Answer:
A) The brand of cars driven by Harvard instructors
C) The types of plants growing in yellowstone
Step-by-step explanation:
Categorical and numerical variables:
Numerical data can be expressed with the help of numerical.Categorical data cannot be expressed by numerical.They do no have any particular value.They are the non-parametric data.Categorical data are also known as qualitative data.A) The brand of cars driven by Harvard instructors
Since the brands cannot be expressed in numerical, it is a categorical variable.
B) The age of randomly selected biology students.
The age of student are expressed in whole numbers or decimals. Thus, it is a numerical data.
C) The types of plants growing in yellowstone
Again, the type of plants cannot be expressed with the help of numerical. Thus, it is a categorical variable.
Variables A ('The brand of cars driven by Harvard instructors') and C ('The types of plants growing in Yellowstone') are categorical. Variable B ('The age of randomly selected biology students') is numerical.
Explanation:In statistical terms, variables are characteristics or properties that can take different values or categories. They can be either categorical (qualitative) or numerical (quantitative).
For the variables provided:
A. The brand of cars driven by Harvard instructors - This is a categorical variable, as car brands are qualitative characteristics that can be placed in different categories.B. The age of randomly selected biology students - This is a numerical variable, as ages are measured in numbers and can be quantitatively analyzed.C. The types of plants growing in Yellowstone - This is another categorical variable, as plant types are qualitative characteristics that can be categorized.Learn more about Categorical Variables here:https://brainly.com/question/34199292
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Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purchasers who select the Swedish model.
(a) What is the pmf of X?
(b) Compute P(X > 10).
(c) Compute P(6 ≤ X ≤ 10).
(d) Compute μ and σ2.
Answer:
a)
[tex] P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k} [/tex]
For any integer k between 0 and 15, and 0 for other values of k.
b)
[tex]P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865[/tex]
c) P(6 ≤ X ≤ 10) = 0.2737
d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125
Step-by-step explanation:
X is a binomial random variable with parameters n = 15, p = 0.75. Therefore
a)
[tex] P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k} [/tex]
For any integer k between 0 and 15, and 0 for other values of k.
b)
P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)
[tex]P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252[/tex]
[tex]P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252[/tex]
[tex]P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559[/tex]
[tex]P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668[/tex]
[tex]P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134[/tex]
Thus,
[tex]P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865[/tex]
c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)
[tex]P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034[/tex]
[tex]P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131[/tex]
[tex]P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393[/tex]
[tex]P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918[/tex]
[tex]P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652[/tex]
Thereofre,
[tex]P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737[/tex]
d) μ = n*p = 15*0.75 = 11.25
σ² = np(1-p) = 11.25*0.25 = 2.8125
Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the probability that at most 10 of the next 150 items produced are unacceptable.
Answer:
The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.
Step-by-step explanation:
Let X = number of items with unacceptable quality.
The probability of an item being unacceptable is, P (X) = p = 0.05.
The sample of items selected is of size, n = 150.
The random variable X follows a Binomial distribution with parameters n = 150 and p = 0.05.
According to the Central limit theorem, if a sample of large size (n > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.
The mean of this sampling distribution is: [tex]\mu_{\hat p}= p=0.05[/tex]
The standard deviation of this sampling distribution is: [tex]\sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178[/tex]
If 10 of the 150 items produced are unacceptable then the probability of this event is:
[tex]\hat p=\frac{10}{150}=0.067[/tex]
Compute the value of [tex]P(\hat p\leq 0.067)[/tex] as follows:
[tex]P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315[/tex]
*Use a z-table for the probability.
Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.
Final answer:
Using the normal approximation to the binomial distribution, the probability that at most 10 of the next 150 items produced are unacceptable is approximately 86.43%.
Explanation:
Approximating the Probability of Defective Items:
To approximate the probability that at most 10 of the next 150 items produced are unacceptable when each item is of acceptable quality independently with probability 0.95, we use the binomial probability formula or normal approximation. However, since the number of trials is large (n = 150), we can use the normal approximation to the binomial distribution to simplify the calculation.
First, we find the mean (μ) and standard deviation (σ) of the binomial distribution:
Mean μ = n * p = 150 * 0.05 = 7.5Standard Deviation σ = sqrt(n * p * (1 - p)) = sqrt(150 * 0.05 * 0.95) ≈ 2.72Next, we convert the binomial problem to a normal distribution problem by finding the z-score for 10.5 (since we are looking for "at most" 10, we use 10 + 0.5 for continuity correction).
The z-score is calculated as follows:
Z = (x - μ) / σ = (10.5 - 7.5) / 2.72 ≈ 1.10Finally, we look up the z-score in a standard normal distribution table, or use a calculator to find the cumulative probability for Z ≤ 1.10, which is approximately 0.8643. Therefore, the probability that at most 10 of the next 150 items are unacceptable is roughly 86.43%.
Let X represent the time it takes from when someone enters the line for a roller coaster until they exit on the other side. Consider the probability model defined by the cumulative distribution function given below.
0 x < 3
F(x) = (x-3)/1.13 3 < x < 4.13
1 x > 4.13
A. What is E(X)? Give your answer to three decimal places.
B. What is the value c such that P(X < c) = 0.75? Give your answer to four decimal places.
C. What is the probability that X falls within 0.28 minutes of its mean? Give your answer to four decimal places.
Answer:
a) E(x)=3.565
b) c=3.8475 --> P(X < 3.8475) = 0.75
c) The probability that X falls above or below 0.28 min from the mean is P=0.4954.
Step-by-step explanation:
We have the cumulative distribution function as information.
a) To calculate the expected value, we can calculate the value of x in which F(x) equals 0.5. This happens for x=3.565.
[tex]F(x)=\frac{x-3}{1.13} =0.5\\\\x-3=0.5*1.13=0.565\\\\x=0.565+3=3.565[/tex]
b) What is the value c such that P(X < c) = 0.75?
In this case, we have to calculate x to have F(x)=0.75
[tex]F(x)=\frac{x-3}{1.13} =0.75\\\\x-3=0.75*1.13=0.8475\\\\x=0.8475+3=3.8475[/tex]
This happens for x=3.8475.
c) We have to calculate the probability that X falls above or below 0.28 min from the mean (x=3.565).
This is the probability that the time is between 3.285 and 3.845
[tex]x_1=3.565-0.28=3.285\\\\x_2=3.565+0.28=3.845[/tex]
We can calculate this as:
[tex]P(3.285<x<3.845)=F(3.845)-F(3.285)\\\\F(3.845)=\frac{3.845-3}{1.13}=\frac{0.845}{1.13}= 0.7478\\\\F(3.285)=\frac{3.285-3}{1.13}=\frac{0.285}{1.13}= 0.2522\\\\\\P(3.285<x<3.845)=F(3.845)-F(3.285)=0.7478-0.2522=0.4956\\\\[/tex]
The probability that X falls above or below 0.28 min from the mean is P=0.4954.
A woman who has recovered from a serious illness begins a diet regimen designed to get her back to a healthy weight. She currently weighs 103 pounds. She hopes each week to multiply her weight by 1.08 each week. (a) Find a formula for an exponential function that gives the woman's weight w, in pounds, after t weeks on the regimen. (b) How long will it be before she reaches her normal weight of 135 pounds?
Answer:
a.) w = 103 * 1.08^t
b.) 3.5weeks
Step-by-step explanation:
If Her current weight is 103 pounds and she hopes to multiply her her weight each week by 1.08, then
her weight after 1 week = 103 * 1.08 = 103 * 1.08¹
Her weight after 2 weeks = [weight of week 1] * 1.08 = [103* 1.08] * 1.08 = 103 * 1.08²
Weight after 3 weeks= [weight of week 2] * 1.08 = [103 * 1.08 * 1.08] * 1.08 = 103 * 1.08³
Hence weight (W) after t weeks = 103 * 1.08^t
b.) If W = 135, Then
103 * 1.08^t = 135
1.08^t = 135/103
1.08^t = 1.31
Taking log of both sides,
log 1.08^t = log 1.31
t log 1.08 = log 1.32
t = log 1.32/log 1.08
t = 3.5 weeks.
Hence, it will take her 3½ weeks to get to 135pounds weight.
A sample of 100 wood and 100 graphite tennis rackets are taken from the warehouse. If 88 wood and 1111 graphite are defective and one racket is randomly selected from the sample, find the probability that the racket is wood or defective.
The question is wrong since it is not possible to have 111 defective graphite rackets when the total number of graphite racket is 100.
Question:
A sample of 100 wood and 100 graphite tennis rackets are taken from the warehouse. Assuming that If 88 wood and 90 graphite are defective and one racket is randomly selected from the sample, find the probability that the racket is wood or defective.
Given Information:
Total wood = 100
Total graphite = 100
Defective wood = 88
Non-defective wood = 12
Defective graphite = 90
Non-defective graphite = 10
Required Information:
Probability of racket being selected is wood or defective = ?
Answer:
P(wood or defective) = 0.95
Step-by-step explanation:
The probability of selecting a wood racket is
P(wood) = number of wood rackets/total number of rackets
P(wood) = 100/200 = 1/2
The probability of selecting a defective racket is
P(defective) = number of defective rackets/total number of rackets
P(defective) = 88+90/200 = 178/200 = 89/100
There is double counting of wood so we have to subtract the probability of wood and defective
P(wood and defective) = 88/200 = 11/25
P(wood or defective) = P(wood) + P(defective) - P(wood and defective)
P(wood or defective) = 1/2 + 89/100 - 11/25
P(wood or defective) = 0.95
Alternatively:
P(defective) = number of defective rackets/total number of rackets
P(defective) = 88+90/200 = 178/200 = 89/100
P(wood and non-defective) = 12/200 = 3/50
There is no double counting here so we dont have to subtract anything
P(wood or defective) = P(wood) + P(wood and non-defective)
P(wood or defective) = 89/100 + 3/50
P(wood or defective) = 0.95
How many license plates can be formed of 4 letters followed by 2 numbers?
Answer:
45,697,600 license plates can be formed of 4 letters followed by 2 numbers
Step-by-step explanation:
There are 4 letters in the plate. In the alphabet, there are 26 letters. So each of the four letters in the plate can have 26 outcomes.
There are 2 digits in the place. There are 10 possible digits.
How many possible plates?
26*26*26*26*10*10 = 45,697,600
45,697,600 license plates can be formed of 4 letters followed by 2 numbers
The graph shows the relationship between men's shoe sizes and their heights. What type of relationship is this?
Answer:
Linear Relationship
Step-by-step explanation:
The size is steadily progressing and so is the height.
Answer:
Step-by-step explanation:
linear relationship
If the pH level of the reservoir is ok, the results at each location will have varying results, with an average pH of 8.5 and a standard deviation of 0.22. If the pH level of the reservoir is ok, what is the probability that the sample average is LESS than 8.47
Final answer:
The probability of the sample average pH being less than 8.47, given a mean of 8.5 and a standard deviation of 0.22, is approximately 44.55%, as calculated using the Z-score and the standard normal distribution.
Explanation:
To determine the probability that the sample average pH is less than 8.47, when the average pH of the reservoir is 8.5 and the standard deviation is 0.22, we can use the concepts of the normal distribution and Z-scores in statistics. A Z-score is a measure of how many standard deviations an element is from the mean. First, we calculate the Z-score for pH 8.47 using the formula:
Z = (X - μ) / σ
Where X is the value of interest (8.47), μ is the mean pH (8.5), and σ is the standard deviation (0.22). Plugging in the numbers:
Z = (8.47 - 8.5) / 0.22 = -0.03 / 0.22 ≈ -0.1364
With a Z-score of approximately -0.1364, we can then consult the standard normal distribution table to find the probability corresponding to this Z-score. The probability associated with a Z-score of -0.1364 is about 0.4455, meaning there is a 44.55% chance that a sample average pH would be less than 8.47, assuming the pH levels are normally distributed.
A random sample of 11 statistics students produced data where x is the third exam score out of 80, and y is the final exam score out of 2 The score on the final exam will likely be lower by about 174 points. b. For each 4.83 point increase in the third exam score, we expect the final exam score t
Answer:
b.
About 44% of the variation in the final exam score can be explained by the students' scores on the 3rd exam. The remaining 56% is due to other factors or unexplained randomness.
Step-by-step explanation:
Hello!
X: Third exam score of a statistics student.
Y: Final exam score of a statistics student.
The estimated regression line is: y = -173.51 + 4.83x
Where
-173.51 is the estimation of the intercept and you can interpret it as the value of the estimated average final exam score when the students scored 0 points on their third exam.
4.83 is the estimation of the slope and you can interpret is as the modification on the estimated average score of the final exam every time the score on the third exam increases 1 point.
R²= represents the coefficient of determination.
Its interpretation is: 44% of the variability of the final exam scores of the statistics students are explained by the scores in the third exam, under the estimated model y = -173.51 + 4.83x.
I hope it helps!
-*-
A random sample of 11 statistics students produced data where x is the third exam score out of 80, and y is the final exam score out of 200. The corresponding regression line has the equation: y = -173.51 + 4.83x, and the value of r2 (the "coefficient of determination") is found to be 0.44. What is the proper interpretation of r2?
a.
Due to the number of points on the two exams, the third exam score will likely be 44% of the final exam score.
b.
About 44% of the variation in the final exam score can be explained by the students' scores on the 3rd exam. The remaining 56% is due to other factors or unexplained randomness.
c.
For each 1 point increase in the 3rd exam score, we expect the final exam score to increase by 0.44 points.
d.
For each 1 point increase in the 3rd exam score, we expect the final exam score to increase by 0.44 percent.
e.
44% of the students scored within 1 standard deviation of the mean on each of the two tests.
The r² value of 0.44 indicates that approximately 44% of the variation in final exam scores can be explained by scores on the third exam, with the remaining variation due to other factors. Option b is the correct interpretation.
The coefficient of determination, denoted as r², explains the proportion of the variance in the dependent variable (y) that is predictable from the independent variable (x).
X is statistic exam score.Y is statistics final exam score.Estimated regression line: y = -173.51 + 4.83xIn this context, the r² value is 0.44, which means that approximately 44% of the variation in the final exam scores (y) can be explained by the variation in the third exam scores (x).
So, the correct interpretation of r² is given by option b: "About 44% of the variation in the final exam score can be explained by the students' scores on the 3rd exam. The remaining 56% is due to other factors or unexplained randomness."
Complete Question:
A random sample of 11 statistics students produced data where x is the third exam score out of 80, and y is the final exam score out of 200. The corresponding regression line has the equation: y = -173.51 + 4.83x, and the value of r2 (the "coefficient of determination") is found to be 0.44. What is the proper interpretation of r2?
a. Due to the number of points on the two exams, the third exam score will likely be 44% of the final exam score.
b. About 44% of the variation in the final exam score can be explained by the students' scores on the 3rd exam. The remaining 56% is due to other factors or unexplained randomness.
c. For each 1 point increase in the 3rd exam score, we expect the final exam score to increase by 0.44 points.
d. For each 1 point increase in the 3rd exam score, we expect the final exam score to increase by 0.44 percent.
e. 44% of the students scored within 1 standard deviation of the mean on each of the two tests.
The weather during a Bloodhound game is somewhat unpredictable. It may be sunny, cloudy or rainy. The probability of sunny weather is 0.42. The probability of cloudy weather is 0.38. The probability of rainy weather is .20. John Jay, mighty center-fielder for the Bloodhounds, is colorblind. Thus, if it is sunny, he has a 0.19 probability of hitting a home run in a game; if it cloudy, he has a 0.12 probability of hitting a home run in a game; if it is rainy, he has a 0.17 probability of hitting a home run in a game. The Bloodhounds play a game, say G.
a. What is the probability that John Jay hits a home run in G
b. What is the probability that John Jay does not hit a home run in G?
c. Find the conditional probability that John Jay hits a home run in G given that it rains?
d. What is the probability that it rains, and John Jay hits a home run?
e. If John Jay hits a home run, what is the probability that it rained?
f. If weather is independent from day to day then what is the probability that it is sunny 3 days in a row.
Answer: a. 0.159
b. 0.841
C. 0.17
d. 0.034
e. 0.2133
F. 0.501
Step-by-step explanation:
Julio filled his gas tank with 6 gallons of premium unleaded gas for $16.98.
How much would it cost to fill an 18 gallon tank?
Answer: it cost $50.94 to fill an 18 gallon tank.
Step-by-step explanation:
Julio filled his gas tank with 6 gallons of premium unleaded gas for $16.98. This means that amount it cost to will fill his gas tank with 1 gallon of premium unleaded gas would be
16.98/6 = $2.83 per gallon
Therefore, the amount of will cost to an 18 gallon tank with premium unleaded gas would be
18 × 2.83 = $50.94
Show that a ball dropped from a height h feet reaches the floor in 14h−−√ seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:
Complete Question
"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem.
Show that a ball dropped from a height h feet reaches the floor in 1/4√h seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:
Answer:
t = ¼√h seconds
Step-by-step explanation:
Given
Height = 15 feet
Show that a ball dropped from a height h feet reaches the floor in 14h−−√ seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:
From this, we understand that
u = Initial Velocity = 0
a = g = acceleration due to gravity = 9.8m/s² = 32ft/s²
h = initial height = 15
Using Newton equation of motion
h = ut + ½at²
Substitute the values
15 = 0 * t + ½ * 32 t²
15 = 16t² ---- make t² the subject of formula
t² = 15/16 ----- square root both sides
t = √15/√16
t = ¼√15
But h = 15
So, t = ¼√h seconds
Or t = 0.25√h seconds
-- Proved
Final answer:
A ball dropped from a height h feet reaches the floor in 14h−−√ seconds. To find the time the ball bounces when it hits the floor for the first, second, third, and fourth times, we can use this result. For example, if h = 1.5 meters, the time it takes for the ball to bounce for the first, second, third, and fourth times would be approximately 6.93 seconds, 7.95 seconds, 8.96 seconds, and 9.98 seconds, respectively.
Explanation:
Given that a ball dropped from a height h feet reaches the floor in 14h√ seconds, we can use this result to find the time the ball bounces when it hits the floor for the first, second, third, and fourth times.
Let's say the time it takes for the ball to reach the floor for the first time is t1. Using the equation 14h√ = t1, we can solve for t1 by squaring both sides of the equation and solving for t1. Similarly, we can find the time for the second, third, and fourth bounces.
For example, if h = 1.5 meters, the time it takes for the ball to bounce for the first, second, third, and fourth times would be approximately 6.93 seconds, 7.95 seconds, 8.96 seconds, and 9.98 seconds, respectively.
At a campground, a rectangular fire pit is 7 feet by 6 feet. What is the area of the largest circular fire that can be made in this fire pit? Round to the nearest square inch.
The area of the circular fire pit is 4096 square inches.
Explanation:
Given that the rectangular fire pit is 7 feet by 6 feet.
We need to determine the area of the largest circular fire that can be made in this fire pit.
The diameter of the circular fire is 6 feet
The radius is given by
[tex]r=\frac{6}{2} =3[/tex]
Radius is 3 feet.
The area of the largest circular fire pit can be determined using the formula,
[tex]Area=\pi r^2[/tex]
Substituting the values in the formula, we have,
[tex]Area = (3.14)(3)^2[/tex]
[tex]=(3.14)(9)[/tex]
[tex]Area= 28.26 \ ft^2[/tex]
We need to convert feet to inches by multiplying by 12, we get,
[tex]Area = 28.26\times (12)^2[/tex]
[tex]Area = 4096.44 \ in^2[/tex]
Rounding off to the nearest square inch, we get,
[tex]Area= 4096 \ in^2[/tex]
Thus, the area of the circular fire pit is 4096 square inches.
An experimenter has conducted a single‐factor experiment with four levels of the factor, and each factor level has been replicated six times. The computed value of the F F ‐statistic is F 0 = 3.26 F 0 = 3.26 . Find bounds on the P P ‐value.
Answer:
Step-by-step explanation:
You can find your answer in attached document.
Final answer:
the null hypothesis is retained and the P-value for this result is greater than 0.05, constituting a lower bound on the P-value.
Explanation:
The experiment involves a single-factor ANOVA with four levels of the factor and six replications per level. The computed value of the F-statistic is F0 = 3.26. Based on the provided information, the critical value for a significance level of α = 0.05 with 3 and 18 degrees of freedom (four levels minus one for the numerator and 24 minus four for the denominator) is 9.197. Since the obtained F0 is less than the critical value, the null hypothesis is retained. Therefore, there is no evidence to suggest a significant difference in variances at the 0.05 level.
To find bounds on the P-value, we look at standard F-distribution tables or use software to find the exact probability. However, since F0 < F(0.05, 3, 18), we know that the P-value must be greater than 0.05, giving us a lower bound. An upper bound is more challenging to state without additional tables or software but is less than 1.0 as a P-value cannot exceed this.
At a restaurant that sells appetizers: • 8% of the appetizers cost $1 each, • 20% of the appetizers cost $3 each, • 32% of the appetizers cost $5 each, • 40% of the appetizers cost $10 each, An appetizer is chosen at random, and X is its price. Each appetizer has 7% sales tax. So Y = 1.07X is the amount paid on the bill (in dollars) Find the variance of Y.
Answer:
12.0 (3 sf)
Step-by-step explanation:
E(X) = 0.08(1)+0.2(3)+0.32(5)+0.4(10)
E(X) = 6.28
E(X²) = .08(1²)+.2(3²)+.32(5²)+.4(10²)
E(X²) = 49.88
Var(X) = E(X²) - [E(X)]²
= 49.88 - 6.28² = 10.4416
Var(1.07X) = 1.07² Var(X)
= 1.1449×10.4416 = 11.95458784
12.0 (3 sf)
The variance of the amount paid on the bill (Y) is $10.62.
To find the variance of Y, we need to calculate the expected value of Y first, and then use that to compute the variance.
Step 1: Calculate the expected value of Y (E(Y)).
E(Y) = Σ [P(X) * Y]
where P(X) is the probability of each price category.
E(Y) = (0.08 * $1) + (0.20 * $3) + (0.32 * $5) + (0.40 * $10)
E(Y) = $0.08 + $0.60 + $1.60 + $4.00
E(Y) = $6.28
Step 2: Find the variance of Y.
Variance of Y (Var(Y)) = Σ [P(X) * (Y - E(Y))²]
Var(Y) = (0.08 * ($1 - $6.28)²) + (0.20 * ($3 - $6.28)²) + (0.32 * ($5 - $6.28)²) + (0.40 * ($10 - $6.28)²)
Var(Y) = (0.08 * $27.92) + (0.20 * $10.22) + (0.32 * $1.58) + (0.40 * $14.58)
Var(Y) = $2.24 + $2.04 + $0.51 + $5.83
Var(Y) = $10.62
The variance of Y is $10.62.
The variance measures the spread or dispersion of the values around the expected value, which, in this case, is $6.28.
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Events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.
If E is an event in S with P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8, compute
P(E) =
P(A1|E) =
P(A2|E) =
P(A3|E) =
Answer:
Step-by-step explanation:
Given that events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.
i.e. A1, A2, and A3 are mutually exclusive and exhaustive
E is an event such that
P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8,
[tex]P(E) = P(A_1E)+P(A_2E)+P(A_3E)\\= \Sigma P(E/A_1) P(A_1) \\= 0.1(0.3)+0.5(0.6)+0.8(0.2)\\= 0.03+0.3+0.16\\= 0.49[/tex]
[tex]P(A_1/E) = P(A_1E)/P(E) = \frac{0.3(0.1)}{0.49} \\=0.061224[/tex]
[tex]P(A_2/E) = P(A_2E)/P(E) = \frac{0.5)(0.6)}{0.49} \\=0.61224[/tex]
[tex]P(A_3/E) = P(A_3E)/P(E) = \frac{0.2)(0.8)}{0.49} \\=0.3265[/tex]
The probability of event E is 0.49. The probability of event A1 given E is approximately 0.0612. The probability of event A2 given E is approximately 0.6122. The probability of event A3 given E is approximately 0.3265.
Explanation:Probability of event E:
P(E) = P(E|A1) * P(A1) + P(E|A2) * P(A2) + P(E|A3) * P(A3)
P(E) = 0.1 * 0.3 + 0.6 * 0.5 + 0.8 * 0.2 = 0.03 + 0.3 + 0.16 = 0.49
Probability of event A1 given E:
P(A1|E) = [P(E|A1) * P(A1)] / P(E) = (0.1 * 0.3) / 0.49 = 0.03 / 0.49 ≈ 0.0612
Probability of event A2 given E:
P(A2|E) = [P(E|A2) * P(A2)] / P(E) = (0.6 * 0.5) / 0.49 = 0.3 / 0.49 ≈ 0.6122
Probability of event A3 given E:
P(A3|E) = [P(E|A3) * P(A3)] / P(E) = (0.8 * 0.2) / 0.49 = 0.16 / 0.49 ≈ 0.3265
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bars(s).
A hot dog vendor at the zoo recorded the average temperature in degrees, x, and the average number of hot dogs she sold, y.
The equation for the line of best fit for this situation is shown below.
y=3/10x+8
Based on the line of best fit, complete the given statements.
The expected number of hot dogs sold when the temperature is 50° would be___hot dogs.
If the vendor sold 35 hot dogs, the temperature is expected to be ___degrees.
Based on the line of best fit, for every 10-degree increase in temperature, she should sell____more hot dogs.
Answer:
(a)23 (b)90 (c)3
Step-by-step explanation:
The equation for the line of best fit for this situation is given as
[tex]y=\frac{3}{10}x+8[/tex]
where x=average temperature in degrees
y=average number of hot dogs she sold,
(a) The expected number of hot dogs sold when the temperature is 50° would be___hot dogs.
When x=50°
[tex]y=\frac{3}{10}X50+8=15+8=23[/tex]
When the temperature is 50°, the expected number of hot dogs sold would be 23.
(b)If the vendor sold 35 hot dogs, the temperature is expected to be ___degrees.
If y=35
[tex]35=\frac{3}{10}x+8\\35-8=\frac{3}{10}x\\27=\frac{3}{10}x[/tex]
Multiply both sides by 10/3
[tex]27 X \frac{10}{3}= \frac{3}{10}x X \frac{10}{3}\\x=90^{0}[/tex]
If the vendor sold 35 hot dogs, the temperature is expected to be 90 degrees.
(c) Based on the line of best fit, for every 10-degree increase in temperature, she should sell 3 more hot dogs.
The monthly amounts spent for food by families of four receiving food stamps approximates a symmetrical, normal distribution. The sample mean is $150 and the standard deviation is $20. Using the Empirical rule, about 95% of the monthly food expenditures are between what two amounts? 20) ______ A) $85 and $105 B) $100 and $200 C) $205 and $220 D) $110 and $190
Answer:
D) $110 and $190
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 150
Standard deviation = 20
95% of the monthly food expenditures are between what two amounts?
By the Empirical Rule, within 2 standard deviations of the mean
150 - 2*20 = $110
150 + 2*20 = $190
So the correct answer is:
D) $110 and $190
Answer: D) $110 and $190
Step-by-step explanation:
The Empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below
68% of data falls within the first standard deviation from the mean.
95% fall within two standard deviations.
99.7% fall within three standard deviations.
From the information given, the mean is $150 and the standard deviation is $20.
2 standard deviations = 2 × 20 = 40
150 - 40 = $110
150 + 40 = 190
Therefore, about 95% of the monthly food expenditures are between $110 and $190
An airline wants to evaluate the depth perception of its pilots over the age of fifty. A random sample of n = 14 airline pilots over the age of fifty are asked to judge the distance between two markers placed 20 feet apart at the opposite end of the laboratory. The population standard deviation is 1.4. The sample data listed here are the pilots’ error (recorded in feet) in judging the distance.2.9 2.6 2.9 2.6 2.4 1.3 2.3
2.2 2.5 2.3 2.8 2.5 2.7 2.6
Is there evidence that the mean error in depth perception for the company’s pilots over the age of fifty is greater than 2.1? Use α = 0.05 and software to get the results, and paste in the appropriate output.What is the 90% confidence interval for the mean error in depth perception?
Answer:
Step-by-step explanation:
Hello!
1) The objective is to evaluate the depth perception of pilots over the age of fifty.
To do so a sample of n=14 airline pilots over the age of fifty was taken, each of them was asked to judge the distance between two markers placed 20 feet apart, the pilot's error in judging the distance was recorded:
2.9, 2.6, 2.9, 2.6, 2.4, 1.3, 2.3, 2.2, 2.5, 2.3, 2.8, 2.5, 2.7, 2.6
Then the variable of interest is X: error in judging the distance between two markers placed 20 feet apart of one pilot. (feet)
Assuming that this variable has a normal distribution, with a standard deviation of σ= 1.4 feet
The hypothesis is that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1, symbolically: μ > 2.1
H₀: μ ≤ 2.1
H₁: μ > 2.1
α: 0.05
Since the variable has a normal distribution and the population standard deviation is known, the statistic to use for this test is the standard normal:
[tex]Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)[/tex]
The sample mean is
X[bar]= ∑X/n= 34.90/14= 2.49
[tex]Z_{H_0}= \frac{2.49-2.1}{\frac{1.4}{\sqrt{14} } } = 1.04[/tex]
The p-value for this test is 0.14917
Using the p-value approach, since it is greater than the significance level, the decision is to not reject the null hypothesis.
Then there is no evidence that the mean error in-depth perception for the company's pilots over the age of fifty is greater than 2.1 feet.
2) To construct the 90% confidence interval you have to use the same distribution as before, the formula for the interval under the standard deviation is:
[X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * (δ/√n)]
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.645[/tex]
[2.49 ± 1.645 * (1.4/√14)]
[1.87; 3.11]
With a 90% confidence, you'd expect that the true mean of the error in-depth perception of the airline's pilots over fifty years old is contained in the interval [1.87; 3.11].
I hope you have a SUPER day!
The One-Sample T Test can be used to determine whether the mean error in depth perception for pilots over the age of fifty is greater than 2.1. The p-value from this test can inform whether to reject the null hypothesis. The 90% confidence interval can be determined by calculating the standard error and using a t-value associated with a 90% confidence level, the obtained margin of error is subtracted and added to the mean to generate the confidence interval.
Explanation:The subject of the question is statistical testing, specifically One-Sample T Test and Confidence Interval computing. The goal is to determine whether the mean error in depth perception of pilots over the age of fifty is greater than 2.1 and to calculate the 90% confidence interval for the mean error in depth perception.
To perform this test, one would organize the provided data set in a software like R or Excel, compute the sample mean and use the One-Sample T Test function to compare the sample mean to the hypothesized mean of 2.1. The null hypothesis would be that the mean error is not greater than 2.1, and the alternative hypothesis is that the mean error is greater than 2.1. If the p-value received from the One-Sample T Test is less than 0.05, we reject the null hypothesis and conclude that the mean error is greater than 2.1.
As for the 90% Confidence Interval, it is done in statistical software by calculating the standard error, which accounts for both the standard deviation and the sample size, and multiplying it by the relevant t-value (associated with a 90% confidence level and degrees of freedom which equals sample size minus 1). This result then serves as the margin of error which is subtracted and added to the mean to generate the confidence interval.
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The College Board originally scaled SAT scores so that the scores for each section were approximately normally distributed with a mean of 500 and a standard deviation of 100. Assuming scores follow a bell-shaped distribution, use the empirical rule to find the percentage of students who scored greater than 700.
Answer:
Percentage of students who scored greater than 700 = 97.72%
Step-by-step explanation:
We are given that the College Board originally scaled SAT scores so that the scores for each section were approximately normally distributed with a mean of 500 and a standard deviation of 100.
Let X = percentage of students who scored greater than 700.
Since, X ~ N([tex]\mu, \sigma^{2}[/tex])
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1) where, [tex]\mu[/tex] = 500 and [tex]\sigma[/tex] = 100
So, P(percentage of students who scored greater than 700) = P(X > 700)
P(X > 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{700-500}{100}[/tex] ) = P(Z < 2) = 0.97725 or 97.72% Therefore, percentage of students who scored greater than 700 is 97.72%.
If all other factors are held constant, increasing the sample size will do the following.
decrease the width of the confidence interval
increase the standard error
None of the other choices are correct.
increase the width of the confidence interval
Answer:
Decrease the width of the confidence interval
Step-by-step explanation:
Sample size is in the denominator, so increasing n would decrease the width for the same level of confidence
Final answer:
Increasing the sample size will decrease the width of the confidence interval and decrease the standard error, leading to a more precise estimate of the population mean with the same level of confidence.
Explanation:
If all other factors are held constant, increasing the sample size will decrease the width of the confidence interval. This is because a larger sample size reduces the variability within the sample. The standard error, which is inversely proportional to the square root of the sample size, will also decrease as a result. Thus, we do not need as wide an interval to capture the true population mean with the same level of confidence when the sample size is larger.
Another related concept is that as the confidence level increases, the error bound increases, making the confidence interval wider. However, this effect is separate from changes in the sample size. Also, it's important to note that the standard deviation of the sampling distribution of the means will decrease as the sample size increases, leading to a more precise estimate of the population mean. Therefore, increasing the sample size, while keeping the confidence level constant, leaves us more confident about our estimate being closer to the true population mean.
Which expression is not a perfect square trinomial?
Answer:
121+11y+y^2 not perfect square trinomial because the second member should be twice the value of the products of the first and second monomers.
Step-by-step explanation:
(A+B) ^2=A^2 +2*A*B+B^2
Because elderly people may have difficulty standing to have their height measured, a study looked at the relationship between overall height and height to the knee. Here are data (in centimeters) for five elderly men:
col1 Knee Height x 56 44 41 44 55
col2 Height y 190 150 145 165 172
What is the equation of the least-squares regression line for predicting height from knee height?
The equation for the least-squares regression line can be found by calculating the slope and y-intercept using the given data on knee height and overall height. The regression line is used for predicting the height from knee height.
Explanation:To find the equation of the least-squares regression line, you first need to calculate the slope (b1) and y-intercept (b0) using the given data. The least-squares regression line is essentially a line of best fit that minimizes the sum of the squared residuals.
The formula for the slope (b1) of the regression line is: b1 = (∑xy - n * mean_x * mean_y) / (∑x^2 - n * mean_x^2) And the y-intercept (b0) is calculated as: b0 = mean_y - b1 * mean_x
Following these formulas and plugging in the given data (for knee height x and height y), we can find b1 and b0. Once we've done that, we can write the equation for the least-squares regression line in the form y = b0 + b1*x.
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The least-squares regression line equation is estimated by determining the slope (b1) and y-intercept (b0) of the line. These values are calculated from the given data sets for height and knee height using the formulas: b1 = [N(Σxy) - (Σx)(Σy)] / [N(Σx²) - (Σx)²] and b0 = (Σy - b1(Σx)) / N
Explanation:The subject matter of this question is statistics, specifically, it's about finding the equation of a least-squares regression line. The least-squares regression line is a tool used in statistics to show the best possible mathematical relationship between two variables. In this case, the variables are height and knee height.
To calculate the least-squares regression line, we need to calculate the slope (b1) and y-intercept (b0) of the line. The formulas to calculate these are:
b1 = [N(Σxy) - (Σx)(Σy)] / [N(Σx²) - (Σx)²]b0 = (Σy - b1(Σx)) / N
Where:
N = number of observations (5 in this case)
Σxy = sum of the product of x and y
Σx = sum of x
Σy = sum of y
Σx² = sum of squares of x
After calculating the values for b0 and b1, the equation for the least-squares regression line would be: y = b0 + b1*x. You would need to calculate these values using the provided datasets for height (x) and knee height (y).
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You flip a coin 4 times that has been weighted such that heads comes up twice as often as tails . What is the probability that all 4 of them are heads?
Answer:
0.1975
Step-by-step explanation:
Let the probability of getting heads on flipping the coin = p
Then the probability of getting tails on flipping the coin = 1-p
It is given that probability of heads is twice the probability of tails.
[tex]\[p= 2* (1-p)\][/tex]
[tex]\[=> p= 2 - 2p\][/tex]
[tex]\[=> 3p= 2 \][/tex]
[tex]\[=> p= \frac{2}{3} \][/tex]
So that probability of getting a head on single coin flip = [tex]\[\frac{2}{3}\][/tex]
This means that the probability of getting heads on 4 coin flips =
[tex]\[ p^{4} \][/tex]
[tex]\[= (\frac{2}{3})^{4} \][/tex]
[tex]\[= 0.1975 \][/tex]
Probability of an event is the measure of its chance of occurrence. The probability of all 4 tossed coins in given context coming out as heads is 0.197 approximately.
How to find that a given condition can be modeled by binomial distribution?Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
For the given case, let model the condition as:
Success = getting head on given biased coin
Probability of success = p = 2/3 (as head comes twice as often as tails, so probability of heads = twice probability of q = x say,
then 2x + x = 1(total probability is 1), or x = 1/3 = probability of tails,
thus, probability of heads= 2/3)
Failure = getting tail on given biased coin
Probability of failure = q = 1-p = 1-2/3 = 1/3
All coins' results are independent, thus, they are Bernoulli trials.
The count of Bernoulli trials is n = 4
Let random variable X tracks the number of heads obtained on tossing these 4 given biased coins.
Then,
[tex]X \sim B(4,2/3)[/tex]
The needed probability is
P(X = 4)
Using the probability function of binomial distribution, we get:
[tex]P(X = 4) = \: ^4C_4(2/3)^4(1/3)^0 = \dfrac{16}{81} \approx 0.197[/tex]
Thus, The probability of all 4 tossed coins in given context coming out as heads is 0.197 approximately.
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[10 points] Let v1, v2 and v3 be three linearly independent vectors in R 3 . (a) Find the rank of the matrix A = (v1 − v2) (v2 − v3) (v3 − v1) . (b) Find the rank of the matrix B = (v1 + v2) (v2 + v3) (v3 + v1) .
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.I hope my explanation will help you in understanding this particular question.
Step-by-step explanation:
Two fair dice are tossed, and the up face on each die is recorded. Find the probability of observing each of the following events: A:{ The sum of the numbers is odd } B:{ The sum of the numbers is 10 or more } C:{ A 3 appears on each of the two dice }
The probability of event A is 1/2, the probability of event B is 5/18, and the probability of event C is 1/36.
Explanation:To find the probability of each event, we need to analyze the possible outcomes and count the favorable outcomes for each event.
a) Event A: The sum of the numbers is odd.
Out of the 36 possible outcomes (6 outcomes for the first die and 6 outcomes for the second die), 18 outcomes have an odd sum. Therefore, the probability of event A is 18/36 = 1/2.
b) Event B: The sum of the numbers is 10 or more.
Out of the 36 possible outcomes, 10 outcomes have a sum of 10 or more. Therefore, the probability of event B is 10/36 = 5/18.
c) Event C: A 3 appears on each of the two dice.
Out of the 36 possible outcomes, only 1 outcome has a 3 on each die. Therefore, the probability of event C is 1/36.
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A. The probability that the sum of the numbers is odd is [tex]\(\frac{1}{2}\)[/tex].
B. The probability that the sum of the numbers is 10 or more is [tex]\(\frac{1}{6}\)[/tex].
C. The probability that a 3 appears on each of the two dice is [tex]\(\frac{1}{36}\)[/tex].
First, note that each die has 6 faces, so the total number of possible outcomes when two dice are tossed is:
[tex]\[ 6 \times 6 = 36 \][/tex]
A. The sum of two numbers is odd if one number is even and the other is odd.
- Dice faces: {1, 2, 3, 4, 5, 6}
- Odd faces: {1, 3, 5}
- Even faces: {2, 4, 6}
For each of the 3 odd faces on the first die, the second die can show any of the 3 even faces. Similarly, for each of the 3 even faces on the first die, the second die can show any of the 3 odd faces.
So, the number of favorable outcomes:
[tex]\[ 3 \times 3 + 3 \times 3 = 9 + 9 = 18 \][/tex]
The probability of event A is
[tex]\[ \frac{18}{36} = \frac{1}{2} \][/tex]
B. Let's list the pairs of dice faces whose sums are 10 or more:
- Sum = 10: (4, 6), (5, 5), (6, 4)
- Sum = 11: (5, 6), (6, 5)
- Sum = 12: (6, 6)
Number of favorable outcomes:
[tex]\[ 3 + 2 + 1 = 6 \][/tex]
The probability of event B is
[tex]\[ \frac{6}{36} = \frac{1}{6} \][/tex]
C. This event means both dice show 3:
- Outcome: (3, 3)
Number of favorable outcomes: 1
The probability of event C is [tex]\[ \frac{1}{36} \][/tex]
If your score on your next statistics test is converted to a z score, which of these z scores would you prefer: minus2.00, minus1.00, 0, 1.00, 2.00? Why?
Answer:
You would prefer a z-score of 2, because the higher the z-score, the higher your grade was relative to your classmates.
Step-by-step explanation:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For example
If the z-score of your grade was -2, it means that your grade was 2 standard deviations below the average grade.
Otherwise, if the z-score of your grade was 2, it means that your grade was 2 standard deviations above the average higher.
The higher the z-score, the better.
So you would prefer a z-score of 2, because the higher the z-score, the higher your grade was relative to your classmates.
Velvetleaf is a particularly annoying weed in cornfields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right. How many seeds do velvetleaf plants produce? (Use 96% confidence). Here are counts from 28 plants that came up in a cornfield when no herbicide was used:
2450 2504 2114 1110 2137 8015 1623 1531 2008 1716721 863 1136 2819 1911 2101 1051 218 1711 1642228 363 5973 1050 1961 1809 130 880
Answer:
96% CI for the production of seeds
[tex]1177.5\leq\mu\leq2520.5[/tex]
Step-by-step explanation:
We have a sample of size n=28. With these data we can calculate the mean and standard deviation of the sample.
Sample = [2450, 2504, 2114, 1110, 2137, 8015, 1623, 1531, 2008, 1716, 721, 863, 1136, 2819, 1911, 2101, 1051, 218, 1711, 1642, 228, 363, 5973, 1050, 1961, 1809, 130, 880 ]
Sample mean = 1849
Sample standard deviation = 1647
To calculate a 96% confidence interval, we use the t-statistic with df=27.
The t-value for this condition is t=2.1578.
[tex]M\pm t_{27}*s/\sqrt{n}\\\\1849\pm2.1578*1647/\sqrt{28}\\\\1849\pm671.5\\\\\\1849-671.5\leq\mu\leq 1849+671.5\\\\\\1177.5\leq\mu\leq2520.5[/tex]
Then, the 96% interval is
[tex]1177.5\leq\mu\leq2520.5[/tex]
The 96% interval is [117.5,2520.5] and this can be determined by using the t-statistics and the given data.
Given :
Sample Size = 28Sample = [2450, 2504, 2114, 1110, 2137, 8015, 1623, 1531, 2008, 1716, 721, 863, 1136, 2819, 1911, 2101, 1051, 218, 1711, 1642, 228, 363, 5973, 1050, 1961, 1809, 130, 880]The sample mean for this is 1849 and the standard deviation is 1647.
Use t-statistic with df = 27 to calculate 96% confidence interval. So, the t-value is 2.1578.
[tex]\rm M\pm t_{27}\times \dfrac{s}{\sqrt{n} }[/tex]
[tex]1849\pm 2.1578 \times \dfrac{1647}{\sqrt{28} }[/tex]
[tex]1849\pm 671.5[/tex]
[tex]1849-671.5\leq \mu \leq 1849+671.5[/tex]
[tex]117.5\leq \mu \leq 2520.5[/tex]
Therefore, the 96% interval is [117.5,2520.5].
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You bike
5
miles the first day of your training,
5.4 miles the second day,
6.2 miles the third day, and
7.8 miles the fourth day. If you continue this pattern, how many miles do you bike the seventh day?
You will bike 30.2 miles in the seventh day according to the prediction.
Explanation:Here we have the following data, You bike
5 miles the first day of your training, 5.4 miles the second day, 6.2 miles the third day, and 7.8 miles the fourth day.So we can know some facts:
From the first day to the second day the number of miles increases:[tex]5.4-5=0.4mi[/tex]
From the second day to the third day the number of miles increases:[tex]6.2-5.4=0.8mi[/tex]
From the third day to the fourth day the number of miles increases:[tex]7.8-6.2=1.6mi[/tex]
By taking a look at the pattern, we can see that each day you increases the number of miles by a factor of 2 compared to the previous day. So:
From the fourth day to the fifth day the number of miles increases:[tex]x_{5}-7.8=3.2mi \\ \\ x_{5}=7.8+3.2=11mi, \ \text{Day 5}[/tex]
From the fifth day to the sixth day the number of miles increases:[tex]x_{6}-11=6.4mi \\ \\ x_{6}=6.4+11=17.4mi, \ \text{Day 6}[/tex]
Finally:
From the sixth day to the seventh day the number of miles increases:[tex]x_{7}-17.4=12.8mi \\ \\ x_{7}=12.8+17.4=30.2mi, \ \text{Day 7}[/tex]
By noting the pattern that the daily increase doubles each time, we can predict that the total distance biked on the seventh day would be 30.2 miles.
Explanation:To predict the number of miles biked on the seventh day, we need to first determine the pattern in the increase of biking distances over the days given. The distances biked in the consecutive days are: 5, 5.4, 6.2, and 7.8 miles. We can see that each day, the distance increases at varying amounts:
From day 1 to 2: 5.4 - 5 = 0.4 milesFrom day 2 to 3: 6.2 - 5.4 = 0.8 milesFrom day 3 to 4: 7.8 - 6.2 = 1.6 milesThe increase pattern appears to be that each day the distance increases by double the amount of the previous day. Therefore, we can predict the increase and the total distance for the next days:
From day 4 to 5: 1.6 * 2 = 3.2 miles increase, Total = 7.8 + 3.2 = 11 milesFrom day 5 to 6: 3.2 * 2 = 6.4 miles increase, Total = 11 + 6.4 = 17.4 milesFrom day 6 to 7: 6.4 * 2 = 12.8 miles increase, Total = 17.4 + 12.8 = 30.2 milesIf the pattern continues, the student would bike 30.2 miles on the seventh day.
Shari buys a house for $240,000. She makes a down payment of 20% and finances the rest with a 15 year mortgage. She agrees to make equal payments at the end of each month. If the annual interest rate is 1.2% and interest is compounded monthly, what is Shari's regular payment? To solve this question, we use the formula P equals R open parentheses fraction numerator 1 minus (1 plus i )to the power of negative n end exponent over denominator i end fraction close parentheses. Fill in the following blanks for the given information:
Answer:
$1166.08 is the monthly payment for the mortgage per month.
Step-by-step explanation:
The meaning of this stated formula on the statement is the present annuity formula because we will have future monthly payments on the mortgage of the house in which they pay off the present value of the house which is $240000 x 80% = $ 192000 as this amount will excludes the down payment of 20% that is made.
We are given Pv the present value which excludes the down payment $192000.
We have the interest rate i which is 1.2%/12 as it is compounded monthly.
n is the number of payments made over a period which is 12 x 15 years= 180 payments as it is compounded monthly.
no we substitute the above mentioned information to the present value annuity formula stated to calculate R the monthly payment:
Pv = R[(1-(1+i)^-n)/i]
$192000 = R[(1-(1+(1.2%/12))^-180)/ (1.2%/12)] divide both sides by the coefficient of R
$192000/[(1-(1+(1.2%/12))^-180)/(1.2%/12)] = R
$1166.08 =R which this is the amount that will be paid for the mortgage every month for 15 years.