values therefore l 3x-7 l is the absolute value of that and same thing for 2x-5
Answer:
1) Solutions are x = 3 and x = 5/3
2) Solution are x ≤ 13/2 and x ≤ -3/2
Step-by-step explanation:
1) Given absolute inequality,
|3x-7| = 2
⇒ 3x - 7 = ± 2
⇒ 3x = 7 ± 2
[tex]\implies x=\frac{7\pm 2}{3}[/tex]
[tex]x=\frac{7+2}{3}\text{ or }x=\frac{7-2}{3}[/tex]
[tex]\implies x = 3\text{ or }x=\frac{5}{3}[/tex]
2) l 2x-5 l ≤ 8
⇒ 2x-5 ≤ ±8
⇒ 2x ≤ 5 ± 8
[tex]\implies x\leq \frac{5\pm 8}{2}[/tex]
[tex]x\leq \frac{5+8}{2}\text{ or }x\leq \frac{5-8}{2}[/tex]
[tex]\implies x \leq \frac{13}{2}\text{ or }x\leq-\frac{3}{2}[/tex]
Answer:
Answer:
1) Solutions are x = 3 and x = 5/3
2) Solution are x ≤ 13/2 and x ≤ -3/2
Step-by-step explanation:
1) Given absolute inequality,
|3x-7| = 2
⇒ 3x - 7 = ± 2
⇒ 3x = 7 ± 2
2) l 2x-5 l ≤ 8
⇒ 2x-5 ≤ ±8
⇒ 2x ≤ 5 ± 8
Using either the critical value rule or the p-value rule, if a one-sided null hypothesis is rejected at a given significance level, then the corresponding two-sided null hypothesis (i.e., the same sample size, the same standard deviation, and the same mean) will ______________ be rejected at the same significance level.
Rejecting a one-sided null hypothesis at a given significance level does not necessarily mean that the corresponding two-sided null hypothesis will also be rejected at the same significance level because one-sided tests and two-sided tests have different rejection regions.
Explanation:Using either the critical value rule or the p-value rule, if a one-sided null hypothesis is rejected at a given significance level, then the corresponding two-sided null hypothesis (i.e., the same sample size, the same standard deviation, and the same mean) will not necessarily be rejected at the same significance level.
One-sided tests and two-sided tests have different rejection regions. For a one-sided test, the rejection region is all on one side of the sampling distribution, while for a two-sided test, the rejection regions are on both sides. If the test statistic falls in the rejection region for a one-sided test, it does not necessarily mean it will fall in the rejection region for the two-sided test, even at the same significance level.
Thus, even if you reject a one-sided null hypothesis at a given significance level, you cannot automatically reject the two-sided null hypothesis at the same level. You need to perform the appropriate statistical test.
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A survey of 100 families showed that 35 had a dog: 28 had a cat: 10 had 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. How many had a parakeet only? A. 20 B. 15 C. 5 D. 10
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.
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.
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%.
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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The resistance,R(in ohms), Of a wire varies directly with the length, L(in cm) of the wire, and inversely with the cross-sectional area, A (in cm2). A 500 cm piece of wire with a radius of 0.2 cm has a resistance of 0.025 ohm. Find an equation that relates these variables.
Answer:
R = 2π×10⁻⁶ L / A
Step-by-step explanation:
Resistance varies directly with length and inversely with area, so:
R = kL/A
A round wire has cross-sectional area of:
A = πr²
Substituting:
R = kL/(πr²)
Given that R = 0.025 Ω when L = 500 cm and r = 0.2 cm:
0.025 = k (500) / (π (0.2²))
k = 2π×10⁻⁶
Therefore:
R = 2π×10⁻⁶ L / A
Answer:
0.00002 in the first blank. in the second blank its A
Step-by-step explanation:
Question 5: What is the radius of the circle with general form x2+y2-4y=21?
Question 5 options:
√21
4.58
5
25
Answer:
r=5 units
Step-by-step explanation:
we have
[tex]x^{2}+y^{2} -4y=21[/tex]
Convert the equation of the circle in standard form
Complete the square twice. Remember to balance the equation by adding the same constants to each side
[tex]x^{2}+(y^{2} -4y+4)=21+4[/tex]
[tex]x^{2}+(y^{2} -4y+4)=25[/tex]
Rewrite as perfect squares
[tex]x^{2}+(y-2)^{2}=25[/tex]
center (0,2)
radius r=5 units