Answer:
A. 18.21 m/s or 65.6 km/h
B. 135.5 km/h
Step-by-step explanation:
A. The vertical speed of the grenade is ...
vv = v·sin(45°) = v/√2
The time in air is given by the equation for ballistic motion:
h = -4.9t² +vv·t = 0
t(vv -4.9t) = 0 . . . . . factor out t
v/√2 = 4.9t
v = t·4.9√2 . . . . . . solve for v
__
The horizontal distance the grenade travels in t seconds is ...
d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters
1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...
x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds
We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...
d = x
4.9t² = (70/9)t +13.4
4.9(t² -(100/63)t) = 13.4
4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486
t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds
This corresponds to a launch speed of ...
v = (4.9√2)t ≈ 18.21 . . . . meters/second
__
B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...
(81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h
The magnitude is the Pythagorean sum:
ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h
The initial velocity Indy must throw the grenade at is found using the equations of motion, and it must be around 14.5 m/s. The magnitude of the velocity of the grenade relative to earth can be found by adding this initial velocity to the velocity of Indy's car, resulting in approximately 37 m/s.
Explanation:This is a physics problem involving the concept of relative velocity. We know that when Indy lets go of the grenade, it will move at the same speed as his car (81.0 km/h) relative to the ground plus the additional speed he gives it. Let's assume Indy throws the grenade when the enemy car is exactly at this distance.
A. If the grenade is thrown at an angle of 45 degrees above the horizontal, we can apply the equations of motion to find the initial velocity of the grenade. Using the equation, Range = [v^2 * sin(2*angle)]/g, we can solve for v, which should be around 14.5 m/s.
B. The magnitude of the velocity of the grenade relative to earth would be the vector sum of the initial velocity of the grenade and the velocity of Indy's car. Converting the car's speed to m/s (81.0 km/h = 22.5 m/s) and using vector addition, we can add 22.5 m/s to the initial speed of the grenade (14.5 m/s) to receive a total of approximately 37 m/s.
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For students who first enrolled in two year public institutions in a recent semester, the proportion who earned a bachelor's degree within six years was 0.398 . The president of a certain college believes that the proportion of students who enroll in her institution have a lower completion rate.
(a) Determine the null and alternative hypotheses.
(b) Explain what it would mean to make a Type I error.
(c) Explain what it would mean to make a Type II error.
Answer:
a) Null hypothesis: [tex]p \geq 0.368[/tex]
Alternative hypothesis: [tex]p<0.368[/tex]
b) A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.
c) A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Part a
On this case we want to test if the proportion of students who enroll in her institution have a lower completion rate (0.398), so the system of hypothesis would be:
Null hypothesis: [tex]p \geq 0.368[/tex]
Alternative hypothesis: [tex]p<0.368[/tex]
Part b
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.
Part c
A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.
The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion. Making a Type I error would mean wrongly rejecting the null hypothesis, while making a Type II error would mean failing to reject the null hypothesis when it is actually false.
Explanation:(a) The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion of 0.398. The alternative hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is lower than 0.398.
(b) Making a Type I error in this context would mean rejecting the null hypothesis when it is actually true. This would imply that the college has a lower completion rate when in fact it does not.
(c) Making a Type II error in this context would mean failing to reject the null hypothesis when it is actually false. This would imply that the college has the same completion rate as the national proportion when it actually has a lower completion rate.
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It is known that x=7 is a root of the equation ax^2+bx+2=0, where a<0. Solve the inequality ax^4+bx^2+2>0.
Answer:
|x| < √7
Step-by-step explanation:
The product of the roots of the given quadratic equation is 2/a, so the other root (the one not given) is 2/(7a). It will be negative, since "a" is negative.
The roots of ax^2 +bx +2 = 0 are the values of x^2 that satisfy ax^4 +bx^2 +2 = 0. That is, roots of the latter equation will be the square root of the roots of the former equation.
We know that two of the zeros of the quartic are ±√7, and the other two are complex, as they are the square roots of a negative number. So, the graph of the quartic opens downward (because a < 0), and has real zeros at x=±√7.
The solution to the inequality must be ...
-√7 < x < √7
_____
The graph shows an example of the quadratic (green) and quartic (black). The ripple in the quartic changes amplitude with different values of "a", but the locations of the zeros do not change.
Write and solve the system of linear equations described by the application and then answer the question. One hundred sixty feet of fencing encloses a rectangular garden on three sides. One side of the garden is the side of a barn and requires no fencing. The longer side is parallel to the barn.
If the length of the longer side of the rectangle is twice the width, what are the dimensions of the garden?
Answer:
80 ft x 40 ft
Step-by-step explanation:
Let 'L' be the length of the longer side and 'W' be the length of the shorter side (or the width).
The equations that compose the linear system are:
[tex]L+2W=160\\L=2W[/tex]
Solving the system:
[tex]L+L=160\\L=80\\W=\frac{L}{2}\\W=40[/tex]
The garden is a rectangle with dimensions 80 ft x 40 ft.
The dimensions of the garden are approximately 53.33 feet in width and 106.66 feet in length according to the given linear equations and conditions.
Explanation:Given that the fencing of 160 feet encloses a rectangular garden on three sides and one side of the rectangle is parallel with the barn, we can derive the linear equations based on these parameters. Let's denote the width of the rectangle as x and the length of the rectangle as 2x, as stated the length is twice the width.
The perimeter of this garden will equal the amount of fencing, which is 160 feet. Because only three sides of the rectangle are enclosed by the fence, the equation for the perimeter becomes 2x + x = 160. Simplifying this, we get 3x = 160. Solving for x, we divide both sides by 3, hence, x = 53.33 ft approximately. The width of the garden is thus around 53.33 feet and the length (being twice the width) would be 2x = 2*53.33 = 106.66 ft.
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INTERPRETATIONS OF THE DEFINITE INTEGRAL: Solar photovoltaic (PV) cells are the world’s fastest growing energy source. In year (t) since 2007, PV cells were manufactured worldwide at a rate of S=3.7e^(0.61t) gigawatts per year. Estimate the total solar energy-generating capacity of the PV cells manufactured between 2007 and 2010. Round your answer to 3 decimal places.
Answer:
31.747 GW
Step-by-step explanation:
The total cell capacity (C) manufactured in the time period is the integral of the rate of manufacture. So, we have ...
[tex]C=\displaystyle\int\limits^3_0 {3.7e^{0.61t}} \, dt=\frac{3.7}{0.61}(e^{0.61\cdot 3}-1)\approx 31.747 \quad\text{GW}[/tex]
About 31.747 GW of generating capacity was manufactured in that time interval.
Can some help with this
Answer:
36
Step-by-step explanation:
multiply the width & the height
Please help if so thank you And explain
Answer:
Less.
Step-by-step explanation:
Most tables are rectangle so hope it helps!
A food safety guideline is that the mercury in fish should be below 1 part per million? (ppm). Listed below are the amounts of mercury? (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 99?% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna? sushi?
a. 0.59
b. 0.68
c. 0.10
d. 0.95
e. 1.23
f. 0.59
g. 0.92
Answer:
can you provide the list?
Step-by-step explanation:
Consider a binomial experiment with n = 10 and p = 0.10. (a) Compute f(0). If required, round your answer to four decimal places. (b) Compute f(2). If required, round your answer to four decimal places. (c) Compute P(x ≤ 2). If required, round your answer to four decimal places. (d) Compute P(x ≥ 1). If required, round your answer to four decimal places. (e) Compute E(x). (f) Compute Var(x) and σ. If required, round Var(x) answer to one decimal place and σ answer to four decimal places. Var(x) = σ =
Answer:
Step-by-step explanation:
Hello!
You have X~Bi (n;ρ)
Where:
n=10
ρ= 0.10
For all asked probabilities you need to use a Binomial distribution table. Remember this table has the information of the cummulative probabilities P(X≤x).
a. f(0) ⇒ P(X=0) = 0.3487
b. f(2) ⇒ P(X=2) ⇒ P(X≤2) - P(X≤1) = 0.9298 - 0.7361 = 0.1937
c. P(X≤2) = 0.9298
d. P(X ≥ 1) = 1 - P(X ≤ 1) = 1 - 0.7361 = 0.2639
e. E(X)= nρ = 10*0.10 = 1
f. V(X)= nρ(1-ρ) = 10*0.1*0.9 = 0.9
σ= √V(X) = √0.9 = 0.9487
I hope it helps!
The binomial experiment depicts that f(0) will be 0.3487.
How to compute the binomial experimentFrom the information given, it can be noted that:
n = 10
p = 0.10
q = 1 - 0.10 = 0.90
f(0) ⇒ P(X=0) = 0.3487
f(2) = P(X=2) ⇒ P(X≤2) - P(X≤1)
= 0.9298 - 0.7361
= 0.1937
P(X≤2) = 0.9298
P(X ≥ 1) = 1 - P(X ≤ 1)
= 1 - 0.7361 = 0.2639
E(X) = nρ
= 10 × 0.10 = 1
V(X) = nρ(1-ρ)
= 10 × 0.1 × 0.9
= 0.9
σ = √V(X) = √0.9
= 0.9487
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A commercial farm uses a machine that packages carrots in eighteen ounce portions. A sample of 7 packages of carrots has a standard deviation of 0.19. Construct the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. Round your answers to two decimal places.
Answer: [tex]0.12< \sigma<0.42[/tex]
Step-by-step explanation:
Confidence interval for standard deviation is given by :-
[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]
Given : Confidence level : [tex]1-\alpha=0.95[/tex]
⇒[tex]\alpha=0.05[/tex]
Sample size : n= 7
Degree of freedom = 6 (df= n-1)
sample standard deviation : s= 0.19
Critical values by using chi-square distribution table :
[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]
Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-
[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex]
[tex]\Rightarrow0.12243< \sigma<0.418400[/tex]
[tex]\approx0.12< \sigma<0.42[/tex]
Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. :
[tex]0.12< \sigma<0.42[/tex]
given f(x)= -x^ {2 }+10x-3, find f(-1)
Answer: f(-) = - 12
Step-by-step explanation:
The function is expressed as
f(x)= -x^2+10x-3
To determine f(-1), we will substitute
x = - 1 into the given function. It becomes
f(-1) = (- 1)^2 + 10(-1) - 3
f(-1) = 1 - 10 - 3
f(-1) = - 12
A national newspaper reported that the state with the longest mean life span is Hawaii, where the population mean life span is 76 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents.72 68 81 93 56 19 78 94 83 8477 69 85 97 75 71 86 47 66 27(i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)x-bar = ______ yrs = ______ yr
Answer:
Mean = 71.4 years
Standard Deviation = 20.64 years
Step-by-step explanation:
We are given the following data set:
72, 68, 81, 93, 56, 19, 78, 94, 83, 84, 77, 69, 85, 97, 75, 71, 86, 47, 66, 27
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1428}{20} = 71.4[/tex]
Sum of squares of differences =
0.36 + 11.56 + 92.16 + 466.56 + 237.16 + 2745.76 + 43.56 + 510.76 + 134.56 + 158.76 + 31.36 + 5.76, 184.96 + 655.36 + 12.96 + 0.16 + 213.16 + 595.36 + 29.16 + 1971.36 = 8100.8
[tex]S.D = \sqrt{\displaystyle\frac{8100.8}{19}} = 20.64[/tex]
Mean = 71.4 years
Standard Deviation = 20.64 years
Given the equation y = -6x + 7, what is the y-
intercept?
Answer: 7
Step-by-step explanation: Linear equations can be written in the form y = mx + b where the multiplier, m, represents the slope of the line and the base, b, represents the y-intercept of the line.
The linear equation y = -6x + 7 has a y-intercept equal to 7.
Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of saturated fat for a sample of 6 brands of stick margarine (solid fat) and for a sample of 6 brands of liquid margarine and obtained the following results: Stick:[26.2,25.6,25.5,26.1,26.5,26.7] Liquid:[16.3,16.4,16.3,16.7,16.6,17.7] We want to determine if there a significant difference in the average amount of saturated fat in solid and liquid fats. What is the test statistic?
Answer:
t= 32.327
Would be a significant difference in the average amount of saturated fat in solid and liquid fats.
Step-by-step explanation:
1) Data given and notation
Stick:[26.2,25.6,25.5,26.1,26.5,26.7]
Liquid:[16.3,16.4,16.3,16.7,16.6,17.7]
[tex]\bar X_{stick}[/tex] represent the mean for the sample Stick
[tex]\bar X_{Liquid}[/tex] represent the mean for the sample Liquid
[tex]s_{stick}[/tex] represent the sample standard deviation for the sample Stick
[tex]s_{Liquid}[/tex] represent the sample standard deviation for the sample Liquid
[tex]n_{stick}=6[/tex] sample size for the group Stick
[tex]n_{Liquid}=6[/tex] sample size for the group Liquid
t would represent the statistic (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
Null Hypothesis :[tex]\mu_{stick}=\mu_{Liquid}[/tex]
Alternative Hypothesis :[tex]\mu_{stick} \neq \mu_{Liquid}[/tex]
If we analyze the size for the samples both are < 30 so for this case we can apply a t test to compare means, and the statistic formula is:
[tex]t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
3) Calculate the statistic
First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:
[tex]\bar X_{stick}=26.10[/tex] [tex]s_{stick}=0.477[/tex]
[tex]\bar X_{Liquid}=16.67[/tex] [tex]s_{Liquid}=0.532[/tex]
And with this we can replace in formula (1) like this:
[tex]t=\frac{26.10-16.67}{\sqrt{\frac{0.477^2}{6}+\frac{0.532^2}{6}}}}=32.327[/tex]
4) Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{stick}+n_{liquid}-2=6+6-2=10[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(10)}>32.327)=1.8x10^{-11}[/tex]
So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average amount of saturated fat in solid and liquid fats.
The consumer demand equation for tissues is given by q = (108 − p)2, where p is the price per case of tissues and q is the demand in weekly sales.
(a) Determine the price elasticity of demand E when the price is set at $32. (Round your answer to three decimal places.) E = Interpret your answer. The demand is going by % per 1% increase in price at that price level.
(b) At what price should tissues be sold to maximize the revenue? (Round your answer to the nearest cent.) $
(c) Approximately how many cases of tissues would be demanded at that price? (Round your answer to the nearest whole number.) cases per week
Answer:
a) -0.842
b) $0
c) 11,664 cases
Step-by-step explanation:
a)
The price elasticity of demand E at a given point [tex]\large (p_1,q_1)[/tex] is defined as
[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)[/tex]
and in this case, it would measure the possible response of tissues demand due to small changes in its price when the price is at [tex]\large p_1[/tex]
When the price is set at $32 the demand is
[tex]\large q=(108-32)^2=5,776 [/tex]
cases of tissues, so
[tex]\large (p_1,q_1)=(32,5776)\Rightarrow \frac{p_1}{q_1}=\frac{32}{5776}=0.00554[/tex]
Also, we have
[tex]\large \frac{\text{d}q}{\text{d}p}=-2(108-p)\Rightarrow \frac{\text{d}q}{\text{d}p}(32)=-2(108-32)=-152[/tex]
hence
[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)=0.00554(-152)=-0.842[/tex]
That would mean the demand is going down about 0.842% per 1% increase in price at that price level.
b)
When the price is $108 the demand is 0, so the price should always be less than $108.
On the other hand, the parabola
[tex]\large q=(108-32)^2=5,776 [/tex] is strictly decreasing between 0 and 108, that means the maximum demand would be when the price is 0.
c)
When the price is 0 the demand is
[tex]\large (108)^2=11,664[/tex] cases
The elasticity for p = 32 is E = -0.84.
The maximum revenue is obtained when p = 32, and the demand for that price is q = 5,184.
How to determine the price elasticity?
It is given by:
[tex]E = \frac{p}{q} *\frac{dq}{dp} (p)[/tex]
if p = $32, then:
q(32) = (108 − 32)^2 = 5,776
and:
dq/dp = -2*(108 - p)
Evaluating that in p = 32 we get:
-2*(108 - 32) = -152
Then the elasticity is:
E = (32/5,776)*-152 = -0.84.
How to maximize the revenue?The revenue is equal to the demand times the price, so we get:
R = p*q = p*(108 - p)^2 = p*(p^2 - 216p + 11,664)
To maximize this we need to find the zeros of the derivation, we have:
R' = 3p^2 - 2*216*p + 11,664
The zeros of that equation are given by:
[tex]p = \frac{432 \pm \sqrt{(-432)^2 - 4*3*11,664} }{2*3} \\\\p = (432 \pm 216)/6[/tex]
Notice that we need to use the smaller of these values, so the demand never becomes zero, then we use:
p = (432 - 216)/6 = 36
(the other root gives p = 108, so that is a minimum).
c) For this price, the number of cases demanded is:
q = (108 - 36)^2 = 5,184
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I need help with 4, 5,6, and 7. Please someone help me!
Answer:
Step-by-step explanation:
g My Notes Water flows from the bottom of a storage tank at a rate of r(t) = 200 − 4t liters per minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30 minutes. 4400 Incorrect: Your answer is incorrect. liters
Answer:
4,200 liters
Step-by-step explanation:
The flow rate is given by:
[tex]r(t) = 200 - 4t[/tex]
Integrating the flow rate expression from t=0 to t=30 minutes, yields the total volume that flows out of the tank during that period:
[tex]\int\limits^{30}_0 {r(t)} = \int\limits^{30}_0 {(200 - 4t} )\, dt \\V=(200t - 2t^2)|_0^{30}\\V= (200*30 -2*30^2)-(200*0 -2*0^2)\\V=4,200\ liters[/tex]
4,200 liters of water flow from the tank during the first 30 minutes.
Consider the following functions.G(x) = 4x2; f(x) = 8x(a)
a. Verify that G is an antiderivative of f.G(x) is an antiderivative of f(x) because f '(x) = G(x) for all x.
A. G(x) is an antiderivative of f(x) because G(x) = f(x) for all x.
B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
C. G(x) is an antiderivative of f(x) because G(x) = f(x) + C for all x.
D. G(x) is an antiderivative of f(x) because f(x) = G(x) + C for all x.
b. Find all antiderivatives of f. (Use C for the constant of integration.)
Answer:
(a) B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
(b) Every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x
Step-by-step explanation:
A function F is an antiderivative of the function f if
[tex]F'(x)=f(x)[/tex]
for all x in the domain of f.
(a) If [tex]f(x) = 8x[/tex], then [tex]G(x)=4x^2[/tex] is an antiderivative of f because
[tex]G'(x)=8x=f(x)[/tex]
Therefore, G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
Let F be an antiderivative of f. Then, for each constant C, the function F(x) + C is also an antiderivative of f.
(b) Because
[tex]\frac{d}{dx}(4x^2)=8x[/tex]
then [tex]G(x)=4x^2[/tex] is an antiderivative of [tex]f(x) = 8x[/tex]. Therefore, every antiderivative of 8x is of the form [tex]4x^2+C[/tex] for some constant C, and every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x.
Final answer:
G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x. The general antiderivative of f(x) = 8x is F(x) = 4x^2 + C.
Explanation:
To determine if G(x) is an antiderivative of f(x), we must consider the derivative of G(x), which is G'(x). If G'(x) is the same function as f(x), then by definition G is an antiderivative of f. Starting with G(x) = 4x2, the derivative is G'(x) = 8x, which is exactly the function f(x). Thus, the correct choice is B: G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
Now, to find all antiderivatives of the function f(x) = 8x, we integrate f(x) to get the general antiderivative, which includes a constant of integration, C. The antiderivative of 8x is F(x) = 4x2 + C, representing all antiderivatives of f(x).
An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 3 km east of the refinery. The cost of laying pipe is $400,000 per km over land to a point P on the north bank and $800,000 per km under the river to the tanks. To minimize the cost of the pipeline, how far from the refinery should P be located? (Round your answer to two decimal places.)
Answer:
P is exactly 3km east from the oil refinery.
Step-by-step explanation:
Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:
[tex]\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}[/tex]
So the cost of laying pipe under water with this distance is
[tex]800000\sqrt{(3 - d)^2 + 4}[/tex]
And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:
[tex]800000\sqrt{(3 - d)^2 + 4} + 400000d[/tex]
We can find the minimum value of this by taking the 1st derivative and set it to 0
[tex]800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0[/tex]
We can move the first term over to the right hand side and divide both sides by 400000
[tex]1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}[/tex]
[tex]\sqrt{(3 - d)^2 + 4} = 6 - 2d[/tex]
From here we can square up both sides
[tex](3 - d)^2 + 4 = (6 - 2d)^2[/tex]
[tex]9 - 6d + d^2 + 4 = 36 - 24d + 4d^2[/tex]
[tex]3d^2-18d+27 = 0[/tex]
[tex]d^2 - 6d + 9 = 0[/tex]
[tex](d - 3)^2 = 0[/tex]
[tex]d -3 = 0[/tex]
d = 3
So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.
A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a 98% confidence interval for σ2.
Answer:
The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
Step-by-step explanation:
1) Data given and notation
[tex]s^2 =16[/tex] represent the sample variance
s=4 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=20 the sample size
Confidence=98% or 0.98
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 mean or variance 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.
The Chi square distribution is the distribution of the sum of squared standard normal deviates .
2) Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=36.191[/tex]
[tex]\chi^2_{1- \alpha/2}=7.633[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]
[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]
So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
How to calculate confidence intervalStandard deviation = √variance = √16 = 4
The z score of 98% confidence interval is 2.326
The margin of error (E) is:
[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]
The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
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Match each item in the lef column with the correct item in the right column. p-value = 0.02 a. do not reject H0 at α = 0.1 p-value = 0.07 b. reject H0 at α = 0.1 but not at 0.05 p-value = 0.3 c. reject H0 at α = 0.05 but not at 0.01 p-value = 0.006 d. reject H0 at α = 0.01
Answer:
Step-by-step explanation:
Hello!
You have some p-values and different decisions that can be made using these p-values.
Remember the decision rule for the p-value method.
If p-value > α, you don't reject the null hypothesis.
If p-value ≤ α, you reject the null hypothesis.
With this in mind, I've arranged the decisions with the given p-values.
a. do not reject H₀ at 10% ⇒ p-value: 0.3
b. do reject H₀ at 10% but not at 5% p-value: ⇒ 0.07
c. do reject H₀ at 5% but not at 1% ⇒ p-value: 0.02
d. do reject H₀ at 1% ⇒ p-value: 0.006
I hope it helps!
Listed below are the amounts of mercury (in parts per million or ppm) found in tuna sushi sampled at 7 different stores in New York City. Construct a 95% confidence interval estimate of the standard deviation of the amounts of mercury in the population (tuna sushi).
Answer:
Amount of mercury in population is in between 0.2841 , 1.1531
Step-by-step explanation:
Step by step explanation is given in the attachments
Ashok arrives at Starbucks at a random time in between 9:00 am and 9:20 am and Melina arrives at Starbucks at a random time in between 9:10 am and 9:30 am. Both stay for exactly 15 minutes. What is the probability that the two of them are in the Starbucks at the exact same time?
Answer:
0.5
Step-by-step explanation:
Since Ashok stays for exactly 15 minutes after his arrival, they will meet if he arrives in the 10 minutes between 9:10 and 9:20 independently of the time Melina arrives.
Since there are 20 minutes from 9:00 and 9:20 and Ashok arrives at a random time, the distribution of his arrival is uniform, so the probability that he arrives between 9:10 and 9:20 is
10/20 = 0.5
A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, the beam warmed to 35degreesF and after another 5 min it was 50degreesF. Use Newton's Law of Cooling to estimate the beam's initial temperature.
Answer:
The beam initial temperature is 5 °F.
Step-by-step explanation:
If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that
[tex]T(t)=T_a+(T_0-T_a)e^{-kt}[/tex]
where [tex]T_a[/tex] is the ambient temperature, [tex]T_0[/tex] is the initial temperature, [tex]t[/tex] is the time and [tex]k[/tex] is a constant yet to be determined.
The goal is to determine the initial temperature of the beam, which is to say [tex]T_0[/tex]
We know that the ambient temperature is [tex]T_a=65[/tex], so
[tex]T(t)=65+(T_0-65)e^{-kt}[/tex]
We also know that when [tex]t=5 \:min[/tex] the temperature is [tex]T(5)=35[/tex] and when [tex]t=10 \:min[/tex] the temperature is [tex]T(10)=50[/tex] which gives:
[tex]T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}[/tex]
[tex]T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}[/tex]
Rearranging,
[tex]35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}[/tex]
[tex]50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}[/tex]
If we divide these two equations we get
[tex]\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}[/tex]
[tex]\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}[/tex]
Now, that we know the value of [tex]k[/tex] we can use it to find the initial temperature of the beam,
[tex]35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5[/tex]
so the beam started out at 5 °F.
Using Newton's Law of Cooling, the initial temperature of the metal beam is estimated to be 5°F.
Estimating the Initial Temperature of the Beam Using Newton's Law of Cooling
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature of its surroundings. Mathematically, it can be expressed as:
dT/dt = -k(T - Ts)where:
dT/dt = rate of change of temperatureT = initial temperature of the beamTs = surrounding temperature (ambient temperature)k = cooling constant which depends on the properties of the objectGiven:
Ambient temperature, Ts = 65°FTemperature after 5 minutes, T1 = 35°FTemperature after another 5 minutes, T2 = 50°FStep-by-Step Solution:
First, express Newton's Law of Cooling in the form: T(t) = Ts + (To - Ts) * e⁻ᵏᵗSubstitute the known values and solve for the initial temperature To:From T1: 35 = 65 + (To - 65) * e⁻⁵ᵏFrom T2: 50 = 65 + (To - 65) * e⁻¹⁰ᵏSolving these equations simultaneously will involve two steps:Step 1: Divide equations to eliminate To from the equations.Step 2: Solve for k and then use it to find To.By dividing: (35 - 65) / (50 - 65) = e⁻⁵ᵏ/ e⁻¹⁰ᵏWhich simplifies to: 25/50 = e⁵ᵏHence, ln(1/2) = 5kTherefore, solving for k: k = ln(1/2) / 5 ≈ -0.1386
Substitute back to find To: 35 = 65 + (To - 65) * e⁻⁵ * ⁻⁰.¹³⁸⁶This simplifies to: 35 = 65 + (To - 65) * 1/2Thus, To - 65 = 2 * (35 - 65)Finally, To = 65 - 60 = 5°F
. Among all the income tax forms filed in a certain year, the mean tax paid was $2000, and the standard deviation was $500. In addition, for 10% of the forms, the tax paid was greater than $3000. A random sample of 625 tax forms is drawn. a. What is the probability that the average tax paid on the sample forms is greater than $1980? b. What is the probability that more than 60 of the sampled forms have a tax of greate
To find the probability that the average tax paid on the sample forms is greater than $1980, standardize the sample mean using the Central Limit Theorem. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, use a normal distribution approximation.
Explanation:To solve this problem, we can use the Central Limit Theorem.
a. To find the probability that the average tax paid on the sample forms is greater than $1980, we need to standardize the sample mean. We calculate the z-score by subtracting the population mean from the sample mean and dividing by the population standard deviation divided by the square root of the sample size. Then, we can use a z-table or a calculator to find the probability.
b. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, we can use a normal distribution approximation. We can calculate the z-score for 60 forms by subtracting the population mean from 60 and dividing by the square root of the population mean multiplied by (1 - the population mean) divided by the sample size. Then, we can use a z-table or a calculator to find the probability.
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The point-slope form of the equation of a nonvertical line with slope, m, that passes through the point (x1,y1) is...?
a. Ax+By=c
b. y-y1=m(x-x1)
c. y1=mx1+b
d. Ax1+by1=C
e. y=mx+b
f. y1-y=m(x-x1)
Please explain why, if you can. Thanks! :)
Answer:
b. y-y1 = m(x-x1)
Step-by-step explanation:
It's a matter of definition. There are perhaps a dozen useful forms of equations for a line. Each has its own name (and use). Here are some of them.
slope-intercept form: y = mx + bpoint-slope form: y -y1 = m(x -x1)two-point form: y = (y2-y1)/(x2-x1)(x -x1) +y1intercept form: x/a +y/b = 1standard form: ax +by = cgeneral form: ax +by +c = 0Adding y1 to the point-slope form puts it in an alternate form that is useful for getting to slope-intercept form faster: y = m(x -x1) +y1. I use this when asked to write the equation of a line with given slope through a point, with the result in slope-intercept form.
The equation of the line, in point-slope form, is given by:
[tex]y - y_1 = m(x - x_1)[/tex]
Option b.
-----------------------------------------
The equation of a line, in point-slope form, is given by:
[tex]y - y_1 = m(x - x_1)[/tex]
In which
m is the slope.The point is [tex](x_1,y_1)[/tex].Nonvertical line means that [tex]m \neq 0[/tex]Thus, the correct option is given by option b.A similar problem is given at https://brainly.com/question/24144915
Maximize and minimize quantities given an expression with two variables Question Find the difference between the maximum and minimum of the quantity p2q250, where p and q are two nonnegative numbers such that p+q=10. (Enter your answer as a fraction.)
To maximize and minimize the quantity given by p^2q^250 with the constraint p+q=10, express q in terms of p, derive a single-variable function, optimize using differentiation, find critical points, and evaluate at the endpoints of the domain [0, 10]. The difference between the maximum and minimum values would be the answer.
Explanation:The student asks how to maximize and minimize the quantity p2q250, given the constraint that p + q = 10 and that p and q are nonnegative. To find the maximum and minimum values of this expression, we can use the concept of optimization in algebra. First, express q in terms of p, using the given equation p + q = 10. We get q = 10 - p. Substitute this into the original expression to obtain a single-variable function f(p) = p2(10 - p)250. Now, we can differentiate this function with respect to p and find its critical points.
To find the difference between the maximum and minimum values, evaluate f(p) at the critical points and at the endpoints of the domain (since p and q have to be nonnegative, the domain is [0, 10]). The required difference is the greatest value of f(p) minus the smallest value of f(p), which is the solution to the student's problem. This can be found by tedious algebra and the final answer would be expressed as a fraction.
A pole that is 3.3 meters tall casts a shadow that is 1.69 meters long. At the same time, a nearby building casts a shadow that is 47.25 meters long. How tall is the building? Round to the nearest meter.
Answer:
92 meters
Step-by-step explanation:
This is a problem that can be solved by the concept of "similar triangles", where we have two right angle triangles that share also another common angle" the angle that the rays of the sun form with them (see attached image).
The smaller triangle is formed by the limiting rays of the sun, the pole, and its shadow: it has a height of 3.3 m (the length of the pole) and a base of 1.69 m (the pole's shadow).
The larger triangle is formed by the rays of the sun, the building and its shadow: it has a base of 47.25 m and an unknown height that we named as "x" (our unknown).
The ratio of the bases of such similar triangles must be in the same proportion as the ratio of their heights, so we can create a simple equation that equals such ratios, and then solve for the unknown "x":
[tex]\frac{H}{h} =\frac{B}{b}\\\frac{x}{3.3} =\frac{47.25}{1.69}\\x=\frac{47.25\,*\,3.3}{1.69} \\x=92.26331\, m[/tex]
which we can round to the nearest meter as: 92 m
A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2902 occupants not wearing seat belts, 30 were killed. Among 7866 occupants wearing seat belts, 20 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis? test?
Answer:
Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]
Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]
The p value is a very low value and using the significance given [tex]\alpha[/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 say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .
Step-by-step explanation:
1) Data given and notation
[tex]X_{NSB}=30[/tex] represent the number of people killed not using seat belts
[tex]X_{SB}=20[/tex] represent the number of people killed using seat belts
[tex]n_{NSB}=2902[/tex] sample of people not wearing seat belts
[tex]n_{SB}=7866[/tex] sample of people wearing seat belts
[tex]p_{NSB}=\frac{30}{2902}=0.0103[/tex] represent the proportion of people killed not using seat belts
[tex]p_{SB}=\frac{20}{7866}=0.00254[/tex] represent the proportion of people killed using seat belts
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that seat belts are effective in reducing fatalities (If using seat belts reduce the proportion of deaths we need to see that the proportion of death using seat belts is lower than not using seat belts) , the system of hypothesis would be:
Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]
Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{SB}-p_{NSB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{SB}}+\frac{1}{n_{NSB}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{SB}+X_{NSB}}{n_{SB}+n_{NSB}}=\frac{20+30}{7866+2902}=0.00464[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.00254-0.0103}{\sqrt{0.00464(1-0.00464)(\frac{1}{7866}+\frac{1}{2902})}}=-5.26[/tex]
4) Statistical decision
Using the significance level provided [tex]\alpha=0.05[/tex], the next step would be calculate the p value for this test.
Since is a one side lower test the p value would be:
[tex]p_v =P(Z<-5.26)=7.2x10^{-8}[/tex]
So the p value is a very low value and using the significance given [tex]\alpha[/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 say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .
Sampling Distribution (5 pts) A bank in a small town has 100,000 customers. A national survey on the banking habits of people in U.S. shows that 80% of the people with income higher than 60,000 dollars have both savings and checking accounts and also shows that the average number of banking operations that a person aged 18 and over performs per week is 10. The manager of the bank decides to do a survey among the customers of his bank and takes a simple random sample of 350 customers aged 18 and over. In the sample, the average number of banking transactions per week is 13 with standard deviation equal to 5 a) The average number of times a customer carries out banking transactions per week is give or take or so. Show how you computed your answers. b) Give a 90% and a 99% confidence interval for the average number of banking operations per week for the town residents aged 18 and over. What is the difference in Margin of error, how does it affect your confidence interval. Show your working and interpret in plain English c) Is the apparent difference in banking habits between the nation and the customers of the bank real or just due to chance? Explain for both 90% and 99% confidence levels. d) A 95% confidence interval gives a range of values for the which are plausible according to the observed data. Fill in the blanks. (Possible answer: (A) Population average, (B) Sample average) e) The sample standard deviation measures how far is from sample average. The standard deviation for the sample average measures how far average for typicalISfrom the To fill in the blanks, choose among: (A) number of bank operations, (B) average number of operations, (C) samples, (D) customers aged 18 and over, (E) bank, (F) person with high income.
Answer:
Step-by-step explanation:
(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.
(b) The 90% confidence interval for the average number of banking operations per week is given by:
(Assuming a normal distribution of the number of banking operations)
CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}
CI=13\pm 1.645\times 5/\sqrt{350}
CI=13\pm 0.439645
CI=(12.560355, 13.439645)
The 99% confidence interval for the average number of banking operations per week is given by:
(Assuming a normal distribution of the number of banking operations)
CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}
CI=13\pm 2.576\times 5/\sqrt{350}
CI=13\pm 0.688465
CI=(12.311535, 13.688465)
The difference in the margin of error is 0.688465-0.439645=0.24882
The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).
(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.
(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.
(The confidence interval always gives the range of possible values for the population values)
(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.
The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.
A trucking firm suspects that the mean lifetime of a certain tire it uses is more than 30,000 miles. To check the claim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 29,400 miles with a population standard deviation of 1200 miles. At ΅ = 0.05, test the trucking firm's claim. Justify your decision with work. Write a short parargraph about the results of the test and what you can conclude about the claim
Answer:
We conclude that the lifetime of tires is less than 30,000 miles.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 30,000 miles
Sample mean, [tex]\bar{x}[/tex] = 29,400 miles
Sample size, n = 54
Alpha, α = 0.05
Population standard deviation, σ =1200 miles
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 30000\text{ miles}\\H_A: \mu < 30000\text{ miles}[/tex]
We use One-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{29400 - 30000}{\frac{1200}{\sqrt{54}} } = -3.6742[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]
Since,
[tex]z_{stat} < z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the lifetime of tires is less than 30,000 miles.