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A sample of 8 new models of automobiles provides the following data on highway miles per gallon. Highway Miles Model Per Gallon 1 33.6 2 26.8 3 20.2 4 38.7 5 35.1 6 28.0 7 26.2 8 27.6 a. What is the point estimate for the average highway miles per gallon for all new models of autos? b. Determine the point estimate for the standard deviation of the population.

Answers

Answer 1

Answer:

Answer:

The point estimate for the mean is 29.5 miles per gallon and the standard deviation is 5.9

Step-by-step explanation:

The formula por the point estimate of the mean is:

$mean(x) = (x1+x2+x3+...+xn)/(n)$

And for the standard deviation:

$desv(x)=((1)/(n-1)sum(xi^(2) - n*x^(2) ))^(1/2)$

So for the mean:

$mean(x)=(33.6+26.8+20.2+38.7+35.1+28.0+26.2+27.6 )/(8) \n\nmean(x)=(236.2)/(8) \n\nmean(x)=29.5$

And for the standard deviation:

$desv(x)=((1)/(8-1) ((33.6-29.5)^(2) +(26.8-29.5)^(2) +(20.2-29.5)^(2) +(38.7-29.5)^(2) +(35.1-29.5)^(2) +(28-29.5)^(2) +(26.2-29.5)^(2) +(27.6-29.5)^(2))^(1/2) \n\ndesv(x)=((1)/(8-1) ((4.1)^(2) +(-2.7)^(2) +(-9.3)^(2) +(9.2)^(2) +(5.6)^(2) +(-1.5)^(2) +(-3.3)^(2) +(-1.9)^(2))^(1/2) \n\ndesv(x)=\sqrt{(243.3)/(7)} \n\ndesv(x)=√(34.7) \n\ndesv(x)=5.9$

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16 is what percent of 20?
A bag contains some number of marbles. It is known that 20 of them are red. When 15 marbles are drawn, without replacement, we get 6 red. Assuming E(X)=6 red, what is the total number of marbles in the bag?
The cost for a pack of 9 padlock is $25.20 find the unit price in dollars per padlock if necessary round your answer to the nearest cent.
Another rock in Death Valley at the Racetrack traveled 55.3 meters during the observation period of nine years. a . What was the rock's average speed in feet per year? Round your answer to the nearest whole number. feet per year b . If a strong wind gives a stone an initial velocity of v 0 = 11 m / s , and the friction coefficient of stone on ice is μ = 0.14 , and g = 9.81 m / s 2 . Use the equation v 2 0 = 2 μ g S to find the stopping distance S for the stone. Round your answer to the nearest tenth.

Evaluate the function f(x)=4•7^x for x=-1 and x2 show your work

Answers

When you evaluate the function f (x) = 4 • 7 ^ x for x = -1 you get:
f (-1) = 4 * 7 ^ -1
f(-1) = 4* 1/7
f (-1) = 0.5714
The next part of the question is not clear. If it refers to the function at x = 2 then:
f (2) = 4 * 7 ^ (2)
f(2) =4*49
f (2) = 196
If it refers to it in x ^ 2
f (x ^ 2) = 4 * 7 ^ (x ^ 2)

the equation of line AB is y = -1/4 x - 2. write an equation of a line parallel to line AB in slope intercept form that contains point (-3,2)

Answers

The general equation of a straight line is in the form of;
Y=MX + c,
Where m is the gradient and
C the y-intersept.
Parallel line have the same gradient.
From the equation,
Y=-1/4X - 2,
Gradient = - 1/4.

Now assuming that the second line passes through (x,y) and (-3,2).
(Y-2)/(X--3) =-1/4
4(Y-2)=-1(X+3)
4Y-8 = -X-3
4Y = -X -3+8
4Y = -X +5
Y = -1/4X +5/4

Answer: y = -(1/4)x + 5/4

Explanation:

1) Parallel lines have equal slopes.

2) The slope-intercept form of the equation of the line is y = mx + b, where m is the slope and b is the y-intercept.

3) Hence, the slope of the given equation, i.e. y = (-1/4)x - 2 is m = -1/4

4) Therefore, that is the same slope of any line parallel to it, m = -1/4

5) The slope-point equation of the line is:

y - y₁ = m (x - x₁)

6) Now you replace -1/4 for m, and the point (-3,2)

⇒ y - 2 = [ -1/4) (x - (-3) ]

7) Expand the product and simplify:

y - 2 = -x/4 - 3/4
y = -x/4 -3/4 + 2
y = -x/4 + 5/4

And that is the equation requested

Mike won 7 of his wrestling matches, lost 6 matches, and tied 2 matches. What percent of all of Mike's matches did he win?

Answers

Mike won 46.67% of all of his matches.

What is a percentage?

A ratio or value that may be stated as a fraction of 100 is called a percentage. Moreover, it is indicated by the symbol "%."

The total number of matches that Mike wrestled is:

Total matches = number of matches won + number of matches lost + number of matches tied

Total matches = 7 + 6 + 2

Total matches = 15

To find the percentage of matches that Mike won, we can use the formula:

Percentage = (Number of matches won / Total number of matches) x 100%

Plugging in the values we know, we get:

Percentage = (7 / 15) x 100%

Percentage = 0.4667 x 100%

Percentage = 46.67%

Therefore, the Percentage = 46.67%.

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Answer:

Mike won 46.67% of his matches

Step-by-step explanation:

Mike participated in a total of 7 + 6 + 2 = 15 matches.

The percentage of matches that Mike won can be found by diving the matches he won by the total number of matches and multiplying it by 100 to convert the decimal to a percentage.

(7/15) x 100% ≈ 46.67%

Therefore, Mike won approximately 46.67% of his matches.

Dy/dx=2xy/x²+y² solve

Answers

Answer:

$y=(-1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)\ or \ (1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)$

Step-by-step explanation:

As it is first order nonlinear ordinary differential equation

Let y(x) = x v(x)

2xy/(x²+y²)=2v/(v^2+1)

dy=xdv+vdx

dy/dx=d(dv/dx)+v

x(dv/dx)+v=(2v)/(v^2+1)

dv/dx=[(2v)/(v^2+1)-v]/x

$(dv)/(dx)=(-(v^2-1)v)/(x(v^2+1))$

$(v^2+1)/((v^2-1)v)dv=(-1)/(x)dx$

∫ $(v^2+1)/((v^2-1)v)dv$ = ∫ $(-1)/(x)dx$

u=v^2

du=2vdv

Left hand side:

∫ $(v^2+1)/(v(v^2-1))dv$

=∫ $(u+1)/(2u(u-1))du$

= $(1)/(2)$ ∫ $((2)/(u-1) -(1)/(u) )du$

= $ln(u-1)-(ln(u))/(2) +c$

= $ln(v^2-1)-(ln(v^2))/(2)+c$

Right hand side:

$=-ln(x)$

Solve for v:

$v=-\frac{-c_(1)+\sqrt{c_(2)+4x^2 } }{2x} \ or \ \frac{-c_(1)+\sqrt{c_(2)+4x^2 } }{2x}\n$

$y=(-1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)\ or \ (1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)$

Every morning Jack flips a fair coin ten times. He does this for anentire year. LetXbe the number of days when all the flips come out the same way(all heads or all tails).(a) Give the exact expression for the probabilityP(X >1).(b) Is it appropriate to approximateXby a Poisson distribution

Answers

Answer:

See the attached picture for detailed answer.

Step-by-step explanation:

See the attached picture for detailed answer.

Final answer:

The probability question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.

Explanation:

The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact expression for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.

For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial distribution and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.

For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate approximation. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.

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A consumer group was interested in comparing the operating time of cordless toothbrushes manufactured by two different companies. Group members took a random sample of 18 toothbrushes from Company A and 15 from Company B. Each was charged overnight and the number of hours of use before needing to be recharged was recorded. Company A toothbrushes operated for an average of 119.7 hours with a standard deviation of 1.74 hours; Company B toothbrushes operated for an average of 120.6 hours with a standard deviation of 1.72 hours.1. which of the following statements is true?A. This is a one tailed test of two dependent samplesB. This is a two tailed test of two independent samplesC. This is a one tailed test of two independent samplesD. These samples are matchedE. None of the above

Answers

Answer: Lyrics B

Step-by-step explanation:

The investigation about the operating time of cordless toothbrushes is in first place associated to two tails experiment since investigation call for evalution of values under and above any given value (mean value). On the other side investigation of two different manufactures implies totally independent samples, unless these two companies have a commercial relationship which is not express in the problem ststement. Therefore the answer is lyrics B

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