Answer:
a) The 95% CI for the mean surgery time is (133.05, 140.75).
b) The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) The level of confidence of the interval (133.9, 139.9) is 69%.
d) The sample size should be 219 surgeries.
e) The sample size should be 377 surgeries.
Step-by-step explanation:
We have a sample, of size n=132, for which the mean time was 136.9 minutes with a standard deviation of 22.6 minutes.
a) We have to find a 95% CI for the mean surgery time.
The critical value of z for a 95% CI is z=1.96.
The margin of error of the CI can be calculated as:
[tex]E=z\cdot s/\sqrt{n}=1.96*22.6/\sqrt{132}=44.296/11.489=3.85[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=\bar x-E=136.9-3.85=133.05\\\\UL=\bar x+E=136.9+3.85=140.75[/tex]
The 95% CI for the mean surgery time is (133.05, 140.75).
b) Now, we have to find a 99.5% CI for the mean surgery time.
The critical value of z for a 99.5% CI is z=2.81.
The margin of error of the CI can be calculated as:
[tex]E=z\cdot s/\sqrt{n}=2.81*22.6/\sqrt{132}=63.506/11.489=5.53[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=\bar x-E=136.9-5.53=131.37\\\\UL=\bar x+E=136.9+5.53=142.43[/tex]
The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) We can calculate the level of confidence, calculating the z-score for the margin of error in that interval.
We know that the difference between the upper bound and lower bound is 2 times the margin of error:
[tex]UL-LL=2E\\\\E=\dfrac{UL-LL}{2}=\dfrac{139.9-133.9}{2}=\dfrac{6}{2}=3[/tex]
Then, we can write the equation for the margin of error to know the z-value.
[tex]E=z \cdot s/\sqrt{n}\\\\z= E\cdot \sqrt{n}/s=2*\sqrt{132}/22.6=2*11.5/22.6=1.018[/tex]
The confidence level for this interval is then equal to the probability that the absolute value of z is bigger than 1.018:
[tex]P(-|z|<Z<|z|)=P(-1.018<Z<1.018)=0.69[/tex]
The level of confidence of the interval (133.9, 139.9) is 69%.
d) We have to calculate the sample size n to have a margin of error, for a 95% CI, that is equal to 3.
The critical value for a 95% CI is z=1.96.
Then, the sample size can be calculated as:
[tex]E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=14.77^2=218.015\approx 219[/tex]
The sample size should be 219 surgeries.
e) We have to calculate the sample size n to have a margin of error, for a 99% CI, that is equal to 3.
The critical value for a 99% CI is z=2.576.
Then, the sample size can be calculated as:
[tex]E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=19.41^2=376.59\approx 377[/tex]
The sample size should be 377 surgeries.
A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy-duty fencing selling for $3 per foot, while the remaining two sides will use standard fencing costing $2 per foot. The budget that will be entirely used for fencing is $6000. In order to maximize the area of the plot, how many feet of standard fencing should be purchased for the project?
So, the required length of the sides that have standard fencing is 1500 ft.
Area of the rectangle:The area of rectangle is the region covered by the rectangle in a two-dimensional plane. A rectangle is a type of quadrilateral, a 2d shape that has four sides and four vertices.
Let [tex]x[/tex] ft be the length of the sides that duty fencing and [tex]y[/tex] ft be the length of the sides that have standard fencing.
So, the area will be calculated by the above formula we get,
[tex]Area=xy[/tex]...(1)
Now, the cost of fencing is,
[tex]3x+2y=6000\\y=3000-\frac{3x}{2}[/tex]...(2)
Now, substituting equation (2) in equation (1) we get,
[tex]A=3000x-\frac{3x^2}{2}[/tex]
Now, differentiating the above equation we get,
[tex]\frac{dA}{dx} =3000-3x=0\\x=1000[/tex]
Substituting [tex]x=1000 ft[/tex] in equation (2) we get,
[tex]y=3000-\frac{3\times 1000}{2}\\=1500 ft[/tex]
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The question pertains to maximizing the area of a rectangular land plot within a fence budget. We derive a mathematical equation to fit the budget involving both types of fences and then maximize the equation to find the solution. It involves applications of maximization and differentiation.
Explanation:The subject of this question is Mathematics, specifically problems involving equations and budgets. This problem can be solved by using the concept of maximization of areas in rectangle and budget calculations.
We know the total budget which is $6000, and costs of the two types of fencing. Let's denote the lengths of the heavy-duty fenced sides as H and the lengths of the standard fenced sides as S. The equation that describes the budget is 3H + 2S = 6000.
If we're maximizing area, we want to maximize S*H. Remember that the area of a rectangle is the product of its dimensions (length times width). Since H and S are lengths of sides of a rectangular plot, H*S will give us the area.
Replacing H from the budget equation we get H = (6000 - 2S) / 3. You can plug this into the area equation to get Area = S * (6000 - 2S) / 3. To find the maximum of this function, take its derivative and set it to zero, solve for S.
Doing this results in the required amount of standard fencing for the project, making it a solution to the question asked.
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14.) Although most of us buy milk by the quart or gallon, farmers measure daily production in pounds. Ayrshire cows average 47 pounds of milk a day, with a standard deviation of 6 pounds. For Jersey cows, the mean daily production is 43 pounds, with a standard deviation of 5 pounds. Assume that Normal models describe milk production for these breeds.
Answer:
69.50%
Step-by-step explanation:
Given:
Ayrshire cows:
E(X) =μ = 47
SD(X) = σ = 6 Var (X) = 6^2 = 36
Jersey cows:
E(Y) = μ = 43
SD(Y) = σ = 5 Var(Y) = 52 = 25
Properties mean, variance and standard deviation:
E(X +Y) = E(X) E(Y)
V ar(X +Y) = Var(X) + Var(Y)
SD(X +Y) = √Var(X)+Var(Y)
X — Y represents the difference between Ayrshire and Jersey cows.
E(X — Y) = E(X) — E(Y). 47 — 43 = 4
SD(X — Y) = √Var(X)+ Var(Y) =√36+ 25 = √61 = 7.8102
The z-score is the value decreased by the mean, divided by the standard deviation:
z = x-μ /σ = 0-4/ 7.8102 = -0.51
Determine the corresponding probability using table Z in appendix F.
P(X—Y [tex]\geq[/tex] 0) = P(Z > —0.51) = 1—P(Z < —0.51) = 1-0.3050 = 0.6950 = 69.50%
7. The ruler below has eleven marks and can be used to measure lengths
from one unit to twelve units.
Design a ruler which can be used to measure all the lengths from one
unit to twelve units but this time put the minimum possible number
of marks on the ruler.
Answer:
use 6cm twice and that adds up to 12cm .six is the minimal possible number
A ruler designed to measure lengths from one unit to twelve units with the minimum number of marks would only have marks at every unit. This results in only eleven markings (from 1 to 11), as the 0 marks the start and the twelfth unit marks the end.
Explanation:The subject of the student's question pertains to the design of a ruler with the minimum possible number of marks that can measure lengths from one to twelve units. To achieve this, the ruler should only be marked at every unit. This way, there would be only eleven marks (1-11), as the 0 marks the start of the ruler and the twelfth unit is considered the end of the ruler but does not require a mark. Therefore, only the spaces between these numbers need to be marked to enable measurement all the way up to twelve units.
In practice, such a ruler is less precise than a ruler with more marks as the latter can measure smaller lengths more accurately. However, the task demands the bare minimum for a measure from 1 to 12, thus we have eleven marks.
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Find the output, k, when the input, t, is 3.
k = 13t – 2
k=
Answer:
k= 13*3-2
k= 39-2
k=37 is the required answer
Answer:
k = 37
Step-by-step explanation:
[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]
Is 26, 31, 36, 41, 46 an arithmetic sequence?
Answer:
Yes, it is arithmetic sequence.
Step-by-step explanation:
31-26=5
36-31=5
41-36=5
46-41=5
Difference between consecutive numbers is the same, so we have arithmetic sequence.
Answer:
Yes it is an arithmetic sequence
26+5=31
31+5=36
36+5=41
41+5=46
A B C or D? I need help
Answer:
The answer is (A)
Answer:
The answer is A.
Step-by-step explanation:
This is because the plus sign after the first mixed number gets multiplied to the second mixed number. When a plus sign gets multiplied to something that is negative already, then the answer is negative, so the equation will still stay the same and be -3 1/8 - 2 1/2.
if this answer helped u, please mark as brainliest!
100 POINTS.
PLEASE PROVIDE STEP BY STEP ANSWER
THANK YOU.
Step-by-step explanation:
W(x) = (10x⁴ − 8) (30x + 25)^0.5
A) Take log of both sides.
ln(W) = ln[(10x⁴ − 8) (30x + 25)^0.5]
ln(W) = ln(10x⁴ − 8) + ln[(30x + 25)^0.5]
ln(W) = ln(10x⁴ − 8) + 0.5 ln(30x + 25)
Take derivative.
W' / W = 40x³ / (10x⁴ − 8) + 0.5 (30) / (30x + 25)
W' / W = 20x³ / (5x⁴ − 4) + 3 / (6x + 5)
W' = W [20x³ / (5x⁴ − 4) + 3 / (6x + 5)]
W'(x) = (10x⁴ − 8) (30x + 25)^0.5 [20x³ / (5x⁴ − 4) + 3 / (6x + 5)]
B) Evaluate at x = 0.
W'(0) = (0 − 8) (0 + 25)^0.5 [0 / (0 − 4) + 3 / (0 + 5)]
W'(0) = (-8) (5) (0 + 3/5)
W'(0) = -24
Answer:
[tex]\textsf{A)}\quad W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]
[tex]\textsf{B)}\quad W'(0)=-24[/tex]
Step-by-step explanation:
Part AGiven function:
[tex]W(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5}[/tex]
Take the natural logarithm of both sides:
[tex]\ln \left(W(x)\right) = \ln \left( (10x^4 - 8) \cdot (30x + 25)^{0.5} \right)[/tex]
Use properties of logarithms to simplify:
[tex]\ln \left(W(x)\right) = \ln (10x^4 - 8) +\ln(30x + 25)^{0.5} \\\\\\ \ln \left(W(x)\right) = \ln (10x^4 - 8) +0.5 \ln(30x + 25)[/tex]
Differentiate both sides with respect to x:
[tex]\dfrac{d}{dx} \ln \left(W(x)\right) = \dfrac{d}{dx} \left( \ln (10x^4 - 8) + 0.5 \ln (30x + 25) \right)[/tex]
Apply the chain rule on the right-hand side:
[tex]\dfrac{W'(x)}{W(x)} = \dfrac{1}{10x^4 - 8} \cdot \dfrac{d}{dx} (10x^4 - 8) + 0.5 \cdot \dfrac{1}{30x + 25} \cdot \dfrac{d}{dx} (30x + 25) \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{1}{10x^4 - 8} \cdot (40x^3) + 0.5 \cdot \dfrac{1}{30x + 25} \cdot (30) \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{40x^3}{10x^4 - 8} + \dfrac{15}{30x + 25} \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{2(20x^3)}{2(5x^4 - 4)} + \dfrac{5(3)}{5(6x + 5)} \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5}[/tex]
Multiply both sides by W(x) to solve for W'(x):
[tex]W'(x) = W(x) \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]
Substitute W(x) back in:
[tex]W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]
Therefore, the derivative W'(x) is:
[tex]W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]
[tex]\dotfill[/tex]
Part BTo find W'(0), substitute x = 0 into W'(x):
[tex]W'(0) = (10(0)^4 - 8) \cdot (30(0) + 25)^{0.5} \left(\dfrac{20(0)^3}{5(0)^4 - 4} + \dfrac{3}{6(0) + 5} \right) \\\\\\ W'(0) = (0 - 8) \cdot (0 + 25)^{0.5} \left(\dfrac{0}{0 - 4} + \dfrac{3}{0 + 5} \right) \\\\\\ W'(0) = (- 8) \cdot (25)^{0.5} \cdot \left( \dfrac{3}{5} \right) \\\\\\ W'(0) = (- 8) \cdot 5 \cdot \dfrac{3}{5} \\\\\\ W'(0)=-24[/tex]
Therefore, the value of W'(0) is -24.
In 2010 as part of the General Social Survey, 1295 randomly selected American adults responded to this question: "Some countries are doing more to protect the environment than other countries are. In general, do you think that America is doing more than enough, about the right amount, or too little?" Of the respondents, 473 replied that America is doing about the right amount. What is the 95% confidence interval for the proportion of all American adults who feel that America is doing about the right amount to protect the environment. Group of answer choices (0.352, 0.378) (0.343, 0.387) (0.339, 0.391) (0.331, 0.400)
Answer:
(0.339, 0.391)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 1295, \pi = \frac{473}{1295} = 0.365[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.365 - 1.96\sqrt{\frac{0.365*0.635}{1295}} = 0.339[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.365 + 1.96\sqrt{\frac{0.365*0.635}{1295}} = 0.391[/tex]
So the correct answer is:
(0.339, 0.391)
An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. These classes are open to any of the 86 students in the school. There are 38 in the Spanish class, 27 in the French class, and 16 in the German class. There are 14 students that in both Spanish and French, 6 are in both Spanish and German, and 5 are in both French and German. In addition, there are 2 students taking all 3 classes. If one student is chosen randomly, what is the probability that he or she is taking at least one language class
Answer: The required probability is 0.674.
Step-by-step explanation:
Since we have given that
Number of students in the Spanish class = 38
Number of students in the French class = 27
Number of students in the German class = 16
Number of students in both spanish and French = 14
Number of students in both Spanish and German = 6
Number of students in both French and German = 5
Number of students in all three class = 2
So, it becomes:
[tex]n(S\cup F\cup G)=n(S)+n(F)+n(G)-n(S\cap F)-n(G\cap F)-n(S\cap G)+n(S\cap F\cap G)\\\\n(S\cup F\cup G)=38+27+16-14-5-6+2=58[/tex]
So, Probability that he or she is taking at least one language class is given by
[tex]\dfrac{58}{86}=0.674[/tex]
Hence, the required probability is 0.674.
A swimming pool is circular with a 20-ft diameter. The depth is constant along east-west lines and increases linearly from 1 ft at the south end to 6 ft at the north end. Find the volume of water in the pool. (Round your answer to the nearest whole number.) ft3
Answer:
[tex]1100 \text{feet}^3[/tex]
Step-by-step explanation:
GIVEN: A swimming pool is circular with a [tex]20\text{ feet}[/tex] diameter. The depth is constant along east-west lines and increases linearly from [tex]1\text{ feet}[/tex] at the south end to [tex]6\text{ feet}[/tex] at the north end.
TO FIND: Find the volume of water in the pool.
SOLUTION:
Consider the image attached.
when two similar figures are attached a new cylinder is formed. volume of swimming pool is half of volume of new cylinder formed.
radius of new cylinder [tex]=\frac{\text{diameter}}{2}=\frac{20}{2}=10\text{ feet}[/tex]
height of new cylinder [tex]=6+1=7\text{ feet}[/tex]
volume of cylinder [tex]=\pi r^2h=\frac{22}{7}\times(10)^27[/tex]
[tex]=2200\text{ feet}^3[/tex]
Volume of swimming pool [tex]=\frac{\text{volume of cylinder}}{2}=\frac{2200}{2}[/tex]
[tex]=1100\text{ feet}^3[/tex]
Hence volume of water in the pool is [tex]1100 \text{feet}^3[/tex].
Final answer:
To find the volume of water in a swimming pool with varying depths, calculate the volume by considering different height sections of the pool.The volume of water in the pool is approximately 13,287 ft³.
Explanation:
A swimming pool is circular with a 20-ft diameter. The depth is constant along east-west lines and increases linearly from 1 ft at the south end to 6 ft at the north end.
Given the pool's dimensions, you can calculate the volume of water in the pool by breaking it into sections of different heights.
The volume of water in the pool is approximately 13,287 ft³.
Find the average temperature on that part of the plane 2x + 5y +z = 9 over the square |x| <= 1, y <= 1, where the temperature is given by T(x,y,z) = e^-z. The average value is :_______
Answer:
This question answer is attached in the attachment,
Step-by-step explanation:
What is the height of a cylinder with a volume of 384 pi cubic inches and a radius of 8 inches? Round to the nearest tenth of an inch.
inches___
The height of the cylinder with a volume of 384π cubic inches and a radius of 8 inches is 6 inches.
To find the height of a cylinder with a known volume and radius, we use the formula for the volume of a cylinder, which is V = πr²h, where 'V' is the volume, 'r' is the radius, and 'h' is the height. We are given a volume of 384π cubic inches and a radius of 8 inches. Plugging these values into the formula gives us:
384 π = π (8²) h
384 π = 64π h
h = 384 π ÷ 64π
h = 6 inches
Therefore, the height of the cylinder is 6 inches.
What percentage of the data values falls between the values of 27 and 45 in the data set shown?
A box-and-whisker plot. The number line goes from 25 to 50. The whiskers range from 27 to 48, and the box ranges from 32 to 45. A line divides the box at 36.
25%
50%
75%
100%
Answer:
the answer is 75% hope that helps
Step-by-step explanation:
i did the test and i got it right
The required percentage of the data values falls between the values of 27 and 45 in the data set shown is 75%.
What is a box plot?A straightforward method of expressing statistical data on a plot in which a rectangle is drawn to represent the second and third quartiles, with a vertical line inside to indicate the median value. Horizontal lines on both sides of the rectangle show the lower and upper quartiles.
Based on the box-and-whisker plot, we can see that the majority of the data values fall between the values of 32 and 45, and the median value is 36.
To find the percentage of data values that fall between 27 and 45, we can estimate by visually analyzing the plot. Since the whiskers go from 27 to 48, we can assume that all data values between 27 and 48 are included. Then, we need to estimate how much of the data falls between 32 and 45, which is the range of the box.
Since the box takes up most of the range of values between 32 and 45, we can estimate that around 75% of the data values fall within this range. Therefore, the percentage of data values that fall between 27 and 45 can be estimated as around 75%.
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the four-digit numeral 3AA1 is divisible by 9. What digit does A represent?
Answer:
7
Step-by-step explanation:
Trial and error. Start with 1 and go to 9
The digit A in the numeral 3AA1, which is divisible by 9, represents the number 7. This is deduced by the rule that a number is divisible by 9 if the sum of its digits is also divisible by 9.
To find the digit that A represents in the four-digit numeral 3AA1 that is divisible by 9, we must remember that a number is divisible by 9 if the sum of its digits is divisible by 9. Let's denote A as a single digit, and since we know that 3 and 1 are already part of the sum, the equation we need to solve is:
3 + A + A + 1 = 3 + 2A + 1
Simplifying further:
2A + 4 must be divisible by 9. Since A is a digit, it can only be one of the following numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. We can now test these values:
For A = 0, the sum is 4 (not divisible by 9)
For A = 1, the sum is 6 (not divisible by 9)
For A = 2, the sum is 8 (not divisible by 9)
For A = 3, the sum is 10 (not divisible by 9)
For A = 4, the sum is 12 (not divisible by 9)
For A = 5, the sum is 14 (not divisible by 9)
For A = 6, the sum is 16 (not divisible by 9)
For A = 7, the sum is 18 (divisible by 9)
For A = 8, the sum is 20 (not divisible by 9)
For A = 9, the sum is 22 (not divisible by 9)
The only value for A that makes the sum divisible by 9 is 7. Therefore, A represents the digit 7.
A university planner is interested in determining the percentage of spring semester students who will attend summer school. She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school. (7 points) What is the proportion of semester students returning to summer school
Answer:
The proportion of semester students returning to summer school
p = 0.35
Step-by-step explanation:
Explanation:-
Given data A university planner is interested in determining the percentage of spring semester students who will attend summer school.
She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school
Given the sample size 'n' = 160
let 'x' = 56
Sample proportion of semester students returning to summer school
[tex]p = \frac{x}{n} = \frac{56}{160}[/tex]
p = 0.35
the difference of two numbers is 1. their sum is 19. find the numbers.
Answer:
x = 10
y = 9
Step-by-step explanation:
First, you need to write two equations that represent the information given. I'm going to set the two numbers and x and y.
x - y = 1
x + y = 19
This turns it into a systems of equations. When you add the two equations, keeping everything on each side, you eliminate a variable, and solve for the other.
2x = 20
x = 10
Now, you'd plug x into either original equation.
(10) - y = 1
-y = -9
x = 10
I hope this helped!
Answer:
The numbers are 9 and 10. Add to get a sum of 19, but subtract to get a difference of 1.
Step-by-step explanation:
A department store wants to know what fraction of its customers in a certain market have store credit cards, and what their average balance might be. Of the 120 customers surveyed, 30 had store credit cards. Amongst those customers, the average balance was $600, with a sample standard deviation of $80. Find the 98% confidence interval for the average credit card balance. ( , ) Round your answers to two decimal places.
Answer:
The 98% confidence interval for the average credit card balance is
(564.04, 635.96).
Step-by-step explanation:
We have to calculate the 98% confidence interval on the average credit card balance.
The sample will consist of the n=30 customers that have credit card.
The sample has a mean of $600 and a standard deviation of $80.
As the population standard deviation is estimated from the sample standard deviation, we will use a t statistic.
The degrees of freedom are:
[tex]df=n-1=30-1=29[/tex]
The critical value for a 98% CI and 29 degrees of freedom is t=2.463 (this can be looked up in a t-table).
Then, the margin of error is:
[tex]E=t\cdot s/\sqrt{n}=2.463*80/\sqrt{30}=197.04/5.48=35.96[/tex]
Then, the upper and lower bounds of the confidence interval are:
[tex]LL=\bar X-E=600-35.96=564.04\\\\UL=\bar X+E=600+35.96=635.96[/tex]
The point (-3,1) is on the terminal side of the angle in standard position. What is tanθ
Answer:
[tex] \tan \theta = - \frac{1}{3}[/tex]
Step-by-step explanation:
Since, point (-3, 1) is on the terminal side of the angle in standard position.
[tex] \therefore \: ( - 3, \: 1) = (x, \: y) \\ \therefore \:x = - 3 \: \: and \: \: y = 1 \\ \because \tan \theta = \frac{y}{x} \\ \therefore \:\tan \theta = \frac{1}{ - 3} \\ \\ \huge \red{ \boxed{\therefore \:\tan \theta = - \frac{1}{3} }}[/tex]
8. A well-balanced stock market portfolio will often experience an exponential growth. A
particular investor with a well-balanced stock market portfolio records the portfolio balance
every month, in thousands of dollars, from the date of investment. The roughly exponential
growth can be transformed to a linear model by plotting the natural log of the balances
versus time, in months, where t - 0 represents the date the money was invested. The linear
regression equation for the transformed data is
In (balance)-5.550 + 0.052t.
Using this equation, what is the predicted balance of the portfolio after 2 years (24 months)?
(A) $5,654
(B) $6,798
(C) $285,431
(D) $896,053
(E) $948,464
Answer:
D) $896,053
Step-by-step explanation:
you just bought 4 bags of chips and 3 cans of soda and paid $9.75. As it is being bagged, you notice one bag of chips has expired. Instead of choosing replacements, you decide to get a refund. You get $2.75 back. How much does one bag of chips cost? How much does one can of soda cost?
Answer:
50 cents or 0.5
Step-by-step explanation:
I need help please.
Answer:2
4 th root of 1024 is 5
Since 4 x4x4x4x4=1024
So 1+2x=5
-1 PN both sides
2x= 4
Divide by 2
X=2
Step-by-step explanation:
A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote the number of defectives among the four sold. The purchaser of the items will return the defectives for repair, and the repair cost is given by 2 C X X = + + 3 2 1. Find the expected repair cost
Answer:
The expected repair cost is $3.73.
Step-by-step explanation:
The random variable X is defined as the number of defectives among the 4 items sold.
The probability of a large lot of items containing defectives is, p = 0.09.
An item is defective irrespective of the others.
The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.
The repair cost of the item is given by:
[tex]C=3X^{2}+X+2[/tex]
Compute the expected cost of repair as follows:
[tex]E(C)=E(3X^{2}+X+2)[/tex]
[tex]=3E(X^{2})+E(X)+2[/tex]
Compute the expected value of X as follows:
[tex]E(X)=np[/tex]
[tex]=4\times 0.09\\=0.36[/tex]
The expected value of X is 0.36.
Compute the variance of X as follows:
[tex]V(X)=np(1-p)[/tex]
[tex]=4\times 0.09\times 0.91\\=0.3276\\[/tex]
The variance of X is 0.3276.
The variance can also be computed using the formula:
[tex]V(X)=E(Y^{2})-(E(Y))^{2}[/tex]
Then the formula of [tex]E(Y^{2})[/tex] is:
[tex]E(Y^{2})=V(X)+(E(Y))^{2}[/tex]
Compute the value of [tex]E(Y^{2})[/tex] as follows:
[tex]E(Y^{2})=V(X)+(E(Y))^{2}[/tex]
[tex]=0.3276+(0.36)^{2}\\=0.4572[/tex]
The expected repair cost is:
[tex]E(C)=3E(X^{2})+E(X)+2[/tex]
[tex]=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73[/tex]
Thus, the expected repair cost is $3.73.
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insucient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% con dence interval for the di erence between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.
Answer:
[tex](0.08-0.0888) - 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= -0.0175[/tex]
[tex](0.08-0.0888) + 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= 0.0015[/tex]
And the 95% confidence interval would be given (-0.0175;0.0015).
We are confident at 95% that the difference between the two proportions is between [tex]-0.0175 \leq p_A -p_B \leq 0.0015[/tex]
And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]p_A[/tex] represent the real population proportion for California
[tex]\hat p_A =0.08[/tex] represent the estimated proportion for California
[tex]n_A=11545[/tex] is the sample size required for California
[tex]p_B[/tex] represent the real population proportion for Oregon
[tex]\hat p_B =0.088[/tex] represent the estimated proportion for Brand B
[tex]n_B=4691[/tex] is the sample size required for Oregon
[tex]z[/tex] represent the critical value for the margin of error
Solution to the problem
The sample proportion have the following distribution
[tex]\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.08-0.0888) - 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= -0.0175[/tex]
[tex](0.08-0.0888) + 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= 0.0015[/tex]
And the 95% confidence interval would be given (-0.0175;0.0015).
We are confident at 95% that the difference between the two proportions is between [tex]-0.0175 \leq p_A -p_B \leq 0.0015[/tex]
And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 29 tires surveyed, the mean lifespan was 45,800 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim?
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ ≥ 50000
For the alternative hypothesis,
µ < 50000
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = lifetime of the tyres
µ = mean lifetime
σ = standard deviation
n = number of samples
From the information given,
µ = 50000 miles
x = 45800 miles
σ = 8000
n = 29
z = (50000 - 45800)/(8000/√29) = - 2.83
Looking at the normal distribution table, the probability corresponding to the z score is 0.9977
Since alpha, 0.05 < than the p value, 0.9977, then we would accept the null hypothesis. Therefore, At a 5% level of significance, the data is not highly consistent with the claim.
4 identical taps can fill a tank in 70 minutes. Find the time taken for 7 such taps to fill the same tank.
Answer:
4 taps take 70 minutes
2 taps would take 140 minutes
8 taps would take 35 minutes
time = (4 / # of taps) * 70
time = (4 / 7) * 70 = 40 minutes
Step-by-step explanation:
Four-legged animals run with two different types of motion: trotting and galloping. An animal that is trotting has at least one foot on the ground at all times, whereas an animal that is galloping has all four feet off the ground at some point in its stride. The number of strides per minute at which an animal breaks from a trot to a gallop depends on the weight of the animal.
1. Use the table and the method of this example to find an equation that relates an animal's weight x (in pounds) and its lowest galloping speed y (in strides per minute).
weight, X 25 35 50 75 500 1000
Galloping speed, Y 191.5 184.7 175.8 162.2 124.9 113.2
Answer:
Galloping speed Y=179.984-0.076 weight X
Step-by-step explanation:
The required equation is
y=a+bx
where y is galloping speed and x is weight of animal.
We have to estimate intercept a and slope b.
x y xy x²
25 191.5 4787.5 625
35 184.7 6464.5 1225
50 175.8 8790 2500
75 162.2 12165 5625
500 124.9 62450 250000
1000 113.2 113200 1000000
sumx=1685
sumy=952.3
sumxy=207857
sumx²=1259975
[tex]slope=b=\frac{n(sumxy)-(sumx)(sumy)}{nsumx^{2}-(sumx)^{2} }[/tex]
[tex]slope=b=\frac{6(207857)-(1685)(952.3)}{6*(1259975)-(1685)^{2} }[/tex]
[tex]slope=b=\frac{-357483.5}{4720625}[/tex]
slope=b=-0.07573.
intercept=a=ybar-b*xbar
[tex]ybar=\frac{sumy}{n}[/tex]
[tex]ybar=\frac{952.3}{6}[/tex]
ybar=158.7167
[tex]xbar=\frac{sumx}{n}[/tex]
[tex]xbar=\frac{1685}{6}[/tex]
xbar=280.8333
intercept=a=ybar-b*xbar
intercept=a=158.7167-(-0.07573)*280.8333
intercept=a=158.7167+21.2675
intercept=a=179.9842
rounding intercept and slope to 3 decimal places
intercept=179.984.
slope=-0.076
The required equation is
Galloping speed Y=179.984-0.076 weight X
The required equation is galloping speed Y = 179.984 - 0.076 weight X.
Given that,
Four-legged animals run with two different types of motion: trotting and galloping.
The number of strides per minute at which an animal breaks from a trot to a gallop depends on the weight of the animal.
We have to determine,
An equation that relates an animal's weight x (in pounds) and its lowest galloping speed y (in strides per minute).
According to the question,
A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation.
The required linear equation is,
y = a + bx
Where y is galloping speed and x is weight of animal.
To estimate intercept a and slope b from the table showing below.
x y x.y x²
25 191.5 4787.5 625
35 184.7 6464.5 1225
50 175.8 8790 2500
75 162.2 12165 5625
500 124.9 62450 250000
1000 113.2 113200 1000000
Then,
sum of x = 1685 , sum of y = 952.3 , sum of xy = 207857 , sum of x² = 1259975.
Therefore, Slope of b is given by the formula;
[tex]b = \dfrac{(sum\ of \ xy )- (sumx).(sumy)}{n(sumx^{2})- (sumx)^{2} }\\\\b = \dfrac{6(207857)-1685(952.3)}{6(1259975)-(1685)^{2} }\\\\b = \dfrac{-357483.5}{4720625}\\\\b = -0.7573[/tex]
Then,
[tex]Y_b_a_r = \dfrac{sumx}{n} = \dfrac{952.3}{6} = 158.71\\\\X_b_a_r = \dfrac{sumy}{n} = \dfrac{1685}{6} = 280.83[/tex]
Therefore, The intercept is given by,
[tex]intercept = a = Y_b_a_r-b\times X_b_a_r\\\\intercept = a = 158.7167 - (-0.07573)\times 280.8333\\\\intercept = a = 158.7167 + 21.2675\\\\intercept = a = 179.984[/tex]
On rounding intercept and slope to 3 decimal places,
intercept = 179.984.
slope = -0.076
Hence, The required equation is galloping speed Y = 179.984 - 0.076 weight X.
To know more about Linear equation click the link given below.
https://brainly.com/question/14824234
Suppose samples of size 100 are drawn randomly from a population that has a mean of 20 and a standard deviation of 5. What are the values of the mean and the standard deviation of the distribution of the sample means
Answer:
a) The mean of the sampling distribution of means μₓ = μ = 20
b) The standard deviation of the sample σₓ⁻ = 0.5
Step-by-step explanation:
Explanation:-
Given sample size 'n' =100
Given mean of the Population 'μ' = 20
Given standard deviation of population 'σ' = 5
a) The mean of the sampling distribution of means μₓ = μ
μₓ = 20
b) The standard deviation of the sample σₓ⁻ = [tex]\frac{S.D}{\sqrt{n} }[/tex]
Given standard deviation of population 'σ' = 5
= [tex]\frac{5}{\sqrt{100} } = 0.5[/tex]
Final answer:
The mean of the distribution of the sample means drawn from a population with a mean of 20 and a standard deviation of 5, using samples of size 100, is 20. The standard deviation of this distribution, also known as the standard error, is 0.5.
Explanation:
The question involves understanding the concept of the distribution of sample means, also known as the sampling distribution. When samples of size 100 are drawn from a population with a mean (μ) of 20 and a standard deviation (σ) of 5, the mean of the distribution of the sample means will be the same as the population mean, which is 20. However, the standard deviation of the distribution of sample means, known as the standard error (SE), will be the population standard deviation divided by the square root of the sample size (√n), which in this case is 5/√100 = 0.5.
Therefore, the mean of the distribution of the sample means is 20 and the standard deviation (standard error) of this distribution is 0.5. This is based on the central limit theorem which states that, for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed regardless of the population’s distribution, with these exact parameters of mean and standard deviation.
The exact numbers of each color of Umbrella Corporation's most popular candy, W&Ws, naturally vary from bag to bag, but the company wants to make sure that the process as a whole is producing the fve colors in equal proportions. Which of the following would represent the alternative hypothesis in a chi squared goodness-of-fost conducted to determine if the five colors of W&Ws occur in equal proportions? My: Al of the proportions are different from 0.20 MP1 P2 =3 =p4 P5 - 0.20 0 Hy: At least two of the mean number of colors differ from one another He the number of candies per bag and candy color are dependent Hy: At least one proportion is not equal to 0 20 5.5556
Question in order:
The exact numbers of each color of Umbrella Corporation's most popular candy, W&Ws, naturally vary from bag to bag, but the company wants to make sure that the process as a whole is producing the five colors in equal proportions. Which of the following would represent the alternative hypothesis in a chi-squared goodness-of-fit test conducted to determine if the five colors of W&Ws occur in equal proportions?
a. H1: All of the proportions are different from 0.20
b. HI: p1 = P2 = p3 =P4=p5 = 0.20
c. H1: At least one proportion is not equal to 0.20
d. H1: At least two of the mean number of colors differ from one another.
e. HI: the number of candies per bag and candy color are dependent
Answer:
option C
Step-by-step explanation:
H1: At least one proportion is not equal to 0.20
I swear they got me doing algebra in 8th grade i wanna refund please help me... again
Answer: C represents a function
Step-by-step explanation: It represents a function because there is exactly one range for each domain, and only one. In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power.
Suppose task B follows task A. The likelihood that a problem will occur with task A is 0.5, and the time to correct the problem would be 2 months. The likelihood that a problem will occur with task B is 0.2, and the time to correct the problem would be 3 months. The risk consequence of both activities is 3.5 months 5.7 months 1.6 months cannot be determined. True or false?
Answer:
The risk consequence of both activities is 1.6 months.
Step-by-step explanation:
We can define a randome variable D: total delay time for the project, and calculate its expected value.
This would be the risk consequence of both activities.
The expected delay time for the project is the sum of the expected delay for task A plus the expected delay for task B. It is assumed the likelihood of a problem in any task is independent of the other.
Then, each expected delay for a task is equal to the probability of a problem multiplied by the consequence (delay time).
[tex]E(D)=E(D_a)+E(D_b)=p_aT_a+p_bT_b\\\\E(D)=0.5*2+0.2*3=1+0.6=1.6[/tex]
The risk consequence of both activities is 1.6 months.