Answer:
[tex]\pm \sqrt{x-16}[/tex] is the inverse of [tex]y = x^2 + 16[/tex]
Step-by-step explanation:
Given that:
[tex]y = x^2 + 16[/tex]
Let us proceed step by step to calculate the inverse:
Step 1: Put [tex]y = f(x)[/tex]
[tex]f(x) = y=x^2 + 16[/tex]
Step 2: Interchange [tex]x[/tex] and [tex]y[/tex]:
[tex]x = y^2 + 16[/tex]
Step 3: Solve the equation to find the value of [tex]y[/tex]:
[tex]y^2 =x- 16\\\Rightarrow y =\pm \sqrt{x- 16}[/tex]
Step 4: Replace [tex]y[/tex] with [tex]f^{-1}(x)[/tex]:
[tex]\Rightarrow y =f^{-1}(x)=\pm \sqrt{x- 16}[/tex]
So, the inverse of [tex]y = x^2 + 16[/tex] is [tex]\pm \sqrt{x- 16}[/tex].
The equation which is the inverse of y = x2 + 16 is; f-¹ = y = ±√(x -16)
To evaluate the inverse of the function, y = x2 + 16.
We must first make x the subject of the formula and swap x and y as follows;
x = ±√(y - 16)y = ±√(x - 16)Therefore, the inverse function is;
f-¹ = y = ±√(x -16)Read more on inverse function:
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Let the sample space be
S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Suppose the outcomes are equally likely. Compute the probability of the event E = 1, 2.
Answer:
probability of the event E = 1/5
Step-by-step explanation:
We are given;
Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
Number of terms in sample S is;
n(S) = 10
We are given the event; E = {1, 2}
Thus, number of terms in event E is;
n(E) = 2
Now, Probability = favorable outcomes/total outcomes
Thus, the probability of the event E is;
P(E) = n(E)/n(S)
P(E) = 2/10
P(E) = 1/5
Suppose the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with α = 4 and β = 3.(a) Compute E(X) and V(X). (Round your answers to four decimal places.)E(X) = Correct: Your answer is correct.V(X) = Correct: Your answer is correct.(b) Compute P(X ≤ 0.5). (Round your answer to four decimal places.)
Answer:
(a) The value of E (X) is 4/7.
The value of V (X) is 3/98.
(b) The value of P (X ≤ 0.5) is 0.3438.
Step-by-step explanation:
The random variable X is defined as the proportion of surface area in a randomly selected quadrant that is covered by a certain plant.
The random variable X follows a standard beta distribution with parameters α = 4 and β = 3.
The probability density function of X is as follows:
[tex]f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} ; \hspace{.3in}0 \le x \le 1;\ \alpha, \beta > 0[/tex]
Here, B (α, β) is:
[tex]B(\alpha,\beta)=\frac{(\alpha-1)!\cdot\ (\beta-1)!}{((\alpha+\beta)-1)!}[/tex]
[tex]=\frac{(4-1)!\cdot\ (3-1)!}{((4+3)-1)!}\\\\=\frac{6\times 2}{720}\\\\=\frac{1}{60}[/tex]
So, the pdf of X is:
[tex]f(x) = \frac{x^{4-1}(1-x)^{3-1}}{1/60}=60\cdot\ [x^{3}(1-x)^{2}];\ 0\leq x\leq 1[/tex]
(a)
Compute the value of E (X) as follows:
[tex]E (X)=\frac{\alpha }{\alpha +\beta }[/tex]
[tex]=\frac{4}{4+3}\\\\=\frac{4}{7}[/tex]
The value of E (X) is 4/7.
Compute the value of V (X) as follows:
[tex]V (X)=\frac{\alpha\ \cdot\ \beta}{(\alpha+\beta)^{2}\ \cdot\ (\alpha+\beta+1)}[/tex]
[tex]=\frac{4\cdot\ 3}{(4+3)^{2}\cdot\ (4+3+1)}\\\\=\frac{12}{49\times 8}\\\\=\frac{3}{98}[/tex]
The value of V (X) is 3/98.
(b)
Compute the value of P (X ≤ 0.5) as follows:
[tex]P(X\leq 0.50) = \int\limits^{0.50}_{0}{60\cdot\ [x^{3}(1-x)^{2}]} \, dx[/tex]
[tex]=60\int\limits^{0.50}_{0}{[x^{3}(1+x^{2}-2x)]} \, dx \\\\=60\int\limits^{0.50}_{0}{[x^{3}+x^{5}-2x^{4}]} \, dx \\\\=60\times [\dfrac{x^4}{4}+\dfrac{x^6}{6}-\dfrac{2x^5}{5}]\limits^{0.50}_{0}\\\\=60\times [\dfrac{x^4\left(10x^2-24x+15\right)}{60}]\limits^{0.50}_{0}\\\\=[x^4\left(10x^2-24x+15\right)]\limits^{0.50}_{0}\\\\=0.34375\\\\\approx 0.3438[/tex]
Thus, the value of P (X ≤ 0.5) is 0.3438.
You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on constructing a confidence interval; however, you are not sure how many citizens should be included in the sample. If you want your sample estimate to be within 5 years of the actual mean with a confidence level of 97%, how many citizens should be included in your sample
Question:
You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on constructing a confidence interval; however, you are not sure how many citizens should be included in the sample. If you want your sample estimate to be within 5 years of the actual mean with a confidence level of 97% , how many citizens should be included in your sample? Assume that the standard deviation of the ages of all the citizens in this community is 18 years.
Answer:
61.03
Step-by-step explanation:
Given:
Standard deviation = 18
Sample estimate = 5
Confidence level = 97%
Required:
Find sample size, n.
First find the Z value. Using zscore table
Z-value at a confidence level of 97% = 2.17
To find the sample size, use the formula below:
[tex] n = (Z * \frac{\sigma}{E})^2[/tex]
[tex] n = ( 2.17 * \frac{18}{5})^2 [/tex]
[tex] n = (2.17 * 3.6)^2 [/tex]
[tex] n = (7.812)^2 [/tex]
[tex] n = 61.03 [/tex]
Sample size = 61.03
Which is the better buy? Store A: $250 of 20% off Or Store B $280 at 25% off
Show your work
Answer:
Store A
Step-by-step explanation:
So. What we are going to want to do here is start off by having two stores obviously. And we have the sales that they have. If the discount is 20% rhat means the new price will be 80% of 250. So we take 250 x .8 = 200
If the discount is 25%, that means the new price will be 75% of what it was before hand. So we take 280 x .75 = 210. So the better price is at Store a
The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today’s sample contains 14 defectives. Determine a 88% confidence interval for the proportion defective for the process today. Place your LOWER limit, rounded to 3 decimal places, in the first blank. For example, 0.123 would be a legitimate answer. Place your UPPER limit, rounded to 3 decimal places, in the second blank. For example, 0.345 would be a legitimate entry.
Answer:
The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 160, \pi = \frac{14}{160} = 0.088[/tex]
88% confidence level
So [tex]\alpha = 0.12[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.12}{2} = 0.94[/tex], so [tex]Z = 1.555[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 - 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.053[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.088 + 1.555\sqrt{\frac{0.088*0.912}{160}} = 0.123[/tex]
The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)
An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is the following.
y 0 1 2 3
p(y) 0.50 0.25 0.20 0.05
A) Compute E(Y).
B) Suppose an individual with Y violations incurs a surcharge of $110Y2. Calculate the expected amount of the surcharg.
Answer:
A. The E(Y) is 0.80
B. The expected amount of the surcharges is $165
Step-by-step explanation:
A. In order to calculate the E(Y), we would have to calculate the following formula:
E(Y)=∑yp(y)
E(Y)=(0*0.5)+(1*0.25)+(2*0.20)+(3*0.05)
E(Y)=0+0.25+0.40+0.15
E(Y)=0.80
B. In order to calculate the expected amount of the surcharges we would have to calculate the following formula:
E($110Y∧2)=110E(Y∧2)
=110∑y∧2p(y)
=110((0∧2*0.5)+(1∧2*0.25)+(2∧2*0.20)+(3∧2*0.05))
110(0+0.25+0.80+0.45)
=$165
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 409 gram setting. It is believed that the machine is underfilling the bags. A 42 bag sample had a mean of 404 grams. Assume the population standard deviation is known to be 24. A level of significance of 0.01 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.
Answer:
[tex]z=\frac{404-409}{\frac{24}{\sqrt{42}}}=-1.35[/tex]
The p value for this case is given by:
[tex]p_v =P(z<-1.35)=0.0885[/tex]
For this case the p value is higher than the significance level given so we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is significantly less than 409
Step-by-step explanation:
Information given
[tex]\bar X=404[/tex] represent the sample mean
[tex]\sigma=24[/tex] represent the population standard deviation
[tex]n=42[/tex] sample size
[tex]\mu_o =409[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is less than 409, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 409[/tex]
Alternative hypothesis:[tex]\mu < 409[/tex]
The statistic for this case is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing the info we got:
[tex]z=\frac{404-409}{\frac{24}{\sqrt{42}}}=-1.35[/tex]
The p value for this case is given by:
[tex]p_v =P(z<-1.35)=0.0885[/tex]
For this case the p value is higher than the significance level given so we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is significantly less than 409
Tasha wants to take money out of the ATM for a taxi fare. She wants to do a quick estimate to see if taking $120 out of her bank account will overdraw it. She knows she had $325 in the account this morning when she checked her balance. Today she bought lunch for $19, a dress for $76, a pair of shoes for $53, and a necklace for $23. She also saw a movie with a friend for $12. Rounding each of her expenses to the nearest tens place, estimate how much money Tasha has left in her account before she goes to the ATM. Do not include the $ in your answer.
Answer:145
Step-by-step explanation: $19=20 76=80 53=50 23=20 12=10 total = 180 325-180 =145
Write the expression in simplest form 3(5x) + 8(2x)
Answer:
31x[tex]solution \\ 3(5x) + 8(2x) \\ = 3 \times 5x + 8 \times 2x \\ = 15x + 16x \\ = 31x[/tex]
hope this helps...
Good luck on your assignment...
The expression [tex]3(5x) + 8(2x)[/tex] in simplest form is 31x.
To simplify the expression [tex]3(5x) + 8(2x)[/tex], we can apply the distributive property:
[tex]3(5x) + 8(2x)[/tex]
[tex]= 15x + 16x[/tex]
Combining like terms, we have:
[tex]15x + 16x = 31x[/tex]
Therefore, the expression [tex]3(5x) + 8(2x)[/tex] simplifies to [tex]31x.[/tex]
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Teresa's parents are getting phones that each and 64 GB of storage how many bits of storage come with each phone answer both in scientific in standard notation
Answer:
5.12 x 10¹¹ bit
Step-by-step explanation:
1GB = 8 x 10⁹ bits
so 64gb = 64 x 8 x 10⁹
= 512 x 10⁹
= 5.12 x 10¹¹ bits
scientific notation = 5.12 x 10¹¹ bits
standard Notation = 512 ,000,000,000 bits.
Find the equation of the line.
Use exact numbers.
y=
Answer:
y = 2x+4
Step-by-step explanation:
First we need to find the slope using two points
(-2,0) and (0,4)
m = (y2-y1)/(x2-x1)
m = (4-0)/(0--2)
= 4/+2
= 2
we have the y intercept which is 4
Using the slope intercept form of the line
y = mx+b where m is the slope and b is the y intercept
y = 2x+4
A financial advisor is analyzing a family's estate plan. The amount of money that the family has invested in different real estate properties is normally distributed with a mean of $225,000 and a standard deviation of $50,000. Use a calculator to find how much money separates the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings.
Answer:
The amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.
Step-by-step explanation:
Let the random variable X represent the amount of money that the family has invested in different real estate properties.
The random variable X follows a Normal distribution with parameters μ = $225,000 and σ = $50,000.
It is provided that the family has invested in n = 10 different real estate properties.
Then the mean and standard deviation of amount of money that the family has invested in these 10 different real estate properties is:
[tex]\mu_{\bar x}=\mu=\$225,000\\\\\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{50000}{\sqrt{10}}=15811.39[/tex]
Now the lowest 80% of the amount invested can be represented as follows:
[tex]P(\bar X<\bar x)=0.80\\\\\Rightarrow P(Z<z)=0.80[/tex]
The value of z is 0.84.
*Use a z-table.
Compute the value of the mean amount invested as follows:
[tex]\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}[/tex]
[tex]=225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57[/tex]
Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.
What is the common difference of the sequence 20, 17, 14, 11, 8.... ?
Answer:
-3
Step-by-step explanation:
every sequence goes down by -3
Answer:
take away 3. the common difference is 3
Step-by-step explanation:
take away 3
Any help would be greatly appreciated
Answer:
[tex]\boxed{\sf \ \ \ 49a^8b^6c^2 \ \ \ }[/tex]
Step-by-step explanation:
Hello,
[tex](-7a^4b^3c)^2=(-1)^27^2a^{4*2}b^{3*2}c^2=49a^8b^6c^2[/tex]
as
[tex](-1)^2=1[/tex]
D
С
Micaela tried to rotate the square 180° about the origin.
Is her rotation correct? If not, explain why.
O No, she translated the figure instead of rotating it.
O No, she reflected the figure instead of rotating it.
O No, the vertices of the image and pre-image do not
correspond.
Yes, the rotation is correct.
cu
Answer:
it’s C
Step-by-step explanation:
No, the vertices of the image and pre-image do not correspond
No, the vertices of the image and pre-image do not correspond, Micaela tried to rotate the square 180° about the origin. Hence, option C is correct.
What is rotation about the origin?A figure can be rotated by 90 degrees clockwise by rotating each vertex of the figure 90 degrees clockwise about the origin.
Let's take the vertices of a square with points at (+1,+1), (-1,+1), (-1,-1), and (+1,-1), centered at the origin, can be found in the following positions after rotation:
The vertex (+1,+1) would be rotated to the point (-1,-1).The vertex (-1,+1) would be rotated to the point (+1,-1).The vertex (-1,-1) would be rotated to the point (+1,+1).The vertex (+1,-1) would be rotated to the point (-1,+1).Micaela's rotation must be accurate if it led to the same points. Her rotation is incorrect if the points are different, though.
It is impossible to tell if Micaela's rotation is accurate without more details or a diagram.
Thus, option C is correct.
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The volume of a trianglular prism is 54 cubic units. What is the value of x?
3
5
7
9
Answer:
X is 3 units.
Step-by-step explanation:
Volume of prism is cross sectional area multiplied by length. So 1/2 ×2× x ×2 into 3x, which is equal to 6x^2. So, 6x^2=54. Therefore, x=3.
what 826,497 in standard form answer
Answer:8.2 x 10^5
Step-by-step explanation:
The graphs below have the same shape. What is the equation of the blue
graph?
Answer: b
Explanation:
The -2 outside of the parentheses means it’s at y=-2 and the -4 inside the parentheses means it’s at x= 4 because it’s always the opposite
Subtract -6 4/9-3 2/9-8 2/9
Answer:
[tex]-\frac{161}{9}=\\or\\-16\frac{8}{9}[/tex]
Step-by-step explanation:
[tex]-6\frac{4}{9}-3\frac{2}{9}-8\frac{2}{9}=\\\\-\frac{58}{9}-\frac{29}{9}-\frac{74}{9}=\\\\-\frac{161}{9}=\\\\-16\frac{8}{9}[/tex]
A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola yequals25minusxsquared. What are the dimensions of the rectangle with the maximum area? What is the area?
Answer:
The answer is "[tex]\bold{\frac{32}{3}}\\[/tex]"
Step-by-step explanation:
The rectangle should also be symmetrical to it because of the symmetry to the y-axis The pole of the y-axis. Its lower two vertices are (-x,0). it means that
and (-x,0), and (x,0). Therefore the base measurement of the rectangle is 2x. The top vertices on the parabola are as follows:
The calculation of the height of the rectangle also is clearly 16-x^2, (-x,16,-x^2) and (x,16,-x^2).
The area of the rectangle:
[tex]A(x)=(2x)(16-x^2)\\\\A(x)=32x-2x^3[/tex]
The local extremes of this function are where the first derivative is 0:
[tex]A'(x)=32-6x^2\\\\32-6x^2=0\\\\x= \pm\sqrt{\frac{32}{6}}\\\\x= \pm\frac{4\sqrt{3}}{3}\\\\[/tex]
Simply ignore the negative root because we need a positive length calculation
It wants a maximum, this we want to see if the second derivative's profit at the end is negative.
[tex]A''\frac{4\sqrt{3}}{3} = -12\frac{4\sqrt{3}}{3}<0\\\\2.\frac{4\sqrt{3}}{3}= \frac{8\sqrt{3}}{3}\\\\\vertical \ dimension\\\\16-(\frac{4\sqrt{3}}{3})^2= \frac{32}{3}[/tex]
For the dilation, DO, K = (10, 0) → (5, 0), the scale factor is equal to _____.
Answer:
[tex] \frac{1}{2} [/tex]
Step-by-step explanation:
[tex]scale \: factor = \frac{5}{10} = \frac{1}{2} \\ [/tex]
if rectangle ABCD was reflected over the y-axis, reflected over x axis, and rotated 180°, where would point A' lie?
Answer:
Option C (-4,-1) (In Quadrant III)
Step-by-step explanation:
Coordinate = (-4,1)
=> Reflecting over y-axis will make the coordinate (4,1)
=> Reflecting across x-axis will make the coordinate (4,-1)
=> Rotating 180 will make it (-4,-1)
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. (x, y) = Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)
Answer: (a) Interval where f is increasing: (0.78,+∞);
Interval where f is decreasing: (0,0.78);
(b) Local minimum: (0.78, - 0.09)
(c) Inflection point: (0.56,-0.06)
Interval concave up: (0.56,+∞)
Interval concave down: (0,0.56)
Step-by-step explanation:
(a) To determine the interval where function f is increasing or decreasing, first derive the function:
f'(x) = [tex]\frac{d}{dx}[/tex][[tex]x^{4}ln(x)[/tex]]
Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),
you have:
f'(x) = [tex]4x^{3}ln(x) + x_{4}.\frac{1}{x}[/tex]
f'(x) = [tex]4x^{3}ln(x) + x^{3}[/tex]
f'(x) = [tex]x^{3}[4ln(x) + 1][/tex]
Now, find the critical points: f'(x) = 0
[tex]x^{3}[4ln(x) + 1][/tex] = 0
[tex]x^{3} = 0[/tex]
x = 0
and
[tex]4ln(x) + 1 = 0[/tex]
[tex]ln(x) = \frac{-1}{4}[/tex]
x = [tex]e^{\frac{-1}{4} }[/tex]
x = 0.78
To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:
interval x-value f'(x) result
0<x<0.78 0.5 f'(0.5) = -0.22 decreasing
x>0.78 1 f'(1) = 1 increasing
With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.
Note: As it is a natural logarithm function, there are no negative x-values.
(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:
Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;After 0.78, it increase (has a change of sign) and f is also defined;Then, x=0.78 is a point of minimum and its y-value is:
f(x) = [tex]x^{4}ln(x)[/tex]
f(0.78) = [tex]0.78^{4}ln(0.78)[/tex]
f(0.78) = - 0.092
The point of minimum is (0.78, - 0.092)
(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:
f"(x) = [tex]\frac{d^{2}}{dx^{2}}[/tex] [[tex]x^{3}[4ln(x) + 1][/tex]]
f"(x) = [tex]3x^{2}[4ln(x) + 1] + 4x^{2}[/tex]
f"(x) = [tex]x^{2}[12ln(x) + 7][/tex]
[tex]x^{2}[12ln(x) + 7][/tex] = 0
[tex]x^{2} = 0\\x = 0[/tex]
and
[tex]12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56[/tex]
Substituing x in the function:
f(x) = [tex]x^{4}ln(x)[/tex]
f(0.56) = [tex]0.56^{4} ln(0.56)[/tex]
f(0.56) = - 0.06
The inflection point will be: (0.56, - 0.06)
In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:
f"(x) = [tex]x^{2}[12ln(x) + 7][/tex]
f"(0.1) = [tex]0.1^{2}[12ln(0.1)+7][/tex]
f"(0.1) = - 0.21, i.e. Concave is DOWN.
f"(0.7) = [tex]0.7^{2}[12ln(0.7)+7][/tex]
f"(0.7) = + 1.33, i.e. Concave is UP.
Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin which lands on heads with some unknown probability p that need not be equal to1/2. Consider the following procedure for accomplishing our task:
1. Flip the coin.
2. Flip the coin again.
3. If both flips land on heads or both land on tails, return to step 1. 4. Let the result of the last flip be the result of the experiment.
(a) Show that the result is equally likely to be either heads or tails.
(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?
Answer:
Step-by-step explanation:
Given that;
the following procedure for accomplishing our task are:
1. Flip the coin.
2. Flip the coin again.
From here will know that the coin is first flipped twice
3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails
But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.
Let the probability of head = p
so P(head) = p
the probability of tail be = (1 - p)
This kind of probability follows a conditional distribution and the probability of getting heads is :
[tex]P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})[/tex]
[tex]= \dfrac{P( \{Tails, Heads\}) \cap \{Tails, Heads,( Heads ,Tails)\})}{ {P( \{Tails, Heads,( Heads ,Tails)\}}}[/tex]
[tex]= \dfrac{P( \{Tails, Heads\}) }{ {P( \{Tails, Heads,( Heads ,Tails)\}}}[/tex]
[tex]= \dfrac{P( \{Tails, Heads\}) } { {P( Tails, Heads) +P( Heads ,Tails)}}[/tex]
[tex]=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}[/tex]
[tex]=\dfrac{(1-p)*p}{2(1-p)*p}[/tex]
[tex]=\dfrac{1}{2}[/tex]
Thus; the probability of getting heads is [tex]\dfrac{1}{2}[/tex] which typically implies that the coin is fair
(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?
For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.
The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.
Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of p.
In essence, NO, procedure (b) does not give a fair coin flip outcome.
You are given an n×n board, where n is an even integer and 2≤n≤30. For how many such boards is it possible to cover the board with T-shaped tiles like the one shown? Each cell of the shape is congruent to one cell on the board.
Answer:
7
Step-by-step explanation:
The number of cells in a tile is 4. If colored alternately, there are 3 of one color and 1 of the alternate color. To balance the coloring, an even number of tiles is needed. Hence the board dimensions must be multiples of 4.
In the given range, there are 7 such boards:
4×4, 8×8, 12×12, 16×16, 20×20, 24×24, and 28×28
Please help me find Jebel dhanna in UAE map.
Answer:
The full name of the place is the "Danat Jebel Dhanna". The Jebel Dhanna is currently located in the Abu Dhabi. It is said that it is one of the most best beach in the UAE, they also say that it is the biggest resort, of course, with a bunch of hotels.
hope this helps ;)
best regards,
`FL°°F~` (floof)
In an office complex of 1110 employees, on any given day some are at work and the rest are absent. It is known that if an employee is at work today, there is an 77% chance that she will be at work tomorrow, and if the employee is absent today, there is a 54% chance that she will be absent tomorrow. Suppose that today there are 899 employees at work.
Required:
a. Find the transition matrix for this scenario.
b. Predict the number that will be at work five days from now.
c. Find the steady-state vector.
Answer:
B
Step-by-step explanation:
Given a normal distribution with (mean) μ= 50 and (standard deviation) σ = 4, what is the probability that:__________.
a) x>43
b) x<42
c) x>57.5
d) 42
e) x<40 or x>55
f) 5% of the values are less than what X value?
g) 60% of the values are between what two X values (symmetrically distributed around the mean)?
h) 85% of the values will be above what X value?
Answer:
a) P(x > 43) = 0.9599
b) P(x < 42) = 0.0228
c) P(x > 57.5) = 0.03
d) P(x = 42) = 0.
e) P(x<40 or x>55) = 0.1118
f) 43.42
g) Between 46.64 and 53.36.
h) Above 45.852.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 50, \sigma = 4[/tex]
a) x>43
This is 1 subtracted by the pvalue of Z when X = 43. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{43 - 50}{4}[/tex]
[tex]Z = -1.75[/tex]
[tex]Z = -1.75[/tex] has a pvalue of 0.0401
1 - 0.0401 = 0.9599
P(x > 43) = 0.9599
b) x<42
This is the pvalue of Z when X = 42.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{42 - 50}{4}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
P(x < 42) = 0.0228
c) x>57.5
This is 1 subtracted by the pvalue of Z when X = 57.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57.5 - 50}{4}[/tex]
[tex]Z = 1.88[/tex]
[tex]Z = 1.88[/tex] has a pvalue of 0.97
1 - 0.97 = 0.03
P(x > 57.5) = 0.03
d) P(x = 42)
In the normal distribution, the probability of an exact value is 0. So
P(x = 42) = 0.
e) x<40 or x>55
x < 40 is the pvalue of Z when X = 40. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40 - 50}{4}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062
x > 55 is 1 subtracted by the pvalue of Z when X = 55. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{55 - 50}{4}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.8944
1 - 0.8944 = 0.1056
0.0062 + 0.1056 = 0.1118
P(x<40 or x>55) = 0.1118
f) 5% of the values are less than what X value?
X is the 5th percentile, which is X when Z has a pvalue of 0.05, so X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 50}{4}[/tex]
[tex]X - 50 = -1.645*4[/tex]
[tex]X = 43.42[/tex]
43.42 is the answer.
g) 60% of the values are between what two X values (symmetrically distributed around the mean)?
Between the 50 - (60/2) = 20th percentile and the 50 + (60/2) = 80th percentile.
20th percentile:
X when Z has a pvalue of 0.2. So X when Z = -0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X - 50}{4}[/tex]
[tex]X - 50 = -0.84*4[/tex]
[tex]X = 46.64[/tex]
80th percentile:
X when Z has a pvalue of 0.8. So X when Z = 0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 50}{4}[/tex]
[tex]X - 50 = 0.84*4[/tex]
[tex]X = 53.36[/tex]
Between 46.64 and 53.36.
h) 85% of the values will be above what X value?
Above the 100 - 85 = 15th percentile, which is X when Z has a pvalue of 0.15. So X when Z = -1.037.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.037 = \frac{X - 50}{4}[/tex]
[tex]X - 50 = -1.037*4[/tex]
[tex]X = 45.852[/tex]
Above 45.852.
All the employees of ABC Company are assigned ID numbers. The ID number consists of the first letter of an employee's last name, followed by three numbers. (a) How many possible different ID numbers are there
Answer:
there will be 9 id no. which it contains
Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.
A. 91.68
Answer:
B) 92.03 < μ < 97.97
99% confidence interval for the mean score of all students.
92.03 < μ < 97.97
Step-by-step explanation:
Step(i):-
Given sample mean (x⁻) = 95
standard deviation of the sample (s) = 6.6
Random sample size 'n' = 30
99% confidence interval for the mean score of all students.
[tex]((x^{-} - Z_{0.01} \frac{S}{\sqrt{n} } , (x^{-} + Z_{0.01} \frac{S}{\sqrt{n} })[/tex]
step(ii):-
Degrees of freedom
ν = n-1 = 30-1 =29
[tex]t_{0.01} = 2.462[/tex]
99% confidence interval for the mean score of all students.
[tex]((95 - 2.462 \frac{6.6}{\sqrt{30} } , 95 + 2.462\frac{6.6}{\sqrt{30} } )[/tex]
( 95 - 2.966 , 95 + 2.966)
(92.03 , 97.97)
Final answer:-
99% confidence interval for the mean score of all students.
92.03 < μ < 97.97