If 40% of a school consists of boys, 40 % of the students have blonde hair, and three times as many girls as boys have blond hair, what percentage of the school are blonde haired boys?

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.

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.

Proof:

Given any functions [tex]f_{1}(x),f_{2}(x),f_{3}(x)[/tex] they are linearly dependent if we can find values of [tex]c_{1},c_{2},c_{3}[/tex] such that

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0[/tex]

Using the given functions in the above equation we get

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0\\\\c_{1}x^{2}+c_{2}(1-x^{2})+c_{3}(2+x^{2})=0\\\\\Rightarrow (c_{1}-c_{2}+c_{3})x^{2}+c_{1}+c_{2}+2c_{3}=0[/tex]

This will be satisfied if and only if

[tex]c_1-c_2+c_3=0,c_1+c_2+2c_3=0[/tex]

Solving the equations we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0[/tex]

Since we have 3 variables and 2 equations thus we will get many solutions  

one being if we put [tex]c_3=1[/tex] we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0\\\\c_1=\frac{-3}{2}\\\\c_2=\frac{-1}{2}[/tex]

Thus we have [tex]c_1=\frac{-3}{2},c_2=\frac{-1}{2},c_3=1[/tex] as one solution. Hence the given functions are linearly dependent.

Answer:  The required value of the given expression is (h +1).

Step-by-step explanation:  We are given a function f(x) as follows :

[tex]f(x)=x^2-3x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to evaluate the following :

[tex]E=\dfrac{f(2+h)-f(2)}{h}.[/tex]

Substituting x = 2 + h in equation (i), we get

[tex]f(2+h)\\\\=(2+h)^2-3(2+h)\\\\=(4+4h+h^2)-6-3h\\\\=4+4h+h^2-6-3h\\\\=h^2+h-2[/tex]

and substituting x = 2 in equation (i), we get

[tex]f(2)=2^2-3\times2=4-6=-2.[/tex]

Therefore, the value of expression E can be evaluated as follows :

[tex]E\\\\\\=\dfrac{f(2+h)-f(2)}{h}\\\\\\=\dfrac{(h^2+h-2)-(-2)}{h}\\\\\\=\dfrac{h^2+h-2+2}{h}\\\\\\=\dfrac{h^2+h}{h}\\\\\\=\dfrac{h(h+1)}{h}\\\\=h+1.[/tex]

Thus, the required value of the given expression is (h +1).

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.

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.

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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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.

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.

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Answer:

4/7

Step-by-step explanation:

To find the probability, we must first find the total number of golf balls. So 99+88+44 = 231 is our total. Now we must find the number of golf balls that are not a type A ball. This is just the total number of type B balls and type C balls. So 88+44 = 132. So the probability is 132/231. This simplifies to 12/21 or 4/7.

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.

To write an equation for the total monthly cost of a phone plan with a base cost and a cost per text message:

For the specific case of 217,000 messages sent between two brothers:

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.

Answer:

Step-by-step explanation:

37

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.

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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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.

Answer:

The peak fare price is $2,2

Step-by-step explanation:

Considering

Person 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.

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].

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:

Answer:

[tex]P =0.3998[/tex]

Step-by-step explanation:

Let [tex]{\displaystyle {\overline{x}}}[/tex] be the average of the sample, and the population mean will be [tex]\mu[/tex]

We know that:

[tex]\mu = 4095[/tex] gr

Let [tex]\sigma[/tex] be the standard deviation and n the sample size, then we know that the standard error of the sample is:

[tex]E=\frac{\sigma}{\sqrt{n}}[/tex]

Where

[tex]\sigma=569[/tex]

[tex]n=130[/tex]

In this case we are looking for:

[tex]P(|{\displaystyle{\overline{x}}}- \mu|>42)[/tex]

This is:

[tex]{\displaystyle{\overline{x}}}- \mu>42[/tex] or [tex]{\displaystyle{\overline{x}}}- \mu<-42[/tex]

[tex]P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu<-42)[/tex]

Now we get the z score

[tex]Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z<-\frac{42}{\frac{569}{\sqrt{130}}})[/tex]

[tex]P=P(z>0.8416) + P(z<-0.8416)[/tex]

Looking at the tables for the standard nominal distribution we get

[tex]P =0.1999+0.1999[/tex]

[tex]P =0.3998[/tex]

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:

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.

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%.

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.

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.

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

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

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

Answer: 0.0081

Step-by-step explanation:

Let X be the number of rafts.

Given : The mean number of rafts floating : [tex]\lambda=0.4[/tex] rafts per day .

Then , for 7 days the number of rafts = [tex]\lambda_1=\lambda\times 7=0.4\times7=2.8[/tex] rafts per day .

The formula to calculate the Poisson distribution is given by :_

[tex]P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}[/tex]

Now, the  probability that they will have to wait more than a week is given by :-

[tex]P(X>7)=1-P(X\leq7)\\\\=1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(5)+P(6)+P(7))\\\\=1-(\dfrac{e^{-2.8}2.8^{0}}{0!}+\dfrac{e^{-2.8}2.8^{1}}{1!}+\dfrac{e^{-2.8}2.8^{2}}{2!}+\dfrac{e^{-2.8}2.8^{3}}{3!}+\dfrac{e^{-2.8}2.8^{4}}{4!}+\dfrac{e^{-2.8}2.8^{5}}{5!}+\dfrac{e^{-2.8}2.8^{6}}{6!}+\dfrac{e^{-2.8}2.8^{7}}{7!})\\\\=1-0.991869258012=0.008130741988\approx0.0081[/tex]

Hence, the required probability : 0.0081

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]

Answer:

450 students

Step-by-step explanation:

Let s = number of students

1/2 are Americans  1/2s are Americans

s -1/2s = number of students left

1/2 s = number of students left

1/3 of the students left are Europeans

1/3 (1/2s) = 1/6s  are Europeans

The rest are Australians = 150

Students = Americans + Europeans + Australians

   s          =   1/2s          +    1/6s    + 150

Combine like terms

s    = 1/2*3/3 *s + 1/6s +150

s = 3/6s + 1/6s +150

s = 4/6s + 150

Subtract 4/6s from each side

s -4/6s = 150

6/6s - 4/6s = 150

2/6s = 150

1/3s = 150

Multiply each side by 3

3* 1/3s = 150*3

s = 450

There are 450 students

Answer:

There are 450 students.

Step-by-step explanation:

Let's call

s = total number of students

am = American students

eu = European students

au = Australian students

Half of the students in an International School are Americans. Then,

1/2 s = am [1]

One third of the remaining students are Europeans.

The remaining students are s - 1/2 s = 1/2 s

1/3 × (1/2 s) = 1/6 s = eu  [2]

There are 150 Australians.

au = 150 [3]

The total number of students is the sum of Americans, Europeans and Australians.

am + eu + au = s

Replacing [1], [2] and [3].

1/2 s + 1/6 s + 150 = s

2/3 s + 150 = s

1/3 s = 150

s = 450

Answer:

R indicates Random Assignment

Step-by-step explanation:

In designing an experiment, R is the symbol used for assigning the subjects randomly to different groups through randomization. It is called random assignment and It is the option (d) random assignment. It is achieved by using chance procedure ( random number generator, by flipping coin, throwing ). This ensures that all the subjects have equal chance of placing in all the groups.

Option (a) Symbol O is used for Observations. (So option (a) is wrong answer)

Option (b) Symbols x , y are used as Observational variables (So option (b) is wrong answer)

Option (c) Comparison groups is wrong answer as R is not used for that.

The symbol R in experimental design represents random assignment, a technique used to ensure that groups in an experiment are initially equivalent and thereby allow researchers to make causal inferences.

In experimental design, the symbol R indicates random assignment. Random assignment is a process used to create initial equivalence between the groups in an experiment, which is crucial for allowing researchers to draw causal conclusions. In this process, participants are assigned to different groups or conditions on a random basis, such as by drawing numbers or using a random number generator.

Random assignment minimizes the effects of lurking variables, ensuring that before the experimental manipulation occurs, the groups can be considered equivalent on various potential confounding factors. When individuals are selected into groups denoted by R, we can generally classify these designs as experimental, as opposed to nonexperimental designs where nonrandom assignment (NR) is indicated.

For example, in a study where the size of tableware is manipulated to see its effect on food consumption, the psychologist would use random assignment to ensure that both treatment groups are comparable prior to the intervention, which is the manipulation of the independent variable.

Answer:

Step-by-step explanation:

X= quarts of sauce # 1 ( 17% of sugar)

Y= quarts of sauce # 2( 30 % of sugar )

General Balance

[tex]X+Y=2.6[/tex]      EQ (1)

[tex]Y=2.6-X\\[/tex]

SUGAR BALANCE

[tex]0.17X+0.3Y=2.6*0.24\\[/tex]   EQ (2)

replace Y from eq(1) into eq(2)

[tex]0.17X+0.3(2.6-X)=0.624\\0.17X+0.78-0.3X=0.624\\-0.13X=0.624-0.780\\X=\frac{-0.156}{-0.13} \\X= 1.2\\Y=2.6-1.2=1.4[/tex]

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.