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:  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:

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.  

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.

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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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:

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:

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: 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:

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:

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%.

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.

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: 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.

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.

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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