Answer:
a) [tex]P(X\leq 12)=0.8586[/tex]
b) [tex]P(X\geq 5)=0.9802[/tex]
c) [tex]P(5\leq X\leq 12)=0.8389[/tex]
Step-by-step explanation:
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=20, p=0.48)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We need to check the conditions in order to use the normal approximation.
[tex]np=20*0.48=9.6 \approx 10 \geq 10[/tex]
[tex]n(1-p)=20*(1-0.48)=10.4 \geq 10[/tex]
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
[tex]E(X)=np=20*0.48=9.6[/tex]
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.48(1-0.48)}=2.234[/tex]
Part a
We want this probability:
[tex]P(X\leq 12)[/tex]
We can use the z score given by this formula [tex]Z=\frac{x-\mu}{\sigma}[/tex].
[tex]P(X\leq 12)=P(\frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(Z\leq 1.074)=0.8586[/tex]
Part b
We want this probability:
[tex]P(X\geq 5)[/tex]
We can use again the z score formula and we have:
[tex]P(X\geq 5)=1-P(X<5)=1-P(\frac{X-\mu}{\sigma}< \frac{5-9.6}{2.234})=1-P(Z<- 2.059)=0.9802[/tex]
Part c
We want this probability:
[tex]P(5\leq X\leq 12)=P(\frac{5-9.6}{2.234}\leq \frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(-2.059\leq Z \leq 1.074)[/tex]
[tex]=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389[/tex]
Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households exceeds 0.27. A random sample of 125 households found that 43 consisted of one person. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?
Answer:
To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.
Step-by-step explanation:
To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.
The average Math SAT score of all incoming freshman at a large university is 490. A simple random sample of 200 of these incoming freshmen has an average Math SAT score of 495 with a standard deviation of 60. Set up an approximate 95% confidence interval for the average Math SAT score of all incoming freshmen at the university.Select only one of the boxes below.A. It is appropriate to compute a confidence interval for this problem using the Normal curve.B. Making the requested confidence interval does not make any sense.C. The Normal curve cannot be used to make the requested confidence interval.
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
Step-by-step explanation:
Find z such that 22% of the area under the standard normal curve lies to the right of z. (Round your answer to two decimal places.)
Answer:
z=0.77
Step-by-step explanation:
If 22% of the area under the standard normal curve lies to the right of z, 78% lies to the left.
According to a one-sided z-score table:
If z-score = 0.77 the area under the curve is 77.94%
If z-score = 0.78 the area under the curve is 78.23%
Since only two decimal plates are required, there is no need to actually interpolate the values, just verify which one is closer to 78%
Since 77.94% is closer, a z-score of 0.77 represents a 22% area under the standard normal curve to the right of z.
To find z such that 22% of the area under the standard normal curve lies to the right of z, the value of z is approximately 0.81.
Explanation:To find the value of z such that 22% of the area under the standard normal curve lies to the right of z, we can use a z-table or a calculator. First, we need to find the area under the curve to the left of z, which is 1 - 0.22 = 0.78. We can then look up the z-score that corresponds to this area in the table. The z-score is approximately 0.81. Therefore, the value of z is approximately 0.81.
Consider the following function. f(x) = ln(1 + 2x), a = 4, n = 3, 3.7 ≤ x ≤ 4.3
(a) Approximate f by a Taylor polynomial with degree n at the number a. T3(x) = Correct: Your answer is correct.
(b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given interval. (Round your answer to six decimal places.) |R3(x)| ≤ 0.000003 Incorrect: Your answer is incorrect.
(c) Check your result in part
(b) by graphing |Rn(x)|.
This answer includes instructions for approximating a function using a third degree Taylor polynomial at a specific point, using Taylor's Inequality to estimate the accuracy of the approximation, and checking the results by graphing the function |Rn(x)|.
Explanation:For part (a), to approximate the function f(x) = ln(1 + 2x) using a Taylor polynomial of degree 3 at a = 4, we first need to find the first 4 derivatives of the function evaluated at a=4. Substituting the values will give the Taylor polynomial - T3(x).
For part (b), Taylor's Inequality states that the remainder term Rn(x) is less than or equal to M|X-a|^(n+1) / (n+1)! where M is the max value of the |f^(n+1)(t)| where t lies in the interval between a and x. By substituting the values we can find |R3(x)|.
For the last part (c), by using graphing software, you can graph the function |Rn(x)| in the given interval to visually confirm your earlier result. These graphs will show the magnitude of the error.|
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A right triangle whose hypotenuse is StartRoot 7 EndRoot7 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.
Explanation:The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.
Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.
From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).
The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.
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7) Lindsay has some yellow and red flowers in a vase. The ratio of yellow to red flowers is 4:7. The number of yellow flowers in Lindsay’s vase is shown. How many red flowers are in the vase?
Answer:
[tex]\frac{7}{4}[/tex]×x
Step-by-step explanation:
given ratio of yellow to red flowers is 4:7.
let number of yellow flowers in Lindsay’s vase is x
let number of red flowers in Lindsay’s vase is y
[tex]\frac{x}{y}[/tex]=[tex]\frac{4}{7}[/tex]
number of red flowers such that x is known is y= [tex]\frac{7}{4}[/tex]×x
In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 times he missed both, 152 times he made both, 33 times he only made the first, and 37 times he made the second. Is it plausible that the successive free throws are independent?
Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
__________________________________________________
First shot Made Second shot missed Total
__________________________________________________
Made 152 33 185
Missed 37 8 45
__________________________________________________
Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
_____________________________________________________
First shot Made Second shot missed
_____________________________________________________
Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
_____________________________________________________
On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]
Now with the calculated value we can find the degrees of freedom
[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.
Black perch are ocean fish found on reefs near shore. A graduate student at CSULB is interested in testing whether black perch from two different populations behave differently, on average. In her experiment, she measured routine swimming velocity (to the nearest 0.001 cm/s) for 18 fish from one population, and 21 fish from another. a) Which statistical test would you use to analyze these data?
One sample t-test
Step-by-step explanation:
In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower mean velocity than the one in the hypothesis.
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In a study of crime, the FBI found that 13.2% of all Americans had been victims of crime during a 1-year period. This result was based on a sample of 1,105. Estimate the percentage of U.S. adults who were victims at the 90% confidence level. What is the lower bound of the confidence interval?
Answer: The lower bound of confidence interval would be 0.116.
Step-by-step explanation:
Since we have given that
p = 13.2%= 0.132
n = 1105
At 90% confidence,
z = 1.645
So, Margin of error would be
[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.132\times 0.868}{1152}}}\\\\=0.0164[/tex]
So, the lower bound of the confidence interval would be
[tex]p-\text{margin of error}\\\\=0.132-0.0164\\\\=0.116[/tex]
Hence, the lower bound of confidence interval would be 0.116.
The confidence interval for percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.
Explanation:To estimate the percentage of U.S. adults who were victims of crime at the 90% confidence level, we can use the formula for a confidence interval:
[tex]CI = p +/- Z * \sqrt{((p * (1-p)) / n)[/tex]
where CI is the confidence interval, p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. In this case, p = 0.132, Z = 1.645 (corresponding to a 90% confidence level), and n = 1105. Plugging in these values, we can calculate the confidence interval as:
[tex]CI = 0.132 +/- 1.645 * \sqrt((0.132 * (1-0.132)) / 1105)[/tex]
Simplifying the expression gives us:
CI = 0.132 ± 0.018
Therefore, the confidence interval for the percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.
Which of the following are solutions to the equation sinx cosx = 1/4? Check all that apply.
A. π/3+nπ/2
B. π/12+nπ
C. π/6+nπ/2
D. 5π/12+nπ ...?
Answer:
B and D
Step-by-step explanation:
sin x cos x = 1/4
Multiply both sides by 2:
2 sin x cos x = 1/2
Use double angle formula:
sin(2x) = 1/2
Solve:
2x = π/6 + 2nπ or 2x = 5π/6 + 2nπ
x = π/12 + nπ or x = 5π/12 + nπ
The parametric equations below describe the line segment that joins the points P1(x1,y1) and P2(x2,y2). Find parametric equations to represent the line segment from (-3, 5) to (1, -2). x = x1 + (x2 – x1)t y = y1 + (y2 – y1)t 0 ≤ t ≤ 1
Answer:
x=-3+4t\\
y =5-7t
Step-by-step explanation:
we are given that when two point (x1,y1) and (x2,y2) are joined the parametric equation representing the line segment would be
[tex]x = x_1 + (x_2 - x_1)t y = y_1 + (y_2 - y_1)t, 0 \leq ≤t \leq ≤ 1[/tex]
Our points given are
[tex](-3, 5) to (1, -2)[/tex]
So substitute the values to get
[tex]x=-3+(1+3)t =-3+4t\\y = 5+(-2-3)t =5-7t[/tex]
Hence parametric equations for the line segment joining the points
(-3,5) and (1,-2) is
[tex]x=-3+4t\\y =5-7t[/tex]
Final answer:
The parametric equations for the line segment from (-3, 5) to (1, -2) are x = -3 + 4t and y = 5 - 7t for 0 ≤ t ≤ 1.
Explanation:
The question asks for parametric equations to describe the line segment joining the points P1(-3, 5) and P2(1, -2). According to the formula provided, we have:
x = x1 + (x2 – x1)t
y = y1 + (y2 – y1)t
Substituting the values of P1 and P2 into the equations:
x = -3 + (1 – (-3))t = -3 + 4t
y = 5 + (-2 – 5)t = 5 - 7t
This means for 0 ≤ t ≤ 1, the parametric equations representing the line segment from (-3, 5) to (1, -2) are:
x = -3 + 4t
y = 5 - 7t
At 6:00 PM, a flagpole that is 35 feet tall casts a shadow that is 50 feet long. At the same time, how long will a person's shadow be if they are 4 feet tall?
Please help ASAP!!! :(
Answer:
The length of the person’s shadow is 5.7ft
Explanation:
Length of the flagpole =a= 35ft
Length of the shadow of the flagpole= b=50ft
Length of the person=c= 4ft
Suppose the length of the person’s shadow is=d
According to the rules of trigonometry
[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]
[tex]\frac{a}{b}=\frac{c}{d}[/tex]
[tex]\frac{35}{50}=\frac{4}{d}[/tex]
35d=200
d=[tex]\frac{200}{35}[/tex]
d=5.7ft
Hence, The length of the person’s shadow is 5.7ft.
The median of a continuous random variable having distribution function F is that value m such that F(m) = 1/2 . That is, a random variable is just as likely to be larger than its median as it is to be smaller. The mode of a continuous random variable having pdf f(x) is the value of x for which f(x) attains its maximum. Find the median and the mode of X if X is(a) uniformly distributed over (a, b)(b) normal with parameters μ, σ2(c) exponential with parameter λ
Answer:
Step-by-step explanation:
To find median and mode for
a) In a uniform distribution median would be
(a+b)/2 and mode = any value
b) X is N
we know that in a normal bell shaped curve, mean = median = mode
Hence mode = median = [tex]\mu[/tex]
c) Exponential with parameter lambda
Median = [tex]\frac{ln2}{\lambda }[/tex]
Mode =0
The median of a distribution is the middle value while the mode is the highest occuring value
(a) uniformly distributed over (a, b)The median (M) of a uniform distribution is:
[tex]M = \frac{a +b}2[/tex]
A uniform distribution has no mode
(b) normal with parameters μ, σ2For a normal distribution with the given parameters, we have:
Median = Mean = Mode = μ
Hence, the median and the mode are μ
(c) exponential with parameter λFor an exponential distribution with the given parameter, we have:
[tex]Median = \frac{\ln 2}{\lambda}[/tex]
The mode of an exponential distribution is 0
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which graph represents this function?
f(x) = 1/2 x – 5
O
Answer: Slope 1/2
y-int -5
graph (0,-5),(1,-9/2)
Step-by-step explanation:
Candita uses 1/4 of an ounce of green paint each time she
draws a green line on her painting. She draws a total of 7 green
lines. Which expression represents the amount of green paint that
Candita used drawing green lines on the painting? Check all that apply.
1/4
1/4+1/4
1/4+1/4+1/4+1/4+1/4+1/4+1/4
7/4
7/4+7/4+7/4+7/4
Answer:
Step-by-step explanation:
She uses 1/4 ounces of green paint to draw to draw a green line on her painting. This means if she draws 2 green lines on her painting, she will require ( 1/4 + 1/4 ) ounces of green paint. This means that if she draws a total of 7 green lines, the correct options would be the options that apply and they are just two viz:
1/4+1/4+1/4+1/4+1/4+1/4+1/4 or
7/4 because they are both the same. Adding each term in the longer oftion gives 7/4
Answer:
c and d
Step-by-step explanation:
Peter and Amanda collect seashells on a beach. They both had 0 seashells when they started. Peter collects 5 seashells every minute, and Amanda collects 8 seashells every minute. Which graph and explanation correctly represent the number of shells Peter and Amanda have collected?
Just wondering I know the equation but I don't get the graphs??? which graph would it be
Answer:
;-; I belive it's the 1st or 2nd one
Step-by-step explanation:
hope this helps you.
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Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x apartments is given by the following function. P(x) = -11 x^2 + 1804 x - 43,000 To maximize the monthly rental profit, how many units should be rented out?
units What is the maximum monthly profit realizable? $
Answer:
82 apartments should be rented.
Maximum profit realized will be $30964.
Step-by-step explanation:
Monthly profit realized from renting out x apartments is modeled by
P(x) = -11x² + 1804x - 43000
To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.
P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]
= -22x + 1804
For P'(x) = 0,
-22x + 1804 = 0
22x = 1804
x = 82
Now we will take second derivative,
P"(x) = -22
(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.
For maximum profit,
P(82) = -11(82)² + 1804(82) - 43000
= -73964 + 147928 - 43000
= $30964
Therefore, maximum monthly profit will be $30964.
The number of units that should be rented out to maximize the monthly rental profit is 82 units.
The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:
[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]
This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .
For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0 (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x) is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting a = -11 and b = 1804 :
[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]
Therefore, the number of units that should be rented out to maximize the profit is x = 82 .
Next, we calculate the maximum monthly profit by substituting x = 82 back into the profit function:
[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]
Calculating step by step:
[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]
So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
In summary:
- The number of units that should be rented out to maximize the monthly rental profit is 82 units.
- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
Express the sum of the polynomial 5x^2+6x−17 and the square of the binomial (x+6) as a polynomial in standard form.
Answer:
6x^2 + 18x + 19.
Step-by-step explanation:
5x^2 + 6x - 17 + (x + 6)^2
= 5x^2 + 6x - 17 + x^2 + 12x + 36
= 6x^2 + 18x + 19.
Answer:
6x^2 + 18x + 19
Step-by-step explanation:
The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.
The square of the binomial is (x+6)^2
(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36
The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36
Collecting like terms,
5x^2 + x^2 + 6x + 12x + 36 -17
= 6x^2 + 18x + 19
A tire company measures the tread on newly-produced tires and finds that they are normally distributed with a mean depth of 0.98mm and a standard deviation of 0.35mm. Find the probability that a randomly selected tire will have a depth less than 0.70mm. Would this outcome warrant a refund (meaning that it would be unusual)?
Answer:
0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.98 mm
Standard Deviation, σ = 0.35 mm
We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(depth less than 0.70 mm)
P(x < 0.70)
[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]
Calculating from normal z table, we have:
[tex]P(z<-0.8) = 0.212[/tex]
[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]
Thus, this event is not unusual and will not warrant a refund.
What is the common difference in the sequence 7,12,17,22,27,...
Answer:
5
Step-by-step explanation:
27-22= 22-17 = 17-12 =12-7=d
d= 5
The common difference of the arithmetic sequence 7, 12, 17, 22, 27,... is 5, which is the constant difference between consecutive terms of the sequence.
The sequence given is an arithmetic sequence, which is a sequence of numbers where the difference between consecutive terms is constant. To find the common difference in the arithmetic sequence 7, 12, 17, 22, 27,..., we subtract any term from the preceding term in the sequence. Let's perform this calculation using the first two terms:
Thus, the common difference is 5. We can confirm this by checking the difference between other consecutive terms:
All differences are the same, confirming that the common difference of the sequence is indeed 5.
help please
1 thought 3
A function is a relation that has one output for a given input.
For the first one, there is no one x value with two or more y values so it is a function.
The second example is also a function because a certain value can only have one cube root.
For problem number 3 input "-3" for every instance of x in h(x).
So, h(-3)=2(3^2)-1= 17
A boutique handmade umbrella factory currently sells 37500 umbrellas per year at a cost of 7 dollars each. In the previous year when they raised the price to 15 dollars, they only sold 17500 umbrellas that year. Assuming the amount of umbrellas sold is in a linear relationship with the cost, what is the maximum revenue?
Answer:
$302,500
Step-by-step explanation:
If cost (C) = $7, then Sales (S) = 37,500 units
If cost (C) = $15, then Sales (S) = 17,500 units
The slope of the linear relationship between units sold and cost is:
[tex]m=\frac{37,500-17,500}{7-15}\\m= -2,500[/tex]
The linear equation that describes this relationship is:
[tex]s-s_0 =m(c-c_0)\\s-17500 =-2500(c-15)\\s(c)=-2500c + 55,000[/tex]
The revenue function is given by:
[tex]R(c) = c*s(c)\\R(c)=-2500c^2 + 55,000c[/tex]
The cost at which the derivative of the revenue equals zero is the cost that yields the maximum revenue.
[tex]\frac{d(R(c))}{dc}=0 =-5000c + 55,000\\c=\$11[/tex]
The optimal cost is $11, therefore, the maximum revenue is:
[tex]R(11)=-2500*11^2 + 55,000*11\\R(11)=\$ 302,500[/tex]
Faced with rising fax costs, a firm issued a guideline that transmissions of 7 pages or more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to be 7 or below. The firm examined 24 randomly chosen fax transmissions during the next year, yielding a sample mean of 8.52 with a standard deviation of 3.81 pages. Find the test statistics.
Answer: t = 1.9287
Step-by-step explanation:
Let [tex]\mu[/tex] be the average number of pages should be sent by 2-day mail instead.
As per given we have,
[tex]H_0: \mu \leq7\\\\H_a:\mu>7[/tex]
Sample mean : [tex]\overline{x}=8.52[/tex]
Sample standard deviation : s=3.81
sample size : n= 24
Since , the sample size is less than 30 and populations standard deviation is unknown , so we use t-test.
The test statistic for population mean :-
[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{8.5-7}{\dfrac{3.81}{\sqrt{24}}}\\\\=\dfrac{1.5}{\dfrac{3.81}{4.8990}}\\\\=\dfrac{1.5}{0.777712993333}=1.92873208093\approx1.9287[/tex]
Hence, the test statistics : t = 1.9287
A beverage company works out a demand function for its sale of soda and finds it to be x = D(p) = 3100 - 20p where x = the quantity of sodas sold when the price per can, in cents, is p. At what prices, p, is the elasticity of demand inelastic?
Answer with Step-by-step explanation:
We are given that a demand function
[tex]x=D(p)=3100-20p[/tex]
Where x=The quantity of sodas sold
p=Per can price (in cents)
We have to find the price p for which the demand inelastic.
Differentiate the demand function w.r.t p
[tex]D'(P)=-20[/tex]
Price elasticity of demand=[tex]E(p)=\frac{-pD'(p)}{D(p)}[/tex]
Price elasticity of demand=[tex]E(p)=\frac{-p(-20)}{3100-20p}[/tex]
When demand is inelastic then
E(p)<1
[tex]\frac{20p}{3100-20p}[/tex] <1
Multiply by (3100-20p) on both sides
[tex]20p<3100-20p[/tex]
Adding 20p on both side of inequality
[tex]40p<3100[/tex]
Divide by 40 on both sides
[tex]p<77.5[/tex]
When the value of price is less than 77.5 then the demand of elasticity is inelastic.
The demand function of a beverage company is inelastic when the percentage change in the quantity of demanded soda is less than 1% for a 1% change in price. If prices between points A and B change by 1%, and the quantity demanded changes by only 0.45% (less than 1%), the demand is inelastic.
Explanation:In the given demand function, x = D(p) = 3100 - 20p, the elasticity of demand is considered inelastic if the absolute value of elasticity is less than 1. Elasticity of demand reflects how sensitive the demand for a good is to changes in its price. With a demand curve, we'd consider whether a 1% change in price changes the quantity of the good demanded.
Should the quantity demanded change by less than 1% for a 1% price change (for example, a 0.45% change), we have inelastic demand. This means, if prices between points A and B alter by 1%, the change in the quantity demanded will be only 0.45%. This percentage indicates an inelastic demand because the price change will result in a comparatively smaller percentage change in quantity demanded. Price elasticities of demand are usually negative, showcasing the negative relationship between price and demanded quantity (law of demand). However, while interpreting, the focus lays on the absolute value.
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The average GRE score at the University of Pennsylvania for the incoming class of 2016-2017 was 311. Assume that the standard deviation was 13.
If you select a random sample of 40 students, what is the probability that the sample mean will be greater than 308? Round your answer to three decimal places.
Answer:
[tex]P(\bar X>308)=1-0.0721=0.928[/tex]
Step-by-step explanation:
1) Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=311,\sigma=13)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(311,\frac{13}{\sqrt{40}})[/tex]
2) Calculate the probability
We want this probability:
[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex]
[tex]P(\bar X>308)=1-0.0721=0.9279[/tex] and rounded would be 0.928
A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on five numbers pays 6 to 1 (that is, if one of the five numbers you bet comes up, you get back your $1 plus another $6). How much do you expect to win with a $1 bet on five numbers? HINT [See Example 4.] (Round your answer to the nearest cent.)
On average, when placing a $1 bet on five numbers in roulette, you would expect to lose about 53 cents due to the game's probability structure.
Explanation:In order to calculate the expected winnings in a roulette game, consider a $1 bet on five numbers. The roulette wheel has 38 possibilities (1 to 36, 0, and 00). So, the chance that your bet will win is 5 out of 38. If it wins, the game pays off 6 to 1, which means you get your $1 bet back and win $6 additional, for a total of $7. The expected value of this bet can be calculated as follows: (probability of winning * amount won if bet is successful) - (probability of losing * amount lost if bet is not successful). In other words, E(X) = [5/38 * $7] - [33/38 * $1]. This equates to an expected value of -$0.53 (rounded to the nearest cent). So, on average, with each $1 bet on five numbers, you would expect to lose about 53 cents.
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In the game of roulette with 38 outcomes, if you place a $1 bet on five numbers, your expected winnings are -$0.08, which means you lose 8 cents per game on average.
Explanation:To calculate the expected winnings of a bet in this roulette scenario, you need to know the probability of winning and the associated payout, and also the probability of losing. In a roulette wheel with numbers 1-36 plus 0 and 00, there are a total of 38 possible outcomes. When you place a 5-number bet, the probability of winning is 5 out of 38 or approximately 0.1316, and the probability of losing is 33 out of 38 or approximately 0.8684.
Now we add the expected winnings and losses. The expected gain from a win is the probability of winning times the payout, which is $6 in this case. So, 0.1316 * $6 = $0.79, rounded to the nearest cent. The expected loss from a bet is the probability of loss times the amount of the bet which is $1. So, 0.8684 * -$1 = -$0.87, rounded to the nearest cent. Total expected value or earnings from a single $1 bet on five numbers would be the sum of expected gains and losses, or $0.79 - $0.87 = -$0.08.
So, with a $1 bet on five numbers on this roulette wheel, you can expect, on average, to lose 8 cents per game. This negative figure indicates that this is not a good bet if you're expecting to make money on average.
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Can someone please help me with that one problem!
Answer:
AAS method can be used to prove that the two triangles are congruent.
Step-by-step explanation:
According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.
Hence, the two given triangles are congruent by AAS rule.
Hence, AAS method can be used to prove that the two triangles are congruent.
Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the surface integral (you don't need to evaluate it). ∫σ∫f(x,y,z)dS=∫R∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = 2pi/3 ∫ 0 7 ∫ 0 u^2+v^2 ∥∥∥ ∥∥∥dudv
The setup for the surface integral is achieved by using the parameterized surface and the given function along with the norm of the cross product of partial derivatives of the parameterized surface. The integral is evaluated to represent the net flux through the surface that denotes the volume enclosed by the parameterized surface.
Explanation:The surface integral of a given function can be calculated using the parameterization of the surface and the given function. In this case, our surface is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩ and our function is f(x,y,z)=x²+y²+z². We first need to calculate the partial derivatives ∂r/∂u and ∂r/∂v, and then cross them to find the norm of the cross product, which gives us the differential area element for the parameterized surface. With these, we can set up the surface integral as follows: ∫σ∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = ∫0 to 2π/3 ∫0 to 7 [u²cos²(3v)+u²sin²(3v)+v²]∥∂r/∂u×∂r/∂v∥ du dv.
Please note that the evaluated result of the integral provides the net flux through the rectangular surface which represents the volume enclosed by the parameterized surface.
The above set up integral needs to be solved to find the exact value which is not required in this scenario. However, the integral can be solved using suitable integral calculus methods if needed.
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A group of 56 computer science students were taught introductory computer programming class with an innovative teaching method that used a graphical interface and drag-and-drop methods of creating computer programs. At the end of the class, 43 of these students said that they felt confident in their ability to write computer programs. Another group of 41 students were taught the same material using a standard method. At the end of class, 25 of these students said they felt confident. Assume that each class contained a simple random sample of students. Let pX represent the population proportion of students taught by the innovative method who felt confident and let pY represent the population proportion of students taught by the standard method who felt confident. Find a 99% confidence interval for the difference pX−pY . Round the answers to four decimal places.
Answer:
(-0.0861, 0.4023)
Step-by-step explanation:
We have large sample sizes [tex]n_{x} = 56[/tex] and [tex]n_{y} = 41[/tex]. A [tex]100(1-\alpha)[/tex]% confidence interval for the difference [tex]p_{x}-p_{y}[/tex] is given by [tex](\hat{p}_{x}-\hat{p}_{y})\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_{x}(1-\hat{p}_{x})}{n_{x}}+\frac{\hat{p}_{y}(1-\hat{p}_{y})}{n_{y}}}[/tex]. [tex]\hat{p}_{x}=43/56 = 0.7679[/tex] and [tex]\hat{p}_{y}=25/41=0.6098[/tex]. Because we want a 99% confidence interval for the difference [tex]p_{x}-p_{y}[/tex], we have that [tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5758[/tex] (The area above 2.5758 and below the curve of the stardard normal density is 0.005) and the confidence interval is [tex](0.7679-0.6098)\pm (2.5758)\sqrt{\frac{0.7679(1-0.7679)}{56}+\frac{0.6098(1-0.6098)}{41}}[/tex] = (-0.0861, 0.4023).
Membership in an elite organization requires a test score in the upper 30% range. If the mean is equal to 115 and the standard deviation is equal to 12, find the lowest acceptable score that would enable a candidate to apply for membership. Assume the variable is normally distributed. (Show Work)
Answer:
Lowest acceptable score = 121.3
Step-by-step explanation:
Mean test score (μ) = 115
Standard deviation (σ) = 12
The z-score for any given test score 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:
[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]