Answer:
3/8
Step-by-step explanation:
Since we have given that
Measure of an angle = 135°
As we know that
Total angle formed by a circle = 360°
So, Fraction of a circle when the an angle 135° turn is given by
135/360=
In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm. Five connectors were masked and then plated with total immersion plating. The av- erage gold thickness was 1.0 μm, with a standard deviation of 0.15 μm. Find a 99% confidence interval for the difference between the mean thicknesses produced by the two methods.
Answer:
99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Step-by-step explanation:
We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.
Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.
Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1+_n__2-2[/tex]
where, [tex]\bar X_1[/tex] = average gold thickness of control-immersion tip plating = 1.5 μm
[tex]\bar X_2[/tex] = average gold thickness of total immersion plating = 1.0 μm
[tex]s_1[/tex] = sample standard deviation of control-immersion tip plating = 0.25 μm
[tex]s_2[/tex] = sample standard deviation of total immersion plating = 0.15 μm
[tex]n_1[/tex] = sample of printed circuit edge connectors plated with control-immersion tip plating = 7
[tex]n_2[/tex] = sample of connectors plated with total immersion plating = 5
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }[/tex] = 0.216
Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.
So, 99% confidence interval for the difference between the mean population mean, ([tex]\mu_1-\mu_2[/tex]) is ;
P(-3.169 < [tex]t_1_0[/tex] < 3.169) = 0.99 {As the critical value of t at 10 degree of
freedom are -3.169 & 3.169 with P = 0.5%}
P(-3.169 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 3.169) = 0.99
P( [tex]-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
P( [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =
[ [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]
= [ [tex](1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] , [tex](1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] ]
= [0.099 μm , 0.901 μm]
Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Answer:
The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.
( 0.17971 , 0.82028)
Step-by-step explanation:
Step:-(i)
Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.
Given sample size n₁ = 7
mean of first sample x₁⁻ =1.5 μ m.
Standard deviation of first sample S₁ = 0.25 μ m
Given data Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.
Given second sample size n₂ = 5
The mean of second sample x⁻₂ = 1.0 μ m
Standard deviation of first sample S₂ = 0.15 μ m
Level of significance ∝ =0.01 or 99%
Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10
tabulated value t = 2.764
Step(ii):-
The 99% of confidence intervals for μ₁- μ₂ is determined by
(x₁⁻ - x⁻₂) - z₀.₉₉ Se((x₁⁻ - x⁻₂) , (x₁⁻ - x⁻₂) + z₀.₉₉ Se((x₁⁻ - x⁻₂)
where [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]
[tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{0.25^2 }{7 } +\frac{0.15^2 }{5 } } = 0.115879[/tex]
[1.5-1.0 - 2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]
(0.5-0.32029,0.5+0.32029
( 0.17971 , 0.82028)
Conclusion:-
The 99% of confidence intervals for μ₁- μ₂ is determined by
( 0.17971 , 0.82028)
Traders often buy foreign currency in hope of making money when the currency's value changes. For example, on a particular day, one U.S. dollar could purchase 0.8167 Euros, and one Euro could purchase 145.8038 yen. Let f (x )represent the number of Euros you can buy with x dollars, and let g (x )represent the number of yen you can buy with x Euros.
If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.
If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.
Traders engage in foreign currency trading to potentially profit from fluctuations in currency values. In this example, the exchange rates are as follows: 1 U.S. dollar buys 0.8167 Euros, and 1 Euro buys 145.8038 yen. We are asked to define two functions: f(x) represents the number of Euros you can buy with x dollars, and g(x) represents the number of yen you can buy with x Euros.
1. To determine the number of Euros you can buy with x dollars, multiply the exchange rate between dollars and Euros by x. In this case, the exchange rate is 0.8167 Euros per dollar. Therefore, the function f(x) is f(x) = 0.8167x.
2. To find the number of yen you can buy with x Euros, multiply the exchange rate between Euros and yen by x. In this case, the exchange rate is 145.8038 yen per Euro. The function g(x) is g(x) = 145.8038x.
For example:
- If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.
- If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.
These functions enable traders to evaluate the quantity of foreign currency they can acquire or exchange based on the prevailing exchange rates.
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If you solve this you are a genius.... Mr.SMITH had 4 daughters , each daughter had a brother... How many children does Mr. Smith have?
Answer:
He has 5 children
Step-by-step explanation:One boy is a brother to all of them
A number is an arithmetic value that is used to represent a quantity and calculate it. The number of children that Mr. Smith will have is 5.
What are Numbers?A number is an arithmetic value that is used to represent a quantity and calculate it. Numericals are written symbols that represent numbers, such as "3."
As it is mentioned that Mr. Smith had 4 daughters and each of the daughters had a brother. Now, if Mr. Smith had just a single son then he will be a brother to every daughter. Therefore, the number of children that Mr. Smith will have is 5.
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Find the length of the bold arc. Round to the nearest tenth and use 3.14 or the pi button on your calculator for pi.
Answer:
8π yards
Step-by-step explanation:
A circle subtends a total angle of 360 ° from its center.The length of an arc is directly proportional to the angle it subtends from the circle's center. The arc's length can therefore be calculated as:
C=πd=2πr
Where C is circumference, d is diameter and r is radius. Given r as 16 yards then the arc length which is equivalent to circumference is given as
[tex]C=\frac {90}{360}*2*\pi*16=8\pi[/tex]
Answer:
2,144.7 i just answered the other one is wrong
Step-by-step explanation:
what is the area of the shaded region
Given:
Given that the radius of the circle is 12 cm.
The length of the rectangle is 11 cm.
The width of the rectangle is 5 cm.
We need to determine the area of the shaded region.
Area of the rectangle:
The area of the rectangle can be determined using the formula,
[tex]A_1=length \times width[/tex]
Substituting the values, we have;
[tex]A_1=11\times 5[/tex]
[tex]A_1=55 \ cm^2[/tex]
Thus, the area of the rectangle is 55 square cm.
Area of the circle:
The area of the circle can be determined using the formula,
[tex]A_2=\pi r^2[/tex]
Substituting r = 12, we have;
[tex]A_2= (3.14)(12)^2[/tex]
[tex]A_2=452.16 \ cm^2[/tex]
Thus, the area of the circle is 452.16 square cm.
Area of the shaded region:
The area of the shaded region can be determined by subtracting the area of the rectangle from the area of the circle.
Thus, we have;
Area = Area of the circle - Area of the rectangle.
Substituting the values, we have;
[tex]Area=452.16-55[/tex]
[tex]Area=397.16 \ cm^2[/tex]
Thus, the area of the shaded region is 397.16 square cm.
mackenzie made 33 out of 50 throws. what is the experimental probability that mackenzie will not make the next free throw she attempts
answer:
1.94117647058
Step-by-step explanation:
33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058
A husband and wife, Ed and Rina, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ed and the rest by Rina. They are interested in determining whether they have each loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 35 of the first 53 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina.
State the null and alternative hypotheses to be tested. How strong is the evidence that Ed and Rina have each loaded a different proportion of songs into the player? Make sure to check the conditions for the use of this test. (Round your test statistic to two decimal places and your P-value to four decimal places. Assume a 95% confidence level.) Hypotheses:A) H0: p = 0.5 Ha: p < 0.5B) H0: p = 0.5 Ha: p ≠ 0.5C) H0: p = 0.5 Ha: p > 0.5
The correct hypotheses to be tested are:
H0: p = 0.5 (The proportion of songs loaded by Rina is 0.5, meaning Ed and Rina loaded an equal proportion of songs.)
Ha: p = 0.5 (The proportion of songs loaded by Rina is not equal to 0.5, meaning Ed and Rina loaded different proportions of songs.)
To determine how strong the evidence is that Ed and Rina have each loaded a different proportion of songs into the player, we can perform a hypothesis test. The test statistic for a proportion in a binomial setting is given by:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \][/tex]
where:
- [tex]\(\hat{p}\)[/tex] is the sample proportion of successes (in this case, the proportion of songs by Rina),
- [tex]\(p_0\)[/tex] is the hypothesized proportion under the null hypothesis (0.5),
- [tex]\(n\)[/tex] is the sample size (53 songs).
Given that 35 out of 53 songs were by Rina, we calculate [tex]\(\hat{p}\)[/tex] as:
[tex]\[ \hat{p} = \frac{35}{53} \approx 0.660 \][/tex]
Now we can calculate the test statistic [tex]\(z\)[/tex] :
[tex]\[ z = \frac{0.660 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{53}}} = \frac{0.160}{\sqrt{\frac{0.25}{53}}} \approx \frac{0.160}{\sqrt{0.004717}} \approx \frac{0.160}{0.0686} \approx 2.33 \][/tex]
The P-value for a two-tailed test is the probability of observing a test statistic as extreme as or more extreme than the one observed, under the assumption that the null hypothesis is true. We can find the P-value by looking up the z-score in a standard normal distribution table or using a calculator:
[tex]\[ P\text{-value} = 2 \times P(Z > |2.33|) \approx 2 \times 0.0099 \approx 0.0198 \][/tex]
Since the P-value (0.0198) is less than the significance level of 0.05, we reject the null hypothesis in favour of the alternative hypothesis. This means there is strong evidence to suggest that Ed and Rina have each loaded a different proportion of songs into the player.
Before we conclude, we must check the conditions for the use of this test:
1. The samples are independent. Each song selection is made independently of the others.
2. The number of successes and failures are each at least 10. In this case, Rina's songs (35) and Ed's songs (18) are both greater than 10, so this condition is satisfied.
3. The sample size is less than 10% of the population size. The population size is 3476 songs, and the sample size is 53 songs, which is much less than 10% of the population, so this condition is also satisfied.
Since all conditions are met, we can confidently conclude that there is strong evidence that Ed and Rina have each loaded a different proportion of songs into the player.
What is the value of the interquartile range of the data below?
+
10
20
30
40
50
A(6)
B(12)
C(14)
D(24)
Find the output, k, when the input, t, is 3.
k = 13t – 2
k=[
S
Answer:
k = 37
Step-by-step explanation:
[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]
Find product (7m+8)(4m +1)
Step-by-step explanation:
[tex](7m + 8)(4m + 1) \\ = 7m(4m + 1) + 8(4m + 1) \\ = 28 {m}^{2} + 7m + 32m + 8 \\ \purple { \bold{= 28 {m}^{2} + 39m + 8}}[/tex]
Answer:
You can use foil or the box method.
Step-by-step explanation:
Sorry it looks so bad :P
Can someone please help me fill this out?
Given:
The height of the the tree from where the tree was broken = 13 ft
The distance from the foot of the tree to the broken top = 36 ft
To find the height of the tree.
Formula
By Pythagoras theorem we get,
h² = l²+b²,
where, h be the hypotenuse
b be the base and
l be the other side of the triangle along the right angle.
Now,
Putting, l =13 and b = 36 we get,
[tex]h^{2} =36^{2} +13^{2}[/tex]
or, [tex]h^{2} = 1296+169[/tex]
or, [tex]h^{2} =1465[/tex]
or, [tex]h=\sqrt{1465}[/tex]
or, [tex]h = 38.3[/tex]
Therefore,
The height of the tree is about = 13+38.3 ft = 51.3 ft
Hence,
The height of the tree is about 51.3 ft.
2 points
18) Find the volume of the sphere. Round your answer to the nearest
tenth. (Use Pi: 3.14) *
548.45
538.51
534.45
Answer:
Step-by-step explanation:
Find the volume of a sphere 12 yards a crossed
What is the slope that passes through points -4 and 1?
What is the range of the data below?
50
60
70
80
90
100
A(22)
B(28)
C(41)
D(42)
Answer:
it d (42)
Step-by-step explanation:
Answer:
d: 42
Step-by-step explanation:
To solve the system of linear equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 by using the linear combination method, Henry decided that he should first multiply the first equation by –3 and then add the two equations together to eliminate the x-terms. When he did so, he also eliminated the y-terms and got the equation 0 = 0, so he thought that the system of equations must have an infinite number of solutions. To check his answer, he graphed the equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 with his graphing calculator, but he could only see one line. Why is this? because the system of equations actually has only one solution because the system of equations actually has no solution because the graphs of the two equations overlap each other because the graph of one of the equations does not exist
Answer: C. Because the graphs of the two equations overlap each other.
Step-by-step explanation: I took the test its right! Hope this helps
A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4. Complete parts a through d below.
a. Give the mean and standard deviation of the sampling distribution of the sample proportion, p. The mean of the sampling distribution of p is The standard deviation of the sampling distribution of p is (Round to four decimal places as needed.)
b. Describe the shape of the sampling distribution of p. 0 A The shape of the sampling distribution of p is approximately normal because the sample size is small. The shape of the sampling distribution of p is approximately normal because the sample size is large The shape of the sampling distribution of p is approximately uniform because the sample size is smal ○ C. O D. The shape of the sampling distribution of p is approximately uniform because the sample size is large.
c. Calculate the standard normal z-score corresponding to a value of p=0.41. The standard normal z-score corresponding to a value of p: 041 is . Round to two decimal places as needed.) Finn)
Answer:
a) The mean of the sampling distribution of p is 0.4.
The standard deviation of the sampling distribution of p is 0.0538.
b) The shape of the sampling distribution of p is approximately normal because the sample size is large.
c) z=0.19
Step-by-step explanation:
We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.
a) The mean of the sampling distribution is equal to the mean of the distribution p:
[tex]\mu_p=p=0.4[/tex]
The standard deviation of the sampling distribution is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538[/tex]
b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.
c) The z-score fof p-0.41 can be calculated as:
[tex]z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19[/tex]
The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.
How to solve the sampling distribution?From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.
The standard deviation will be:
= [(✓0.4 × ✓0.6) / ✓83)]
= 0.0538
Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.
The standard normal z-score will be:
= (0.41 - 0.4) / 0.0538
= 0.10/0.0538
= 0.19.
Therefore, the standard normal z-score is 0.19.
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find and list the slope and y-intercept of the equation below. -2y=6(2x-2)
[tex]\text{Solve the equation to make it into slope intercept form}\\\\\text{Slope intercept form: y = mx + b}\\\\\text{Solve:}\\\\-2y=6(2x-2)\\\\\text{Distribute the 6 to the variables inside the parenthesis}\\\\-2y=12x-12\\\\\text{Divide both sides by -2}\\\\y=-6x+6\\\\\text{The slope would be -6 and the y intercept would be 6}\\\\\boxed{\text{Slope: -6 y-intercept: 6}}[/tex]
Write down all of the prime numbers between 15 and 25.
Answer:
17,19,23
that the prime number
Answer:
A prime number is a number whose only factors are one and itself
Prime numbers between 15 and 25:
17, 19, 23
0.73 as a fraction. please
Answer:
73/100
Step-by-step explanation:
.73 is 73 hundredth which is also 73/100.
Answer:
73/100
Step-by-step explanation:
.73
There are 2 numbers after the decimal, which means the denominator is 1 with 2 zeros after it
73/100
This is as simple as it get since 73 is a prime number
Game consoles: A poll surveyed 341 video gamers, and 89 of them said that they prefer playing games on a console, rather than a computer or hand-held device. An executive at a game console manufacturing company claims that less than 28% of gamers prefer consoles. Does the poll provide convincing evidence that the claim is true? Use the =α0.05 level of significance and the P-value method with the TI-84 Plus calculator.
Answer:
Step-by-step explanation:
Hello!
The objective is to test if the population proportion of gamers that prefer consoles is less than 28% as the manufacturer claims.
Of 341 surveyed players, 89 said that they prefer using a console.
The sample resulting sample proportion is p'= 89/341= 0.26
If the company claims is true then p<0.28, this will be the alternative hypothesis of the test.
H₀: p ≥ 0.28
H₁: p < 0.28
α: 0.05
To study the population proportion you have to use the approximation of the standard normal [tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)
[tex]Z_{H_0}= \frac{0.26-0.28}{\sqrt{\frac{0.28*0.72}{341} } }= -0.82[/tex]
This test is one-tailed left, i.e. that you'll reject the null hypothesis to small values of Z, and so is the p-value, you can obtain it looking under the standard normal distribution for the probability of obtaining at most -0.82:
P(Z≤-0.82)= 0.206
Using the p-value approach:
If p-value ≤ α, reject the null hypothesis
If p-value > α, don't reject the null hypothesis
The decision is to not reject the null hypothesis.
Then at a level of 5%, you can conclude that the population proportion of gamers that prefer playing on consoles is at least 28%.
I hope this helps!
A quadratic function has an equation of f(x) = a(x + 3)2 – 4 and goes through
the point (-2, 2). What is the a-value of the equation?
Answer:
6
Step-by-step explanation:
Put the given point into the equation and solve for a.
f(-2) = a(-2 +3)² -4 = a -4 = 2
a = 6 . . . . . add 4
The a-value in the equation is 6.
Type the noun form of the verb.
contribute
Answer:
Contribution
Step-by-step explanation:
The verb would be used as:
I like to contribute to discussions.
The noun would be used as:
My contribution to the discussion was alright.
I hope this helped! (Sorry for the dry examples, I couldn't think of anything else)
(1 point) Supppose A is an invertible n×n matrix and v⃗ is an eigenvector of A with associated eigenvalue 8. Convince yourself that v⃗ is an eigenvector of the following matrices, and find the associated eigenvalues. The matrix A4 has an eigenvalue . The matrix A−1 has an eigenvalue . The matrix A+4In has an eigenvalue . The matrix 5A has an eigenvalue .
By definition, if [tex]\vec v[/tex] is an eigenvector of [tex]\mathbf A[/tex] (with associated eigenvalue 8), then
[tex]\mathbf A\vec v=8\vec v[/tex]
Notice that[tex]\mathbf A^4\vec v=\mathbf A^3\cdot\mathbf A\vec v=8\mathbf A^3\vec v[/tex]
Rinse and repeat to find
[tex]\mathbf A^4\vec v=8^4\vec v[/tex]
so that [tex]\mathbf A^4[/tex] has a corresponding eigenvalue of 4096.
Let [tex]\lambda[/tex] be the eigenvalue of [tex]\mathbf A^{-1}[/tex] corresponding to [tex]\vec v[/tex]. Then[tex]\mathbf A^{-1}\vec v=\lambda\vec v\implies\mathbf A\cdot\mathbf A^{-1}\vec v=\mathbf A\cdot\lambda \vec v\implies\vec v=8\lambda\vec v[/tex]
For this to hold, we require [tex]8\lambda=1[/tex], or [tex]\lambda=\frac18[/tex]. So [tex]\mathbf A^{-1}[/tex] has a corresonding eigenvalue of 1/8.
Expanding gives[tex](\mathbf A+4\mathbf I_n)\vec v=\mathbf A\vec v+4\mathbf I_n\vec v=8\vec v+4\vec v=12\vec v[/tex]
so that [tex]\mathbf A+4\mathbf I_n[/tex] has an associated eigenvalue of 12.
You know the drill:[tex]5\mathbf A\vec v=5\cdot8\vec v=40\vec v[/tex]
so the eigenvalue of [tex]5\mathbf A[/tex] is 40.
The matrix [tex]A^{4}[/tex] has eigen value of 4096.
The matrix [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]
The eigen value of (A+4I) is 12.
The eigen value of [tex]5Av[/tex] is 40.
Eigen values of matrix:Eigenvalues are the set of scalar values that is associated with the set of linear equations most probably in the matrix equations.
The eigenvectors are also termed as characteristic roots.
If v is an eigen vector of matrix A associated with eigen value [tex]\lambda[/tex].
Then, [tex]Av=\lambda v[/tex]
it is given that, v is an eigen vector of A with eign value 8.
so taht, [tex]Av=8v[/tex]
We have to find eigen value for [tex]A^{4}[/tex].
[tex]A^{4}v=A^{3}(8v)=8^{4}v=4096v[/tex]
Thus, [tex]A^{4}[/tex] has eigen value of 4096.
If A has eigen value of 8 . Then [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]
The eigen value of (A+4I) is computed as,
[tex](A+4I_{n})v=Av+4v=8v+4v=12v[/tex]
The eigen value of (A+4I) is 12.
The eigen value of [tex]5Av[/tex] is,
[tex]5Av=5*8v=40v[/tex]
The eigen value of [tex]5Av[/tex] is 40.
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In order to estimate the proportion of students that text while driving, a school administrator selects a simple random sample of students from a list of all students at the school who have parking permits. The students are called to the office, one at a time. The administrator asks each student, “Do you text while driving, even though you are not supposed to?” Based on the survey, the administrator estimates that only 2% of students with parking passes text while driving. What potential bias is present in the design of this survey and what is the likely direction of the bias?
Answer:the answer is (A)
Step-by-step explanation:
Because the survey face-to-face, one at a time, with an administrator of the school the students would likely reply that they do not text and drive even if they actually do. This response bias would cause the sample proportion to underestimate the value of the true proportion of all students at this school that text and drive.
The design of the survey contains response bias, which is likely to underestimate the true proportion of students who text while driving due to students' reluctance to admit to prohibited and risky behaviors to an authority figure.
Explanation:The potential bias present in the design of this survey is response bias.
This type of bias occurs when the way a question is asked leads the respondents to give a certain answer
. In this case, by directly asking the students if they engage in the prohibited behavior of texting while driving, there is a likelihood that students may not answer truthfully due to fear of repercussions, the desire to give socially acceptable answers, or reluctance to admit to risky behaviors in a face-to-face setting with a school authority figure.
The likely direction of the bias is that it would underestimate the true proportion of students who text while driving since students might not want to admit to a behavior that is both dangerous and frowned upon while talking to a school administrator.
A health inspector must visit 3 of 13 restaurants on Monday. In how many way can she pick 3 restaurants?
Step-by-step explanation:
A health inspector must visit 3 of 13 restaurants on Monday.
In how many way can she pick 3 restaurants = 13 C 4 !
nCr ! = [tex]\frac{n!}{r!(n-r)!}[/tex]
13 C 4! = [tex]\frac{13!}{3!(10)!}[/tex]
= [tex]\frac{11.12.13}{2.3}[/tex]
= 286
Health inspector can pick 3 restaurants of 13 restaurants on Monday in 286 ways
Final answer:
There are 286 different ways for a health inspector to pick 3 out of 13 restaurants to visit, calculated using the combination formula C(n, k) = n! / (k!(n - k)!).
Explanation:
The student asked how many different ways a health inspector could pick 3 out of 13 restaurants to visit on Monday. This is a problem of combinatorics, specifically concerning combinations without repetition since the order of choosing restaurants does not matter.
To calculate this, we use the combination formula, which is defined as:
C(n, k) = n! / (k!(n - k)!)
Where:
n is the total number of items to choose from (in this case, 13 restaurants)k is the number of items to choose (in this case, 3 restaurants)Applying the formula:
C(13, 3) = 13! / (3!(13 - 3)!) = 13! / (3! * 10!) = (13*12*11) / (3*2*1) = 286
So, the health inspector has 286 different ways to pick 3 restaurants from a list of 13.
answer like gauss 1+3+5+7+...=999
1+3+5+7+...+999 =
= 1+2+3+4+...+500
+1+2+3+...+499
= 2·(1+2+3+...+499) + 500
= 2·(499·500)/2 + 500
= 499·500 + 500
= 500·(499 + 1)
= 500·500
= 250.000
To find the sum of the series 1 + 3 + 5 + 7 + ... + 999, we recognize this is the sum of the first 500 odd numbers. Using the property that the sum of the first n odd numbers is n², we find the sum to be 500², which is 250000.
Summation of Odd Numbers:
To find the sum of the series 1 + 3 + 5 + 7 + ... + 999 using Gauss's method, we first recognize that this sequence is a series of the first 500 odd numbers.
An interesting property to note is that the sum of the first n odd numbers is n².
For example:
→ 1 = 1 (which is 1¹)
→ 1 + 3 = 4 (which is 2²)
→ 1 + 3 + 5 = 9 (which is 3³)
And so on...
In general, the [tex]n_{th[/tex] odd number can be expressed as 2n - 1.
For 999, to find its position in the series, note that 999 is the 500th odd number (since 2*500 - 1 = 999).
Thus, the sum of the first 500 odd numbers is:
→ Sum = 500²
= 250000
→ So, the sum of the series 1 + 3 + 5 + 7 + ... + 999 is 250000.
A random variable follows the continuous uniform distribution between 160 and 340. Calculate the following quantities for the distribution. a) P(220less than or equalsxless than or equals290) b) P(160less than or equalsxless than or equals250) c) P(xgreater than190) d) What are the mean and standard deviation of this distribution?
Answer:
a) 0.3889
b) 0.5
c) 0.8333
d) The mean is 250 and the standard deviation is 51.96.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability of finding a value of X higher than x is:
[tex]P(X > x) = 1 - \frac{x - a}{b-a}[/tex]
The probability of finding a value of X between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
The mean and the standard deviation are, respectively:
[tex]M = \frac{a+b}{2}[/tex]
[tex]S = \sqrt{\frac{b-a}^{2}{12}}[/tex]
A random variable follows the continuous uniform distribution between 160 and 340.
This means that [tex]a = 160, b = 340[/tex]
a)
[tex]P(220 \leq X \leq 290) = \frac{290 - 220}{340 - 160} = 0.3889[/tex]
b)
[tex]P(160 \leq X \leq 250) = \frac{250 - 160}{340 - 160} = 0.5[/tex]
c)
[tex]P(X > 190) = 1 - \frac{190 - 160}{340 - 160} = 0.8333[/tex]
d)
[tex]M = \frac{160 + 340}{2} = 250[/tex]
[tex]S = \sqrt{\frac{340 - 160}^{2}{12}} = 51.96[/tex]
The mean is 250 and the standard deviation is 51.96.
On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.
a. In words, define the random variable X
b. What is X ~N(,)
c. Find the probability that the child spends less than 2 hours per day unsupervised.
d. What percent of the children spend over 12 hours per day unsupervised?
Given Information:
Mean = μ = 3.2 hours
Standard deviation = σ = 1.7 hours
Required Information:
a. In words, define the random variable X
b. X ~N(,) = ?
c. P(X < 2) = ?
d. P(X > 12) = ?
Answer:
a) X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b) X ~N(μ,σ) = X ~N(3.2, 1.7)
c) P(X < 2) = 23.88%
d) P(X > 12) = 0%
Explanation:
a)
Let X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b)
X ~N(μ,σ) = X ~N(3.2, 1.7)
Where 3.2 is the average number of hours that 4-year-old child spends being unsupervised and 1.7 is the standard deviation.
c)
We want to find out the probability that a child spends less than 2 hours per day unsupervised.
P(X < 2) = P(Z < (x - μ)/σ)
P(X < 2) = P(Z < (2 - 3.2)/1.7)
P(X < 2) = P(Z < (- 1.2)/1.7)
P(X < 2) = P(Z < -0.71)
The z-score corresponding to -0.71 is 0.2388
P(X < 2) = 0.2388
P(X < 2) = 23.88%
Therefore, the probability that a child spends less than 2 hours per day unsupervised is 23.88%
d)
We want to find out the probability that a child spends over 12 hours per day unsupervised.
P(X > 12) = 1 - P(X < 12 )
P(X > 12) = 1 - P(X < (x - μ)/σ)
P(X > 12) = 1 - P(X < (12 - 3.2)/1.7)
P(X > 12) = 1 - P(X < 8.8/1.7)
P(X > 12) = 1 - P(X < 5.18)
The z-score corresponding to 5.18 is 1
P(X > 12) = 1 - 1
P(X > 12) = 0
Therefore, the probability that a child spends over 12 hours per day unsupervised is 0%
The question discusses unsupervised hours of Mercurian children and it's a statistics problem involving normal distribution where the average unsupervised time is 3.2 hours with a standard deviation of 1.7 hours.
Explanation:This question is related to statistics, specifically about normal distribution. The mean unsupervised time for 4-year-olds on Mercury is 3.2 hours, with a standard deviation of 1.7 hours. If we randomly survey a 4-year-old Mercurian child living in a rural area, the time they likely spend alone, denoted as X, will range according to this distribution.
Normal distributions can be defined by two parameters: the mean (μ) and the standard deviation (σ). In this case, μ = 3.2 hours and σ = 1.7 hours. Therefore, we can say that the time X spent by this sampled child alone is normally distributed with these parameters.
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The product of two is 450.the first number is half the second number
Answer:
I don't know you're options but the answer could be 30.
Whats the answer of this question:
Question: 1+1
Answer:
2
Step-by-step explanation:
Answer: 9?
Step-by-step explanation: I think.