A 99% confidence interval for the true proportion of college students believing in ghosts is Option(c) (0.4901, 0.6699)
According to the question,
A professor wanted to know what proportion of college students believe in ghosts.
For that he drawn a sample from the college.
Sample size : n = 200
Proportion of students that believed in ghosts in sample = 58%
=> p = 0.58
Proportion of students who does not believe in ghost : q = 1 - p
=> q = 0.42
We have to create a 99% confidence interval for a true proportion of college students believing in the ghosts
Formula of C.I = [tex]p \pm ( z_{\alpha} \sqrt{\frac{pq}{n}})[/tex]
Where p and q is sample proportion , z is value of CI at given α level of significance and n is sample size
C.I at 99% = 0.58 ± 2.576 √0.58×0.42/200
=> 0.58 ± 2.576 √0.001218
=> 0.58 ± 2.576 × 0.035
=> 0.58 ± 0.08958
C.I = ( 0.58 - 0.08958 , 0.58 + 0.08958)
=> C.I = (0.4901, 0.6699)
Hence, Option (c) is correct
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Answer:
41%
Step-by-step explanation:
Which statement correctly demonstrates using limits to determine a vertical asymptote of g (x) = StartFraction 2 (x + 4) squared Over x squared minus 16 EndFraction
There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = negative infinity
There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = infinity
There is a vertical asymptote at x = –4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = negative infinity
There is a vertical asymptote at x = –4 because Limit of g (x) as x approaches 4 minus = negative infinity and limit of g (x) as x approaches 4 plus = infinity
The correct option that describes the vertical asymptote is; B: There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = infinity
How to find the vertical asymptote of a function?A vertical asymptote of a graph is defined as a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a.
A vertical asymptote is a value of x for which the function is not defined, that is, it is a point which is outside the domain of a function;
In a graph, these vertical asymptotes are given by dashed vertical lines.
An example is a value of x for which the denominator of the function is 0, and the function approaches infinity for these values of x.
We are given the function;
g(x) = 2(x + 4)²/(x² - 16)
Simplifying the denominator gives;
(x² - 16) = (x + 4)(x - 4)
Thus, our function is;
g(x) = 2(x + 4)²/[(x + 4)(x - 4)]
(x + 4 ) will cancel out to give;
g(x) = 2(x + 4)/(x - 4)
Vertical asymptote:
Point in which the denominator is 0, so:
(x - 4) = 0
x = 4
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I need. Help with this question
Derivative of function is 5(2x + 4) * (x² + 4x + 6)⁴.
What is differentiation?
In arithmetic, the derivative of a function of a true variable measures the sensitivity modification of the perform price with relation to a change in its argument. Derivatives are a elementary tool of calculus.
Main body:
by using product rule;
Let = u = x² + 4x + 6
⇒ du/dx = 2x + 4 .
Now y = u⁵
⇒ dy/dx = 5u⁴ .
dy/dx = dy/dx * du/dx = 5u⁴ * (2x + 4)
= 5 * (x² + 4x + 6)⁴ * (2x + 4)
= 5(2x + 4) * (x² + 4x + 6)⁴
hence , derivative of function is 5(2x + 4) * (x² + 4x + 6)⁴.
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A train travels at 80 miles per hour. An equation can be written that compares the time (t) with the distance (d). What is the domain and range?
1. The domain is distance (d) and the range is time (t).
2. The domain is time (t) and the range is distance (d).
3. The domain is time (t) and the range is 80.
4. The domain is 80 and the range is time (t).
The required answer is the domain is time (t) and the range is a distance (d) i.e. Option 2.
What are domain and range?
The value range that can be plugged into a function is known as its domain. In a function like f, this set represents the x values f(x). The collection of values that a function can take on is known as its range. The values that the function outputs when we enter an x value are in this set.
From the given question, and the above definition of domain and range,
the time (t) acts as an x-values or input value and the distance (d) acts as a y-value or output value
Hence, the domain is time (t) and the range is a distance (d) i.e. Option 2.
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two different two-digit whole numbers are selected at random. what is the probability that their product is less than 200. express your answer as a common fraction. (hints: (l) there are 90 different two-digit numbers, (2) the pair {10, 11} produces the smallest product and the pair {11, 18} produces the largest product less than 200).
The probability that the product of the two two-digit numbers is less than 200 is given as follows:
43/8010.
How to calculate the probability?A probability is calculated as the division of the number of desired outcomes in the context of the experiment by the number of total outcomes.
There are 90 different two-digit numbers, hence the number of total outcomes for the product is of:
90 x 89 = 8010.
(the numbers have to be different)
The desired outcomes which result in a product of less than 200 are of given as follows:
10 multiplied by 9 numbers, from 11 to 19.11 multiplied by 10, 12, 13, 14, 15, 16, 17, 18. (8 numbers).12 multiplied by 10, 11, 13, 14, 15, 16. (6 numbers).13 multiplied by 10, 11, 12, 14, 15 (5 numbers).14 multiplied by 10, 11, 12, 13. (4 numbers).15 multiplied by 10, 11, 12, 13. (4 numbers).16 multiplied by 10, 11, 12. (3 numbers).17 multiplied by 10, 11. (2 numbers).18 multiplied by 10, 11. (2 numbers).Hence the number of desired outcomes is given as follows:
9 + 8 + 6 + 5 + 4 + 4 + 3 + 2 + 2 = 43.
Meaning that the probability is of:
43/8010.
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What is the freezing point, in C, of a 2.75 m solution of C8H18 in benzene?
The freezing point of the 2.75 m solution of octane in benzene is -8.56 °C.
The freezing point of a solution is the temperature at which the solution becomes a solid. The freezing point of a solution is lower than the freezing point of the pure solvent because the solute particles interfere with the movement of the solvent molecules, which slows down the freezing process.
To determine the freezing point of a solution, we can use the freezing point depression equation:
ΔTf = Kf x molality
where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent, and molality is the concentration of the solute in the solution expressed in moles of solute per kilogram of solvent.
To find the freezing point of a 2.75 m solution of C8H18 (octane) in benzene, we need to know the freezing point depression constant for benzene, which is 5.12 °C/m. We can then use the equation above to calculate the change in freezing point:
ΔTf = 5.12 °C/m x 2.75 m = 14.06 °C
To find the freezing point of the solution, we need to subtract the change in freezing point from the freezing point of the pure solvent. The freezing point of pure benzene is 5.5 °C, so the freezing point of the 2.75 m solution of octane in benzene is:
5.5 °C - 14.06 °C = -8.56 °C
This means that the freezing point of the 2.75 m solution of octane in benzene is -8.56 °C. At this temperature, the solution will become a solid.
Given: ABCD is a parallelogram with AE = 9x−5, AC = 14x + 34. Find AC
The value of AC according to given equation of Parallelogram is 188 units.
What is parallelogram?
In elementary geometry, a parallelogram may be a quadrilateral with 2 pairs of parallel sides. the alternative or facing sides of a quadrangle square {measure} of equal length
Main body:
according to question:
AE = 9X-5
AC = 14X+34
as E is midpoint of AC so , we can say
2AE = AC
2(9x-5)= 14x+34
18x-10= 14x+34
4x = 44
x = 11
Now we need to find AC = 14x+34
= 14*11+34
= 188 units
Hence the value of AC according to given equation of Parallelogram is 188 units.
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