Final answer:
Keith Bowie should deposit $667.45 into an account today to have enough money in 6 years to pay for the engine overhaul.
Explanation:
To determine the amount Keith Bowie should deposit today in an account earning 6% interest compounded annually, we can use the concept of present value. The present value of a future cash flow is calculated by dividing the future cash flow by (1 + interest rate) raised to the power of the number of years. In this case, we want to find the amount needed that would accumulate to cover the engine overhaul cost in 6 years. Let's assume the cost of the overhaul is $390 with a 10% probability, $570 with a 30% probability, $750 with a 50% probability, and $790 with a 10% probability. We can calculate the present value by multiplying each cash flow amount with its respective probability, discounting it back at the interest rate over 6 years, and adding them all up:
Present Value = (390 * 0.10) / (1 + 0.06)6 + (570 * 0.30) / (1 + 0.06)6 + (750 * 0.50) / (1 + 0.06)6 + (790 * 0.10) / (1 + 0.06)6 = $667.45
Therefore, Keith Bowie should deposit $667.45 into the account today to ensure he has enough money on hand in 6 years to pay for the engine overhaul.
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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A. Create a set of 5 points that are very close together and record the standard deviation. Next, add a sixth point that is far away from the original 5 and record the new standard deviation. What is the impact of the new point on the standard deviation?
Answer: The addition of the new point alters the previous standard deviation greatly
Step-by-step explanation:
Let the initial five points be : 2 3 4 5 and 6. In order to calculate the standard deviation for this data, we will need to calculate the mean first.
Mean = summation of scores/number of scores.
The mean is therefore: (2+3+4+5+6)/5 = 20/5 = 4.
We'll also need the sum of the squares of the deviations of the mean from all the scores.
Since mean = 4, deviation of the mean from the score "2" = score(2) - mean (4)
For score 3, it is -1
For 4, it's 0
For 5 it's 1
For 6 it's 2.
The squares for -2, -1, 0, 1, and 2 respectively will be 4, 1 , 0, 1, 4. Summing them up we have 10 i.e (4+1+0+1+4=10).
Calculating the standard deviation, we apply the formula:
√(summation of (x - deviation of mean)^2)/N
Where N means the number of scores.
The standard deviation = √(10/5) = 1.4142
If we add another score or point that is far away from the original points, say 40, what happens to the standard deviation. Let's calculate to find out.
i.e we now have scores: 2, 3, 4, 5, 6 and 40
We calculate by undergoing same steps.
Firstly mean. The new mean = (2+3+4+5+6+40)/6 = 60/6 = 10.
The mean deviations for the scores : 2, 3, 4, 5, 6 and 40 are -8, -7, -6, -5, -4 and 30 respectively. The squares of these deviations are also 64, 49, 36, 25, 16 and 900 respectively as well. Their sum will then be 1090. i.e. (64+49+36+25+16+900 = 1090).
The new standard deviation is then=
√(1090/6)
= √181.67
= 13.478.
It's clear that the addition of a point that's far away from the original points greatly alters the size of the standard deviation as seen /witnessed in this particular instance where the standard deviation rises from 1.412 to 13.478
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
If we start at the point (1,0) and travel once around the unit circle, we travel a distance of 2 pi units and arrive back where we started at the point (1,0). If we continue around the unit circle a second time, we will repeat all the values of x and y that occurred during our first trip around. Use the this discussion to evaluate the following expressions
sin (2pi + 3pi/2)
Answer:
-1
Step-by-step explanation:
We evaluate [tex]\sin(2\pi+3\pi/2)[/tex]
In [tex]2\pi+3\pi/2[/tex], [tex]2\pi[/tex] is a complete revolution and is the same as 0. So we have
[tex]\sin3\pi/2 = \sin(\pi+\pi/2)[/tex]
One [tex]\pi[/tex] is a half revolution, putting the point at (-1, 0). [tex]\pi/2[/tex] is a quarter of a revolution. A quarter circle from (-1, 0) anticlockwise is (0, -1).
The sine is the y-coordinate of a point along a unit circle.
Hence, [tex]\sin(2\pi+3\pi/2)=-1[/tex]
Using the cyclical nature of the unit circle and the sine function, the mathematical expression sin (2π + 3π/2) can be simplified to sin (3π/2). travelling 3π/2 around the unit circle takes us to the point where sin is -1.
Explanation:The question requires the evaluation of the mathematical expression sin (2π + 3π/2). This can be solved by utilizing the cyclical nature of the unit circle and the sine function. Since the distance around the unit circle is 2π, adding or subtracting multiples of 2π from the angle doesn't change the result of the sin function.
So, we can simplify the function sin (2π + 3π/2) to sin (3π/2). Because every π/2 around the unit circle the sine function repeats, sin (π/2) is 1, sin (2π/2) or sin (π) is 0, sin (3π/2) is -1, and sin (4π/2) or sin (2π) is 0. So sin (3π/2) equals -1.
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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.
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%.
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
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.
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
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
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.
There was a special on sweatshirts is sweatshirt was on sale for $9.69 if a customer bought three striped shirt at the regular price of $12.95 a fourth sweatshirt was free which is the better buy for sweatshirts at 9.69 each or three sweatshirts at $12.95 in a fourth one free?
Answer: it is cheaper to buy for sweatshirts at $9.69 each.
Step-by-step explanation:
The regular price for one sweatshirt is $12.95. if a customer bought three shirts at the regular price of $12.95, a fourth sweatshirt was free
It means that the cost of buying 4 shirts is
12.95 × 3 = $38.85
Due to a special, the price of one sweatshirt was $9.69. It means that the cost of buying 4 shirts at this price is
9.69 × 4 = $38.76
Therefore, it is cheaper to buy for sweatshirts at $9.69 each than to buy at $12.95 each and get a free shirt
Answer: it is cheaper to buy for sweatshirts at $9.69 each.
Suppose we want to see if American children have higher levels of cholesterol than the average child (i.e., in the entire world - the total population). We find that the population average for cholesterol for children all over the world is 190. We test 25 US children and find an average of 201 with a standard deviation of 10. Conduct a hypothesis with a significance level of 0.05.
Answer:
[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]
[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190
Step-by-step explanation:
Data given and notation
[tex]\bar X=201[/tex] represent the mean
[tex]s=10[/tex] represent the sample standard deviation for the sample
[tex]n=25[/tex] sample size
[tex]\mu_o =190[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 190[/tex]
Alternative hypothesis:[tex]\mu > 190[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=25-1=24[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190
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.
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.
If np greater than or equals 5 and nq greater than or equals 5, estimate Upper P (fewer than 3 )with nequals13 and pequals0.4 by using the normal distribution as an approximation to the binomial distribution; if npless than5 or nqless than5, then state that the normal approximation is not suitable.
Answer:
We need to check the conditions in order to use the normal approximation.
[tex]np=13*0.4=5.2 \geq 5[/tex]
[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]
Assuming that each trial is independent and we have a sample obtained from a random sampling method.
Then we can conclude that we can use the normal approximation since all the conditions are satisfied.
Step-by-step explanation:
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=13, p=0.4)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We need to check the conditions in order to use the normal approximation.
[tex]np=13*0.4=5.2 \geq 5[/tex]
[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]
Assuming that each trial is independent and we have a sample obtained from a random sampling method.
Then we can conclude that we can use the normal approximation since all the conditions are satisfied.
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
Please help? (03.04) What are the coordinates of the vertex for f(x) = x^2 + 4x + 10?
Answer:
(-2, 6)
Step-by-step explanation:
f(x) = x² + 4x + 10
f(x) = x² + 4x + 4 + 6
f(x) = (x + 2)² + 6
The vertex is at (-2, 6).
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%.
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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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.
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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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
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
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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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.
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.
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.