Answer:
[tex]\boxed{\sf \ a = 1 \ }[/tex]
Step-by-step explanation:
let s assume that a >=0 so that we can take the square root
if [tex]x-\sqrt{a}[/tex] is a factor of this expression it means that [tex]\sqrt{a}[/tex] is a root of it
it comes
[tex]2*(\sqrt{a})^4-2*a^2*(\sqrt{a})^2-3*\sqrt{a}+2*(\sqrt{a})^3-2(\sqrt{a})^2+3=0[/tex]
So
[tex]2*a^2-2*a^3-3*\sqrt{a}+2*a*\sqrt{a}-2*a+3=0[/tex]
we can notice that 1 is a trivial solution as
2-2-3+2-2+3=0
so the answer is 1
let s double check
if a =1
the expression is
[tex]2x^4-2x^2-3x+2-2+3=2x^4-2x^2-3x+3[/tex]
and we can write
[tex]2x^4-2x^2-3x+3=(x-1)(2x^3+2x^2-3)[/tex]
so 1 is the correct answer
Given: ABCD is a parallelogram.
Diagonals AC, BD intersect at E.
Prove: AE = CE and BE = DE
B.
С
E
A
D
Assemble the proof by dragging tiles to
the Statements and Reasons columns.
Answer:
the gram
Step-by-step explanation:
casegunnell is my gram, follow if your a real g
Following are the calculation to the given points:
When the ABCD is parallelogram:
The properties of parallelogram:
[tex]\to \angle CBD = \angle ADB \\\\[/tex]
[tex]\to \angle BCA = \angle DAC \\\\[/tex]
When the Two-lines are parallel and alternate interior angles are equal:
[tex]\to \Delta BEC \cong \Delta AED \ \ \ \ \ \ \ \{ASA\} \\\\\to \overline {AE} \cong \overline{CE}\\\\ \to \overline{BE} \cong \overline{ED} \\\\[/tex]
When the properties of congruent triangle.
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I need help for the solution
Answer:
[tex]\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }[/tex]
Step-by-step explanation:
it is a long time I have not applied Ito's lemma
I would say the following
for [tex]f(x)=x^2[/tex]
f'(x)=2x
f''(x)=2
so using Ito's lemma we can write that
[tex]dY_t=2V_tdV_t+\phi^2dt[/tex]
[tex]dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt[/tex]
[tex]dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t[/tex]
so it comes
[tex]dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t[/tex]
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
a. What fraction of the calls last between 4.50 and 5.30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
b. What fraction of the calls last more than 5.30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
c. What fraction of the calls last between 5.30 and 6.00 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
d. What fraction of the calls last between 4.00 and 6.00 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
e. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 5 percent of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)
Answer:
(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is 0.1271.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.
(e) The time is 5.65 minutes.
Step-by-step explanation:
We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
Let X = the length of the calls, in minutes.
So, X ~ Normal([tex]\mu=4.5,\sigma^{2} =0.70^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean time = 4.5 minutes
[tex]\sigma[/tex] = standard deviation = 0.7 minutes
(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X [tex]\leq[/tex] 4.50 min)
P(X < 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z < 1.14) = 0.8729
P(X [tex]\leq[/tex] 4.50 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.5-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 0) = 0.50
The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)
P(X > 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z > 1.14) = 1 - P(Z [tex]\leq[/tex] 1.14)
= 1 - 0.8729 = 0.1271
The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 5.30 min)
P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838
P(X [tex]\leq[/tex] 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 1.14) = 0.8729
The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 4.00 min)
P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838
P(X [tex]\leq[/tex] 4.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.0-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] -0.71) = 1 - P(Z < 0.71)
= 1 - 0.7612 = 0.2388
The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.
(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;
P(X > x) = 0.05 {where x is the required time}
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05
P(Z > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05
Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;
[tex]\frac{x-4.5}{0.7}=1.645[/tex]
[tex]{x-4.5}{}=1.645 \times 0.7[/tex]
x = 4.5 + 1.15 = 5.65 minutes.
SO, the time is 5.65 minutes.
can someone help me again please im giving 20 points
Answer:
I answered this one for you already I think.
identify the property being demonstrated
if x/5 = 7, then x=35
a. division
b. multiplication
c. reflexive
d. symmetric
Answer:
[tex] \: \: \: \: \: \: \: \: \: \: \dfrac{x}{5} = 7 \\ \implies \: x = 7 \times 5 \\ \implies \: x = 35[/tex]
So,b. multiplication
Answer:
A. division
Step-by-step explanation:
[tex]x/5=7[/tex]
[tex]x[/tex] is being divided by an integer.
[tex]x=35[/tex]
[tex]35/5=7[/tex]
35 divided by 5 is equal to 7.
The equation of a circle is x2 + y2 = 56.25. Find the radius of the circle?
Answer:
r = 7.5
Step-by-step explanation:
Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!
Consider three consecutive positive integers, such that the sum of the
squares of the two larger integers is 5 more than 40 times the smaller
one. Find the smaller integer.
Answer:
17
Step-by-step explanation:
Let x represent the smaller integer. Then we have ...
(x +1)² +(x +2)² = 40x +5
2x² +6x +5 = 40x +5
x² -17x = 0 . . . . . subtract (40x+5), divide by 2
x(x -17) = 0 . . . . . factor
The solution of interest is x = 17.
The smaller integer is 17.
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
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An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women?
Answer:
5/33649= approx 0.00015
Step-by-step explanation:
Total number of outcomes are C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23
Half of the Committee =3 persons. That mens that number of the women in Commettee=3. 3 women from 6 can be elected C6 3 ways ( outputs)=
6!/3!/3!=4*5*6*/2/3=20
So the probability that 3 members of the commettee are women is
P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015
The probability that precisely half of the members will be women is;
P(3 women) = 0.1213
This question will be solved by hypergeometric distribution which has the formula;
P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)
where;
S is success from population
s is success from sample
N is population size
n is sample size
We are give;
s = 3 women (which is precisely half of the members selected)
S = 6 women
N = 24 men and women
n = 6 people selected
Thus;
P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)
P(3 women) = (20 * 816)/134596
P(3 women) = 0.1213
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Any help would be great
Answer:
2/5
Step-by-step explanation:
The graph of the function f(x) =4 square root x is shown. what is the domain of the function?
Answer:
[0, positive infinity)
Step-by-step explanation:
The domain is all x values a graph inputs. In a square root function, you cannot have negative inputs as it will turn out imaginary numbers. Therefore, your domain is all values of x above and including 0.
Answer: d on Ed
Step-by-step explanation:
Just took the test
The weights of a certain brand of candies are normally distributed with a mean weight of .8551 g and a standard deviation of 0.0518 g. A sample of these candies came from a package containing 467 candies, and the package label stated that the net weight is 399 g. (If every package has 467467 candies, the mean weight of the candies must exceed 399.0/467 = .8544 g for the net contents to weigh at least 399 g.)If 1 candy is randomly selected, find the probability that it weighs more than 0.8544 g.
Answer:
50.40% probability that it weighs more than 0.8544 g.
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 = 0.8551, \sigma = 0.0518[/tex]
If 1 candy is randomly selected, find the probability that it weighs more than 0.8544 g.
This is 1 subtracted by the pvalue of Z when X = 0.8544. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.8544 - 0.8551}{0.0518}[/tex]
[tex]Z = -0.01[/tex]
[tex]Z = -0.01[/tex] has a pvalue of 0.4960
1 - 0.4960 = 0.5040
50.40% probability that it weighs more than 0.8544 g.
A hotel manager believes that 27% of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6%
Answer:
The probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided here is:
p = 0.27
n = 423
As n = 423 > 30, the sampling distribution of sample proportion can be approximated by the Normal distribution.
The mean and standard deviation of the sampling distribution of sample proportion are:
[tex]\mu_{\hat p}=p=0.27\\\\\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.27\times(1-0.27)}{423}}=0.0216[/tex]
Compute the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% as follows:
[tex]P(|\hat p-p|<0.06)=P(p-0.06<\hat p<p+0.06)[/tex]
[tex]=P(0.27-0.06<\hat p<0.27+0.06)\\\\=P(0.21<\hat p<0.33)\\\\=P(\frac{0.21-0.27}{0.0216}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.33-0.27}{0.0216})\\\\=P(-2.78<Z<2.78)\\\\=P(Z<2.78)-P(Z<-2.78)\\\\=0.99728-0.00272\\\\=0.99456\\\\\approx 0.9946[/tex]
*Use a z-table.
Thus, the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.
two sides of a parallelogram meet at an angle of 50 degrees. If the length of one side is 3 meters and the length of the other side is 5 meters, find the length of the longest diagonal and the angles that it forms with each of the given sides.
Answer:
The longer diagonal has a length of 7.3 meters.
The angles are 31.65° and 18.35°
Step-by-step explanation:
If one angle of the parallelogram is 50°, another angle is also 50° and the other two angles are the supplement of this angle. so the other three angles are:
50°, 130° and 130°.
The longer diagonal will be the one opposite to the bigger angle (130°), and this diagonal divides the parallelogram in two triangles.
Using the law of cosines in one of these two triangles, we have:
[tex]diagonal^2 = a^2 + b^2 - 2ab*cos(130\°)[/tex]
[tex]diagonal^2 = 3^2 + 5^2 - 2*3*5*(-0.6428)[/tex]
[tex]diagonal^2 = 53.284[/tex]
[tex]diagonal = 7.3\ meters[/tex]
So the longer diagonal has a length of 7.3 meters.
To find the angles that this diagonal forms with the sides, we can use the law of sines:
[tex]a / sin(A) = b/sin(B)[/tex]
[tex]5 / sin(A) = diagonal / sin(130)[/tex]
[tex]sin(A) = 5 * sin(130) / 7.3[/tex]
[tex]sin(A) = 0.5247[/tex]
[tex]A = 31.65\°[/tex]
The other angle is B = 50 - 31.65 = 18.35°
Please check the image attached for better comprehension.
A car can travel 45 miles on 2 gallons of gasoline. How far can it travel on 5.6
gallons?
Answer:
It can travel 45 / 2 = 22.5 miles per gallon so the answer is 22.5 * 5.6 = 126 miles.
An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly more than 10 years. In order to test this claim, 9 car transmissions are randomly selected and their useful lifetimes are recorded. The sample mean lifetime is 13.5 years and the sample standard deviation is 3.2 years. Assuming that the useful lifetime of a typical car transmission has a normal distribution, based on these sample result, the correct conclusion at 1% significance level for this testing hypotheses problem is: Group of answer choices
Answer:
Step-by-step explanation:
The question is incomplete. The missing information is the group of answer choices. The group of answer choices are
a) none of the above
b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).
c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).
d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).
e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).
Solution:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
H0: µ ≥ 10
For the alternative hypothesis,
µ < 10
This is a left tailed test.
Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.
Since n = 9,
Degrees of freedom, df = n - 1 = 9 - 1 = 8
t = (x - µ)/(s/√n)
Where
x = sample mean = 13.5
µ = population mean = 10
s = samples standard deviation = 3.2
t = (13.5 - 10)/(3.2/√9) = 3.28
Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.
1 - α = 1 - 0.01 = 0.99
The negative critical value is - 2.896
Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.
By using probability value,
We would determine the p value using the t test calculator. It becomes
p = 0.0056
Level of significance = 1%
Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years
The correct option is
a) none of the above
Black walnut trees contain chemicals that inhibit the growth of other plants. In a simple experiment to test whether this is true, you grow several tomato plants in soil with and without decomposing leaves from a black walnut tree. You collect data on plant height as a measure of growth. In this experiment, __________ is the independent variable, __________ is the dependent variable, and __________ is the control.
Answer:
Height of tomato plant is the dependent variable
Presence of walnut leaves in the soil is the independent variable
Tomato plants grown without walnut leaves is the control
Step-by-step explanation:
An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.
A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.
A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.
The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.
Required:
a. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength?
b. If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap?
Answer:
a. 0.0668
b. 0.9545
Step-by-step explanation:
We have the following information:
mean (m) = 10000
standard deviation (sd) = 100
(a)
We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:
P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]
P (x> 10150) = P (z> 1.5)
P (x> 10150) = 1 - P (z <1.5)
P (x> 10150) = 1 - 0.9332 (attached table)
P (x> 10150) = 0.0668
Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668
(b)
We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:
P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]
P (9800 <x <10200) = P (-2 <z <2)
P (9800 <x <10200) = P (z <2) - P (z <-2)
P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)
P (9800 <x <10200) = 0.9545
the proportion of pieces that would expect to scrap is 0.9545
after allowing 20%discount on the marked price of a computer, 13 %vat was leived on it if its price become rs36160. what amount wsaeveid in the vat
Answer: Rs. 40,000
Step-by-step explanation:
Let say Marked price of the Watch
= M Rs
Discount = 20 %
Discount = (20/100)M = 0.2M Rs
Price after Discount = M - 0.2M =
Rs 0.8M
13 % Value added tax
=> VAT = (13/100) * 0.8M = 0.104M Rs
Value of Watch = 0.8M + 0.104M = 0.904M Rs
0.904M = 36160
=> M = 40000
Marked Price of Watch = Rs 40,000
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
Answer:
Option A is correct.
A uniform distribution.
Step-by-step explanation:
Complete Question
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
A) Uniform Distribution
B) Continuous Distribution
C) Poisson Distribution
D) Relative Frequency Distribution
Solution
A uniform distribution is one in which all the variables have the same probability of occurring.
It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.
And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.
Hope this Helps!!!
A manufacturer has determined that the total cost C of operating a factory is
C = 2.5x2 + 75x + 25000,
where x is the number of units produced. At what level of production will the average cost per unit be minimized? (The average cost per unit is C/x.)
x = _____ units
Answer:
x = 100 units
Step-by-step explanation:
C = 2.5x^2 + 75x + 25000
To find average cost per unit, divide C by x:
C/x or Average cost (AC) per unit = [tex]\frac{2.5x^2 + 75x + 25000}{x}[/tex] = 2.5x + 75 + 25000/x ⇔ 2.5x + 75 + 25000x^-1 (equation 1)
To find cost minimising, we use differentiation (differentiate equation 1 in respect to x and set it equal to 0):
d(AC)/dx = 0
d(AC)/dx ⇔ 2.5 - 25000x^-2 = 0
2.5 = 25000x^-2
2.5 = [tex]\frac{25000}{x^2}[/tex]
2.5x^2 = 25000
x^2 = 10000
x = [tex]\sqrt{10000}[/tex]
x = 100 units
Cost functions are used to model the outputs from inputs.
The average cost per unit will be minimized at 100 units of production level.
The cost function is given as:
[tex]\mathbf{C(x) = 2.5x^2 + 75x + 25000}[/tex]
Calculate the average cost function using
[tex]\mathbf{A(x) = \frac{C(x)}{x}}[/tex]
So, we have:
[tex]\mathbf{A(x) = \frac{2.5x^2 + 75x + 25000}{x}}[/tex]
Simplify
[tex]\mathbf{A(x) = 2.5x + 75 + \frac{25000}{x}}[/tex]
Differentiate
[tex]\mathbf{A'(x) = 2.5 + 0 - \frac{25000}{x^2}}[/tex]
[tex]\mathbf{A'(x) = 2.5 - \frac{25000}{x^2}}[/tex]
Set to 0
[tex]\mathbf{2.5 - \frac{25000}{x^2} = 0}[/tex]
Collect like terms
[tex]\mathbf{-\frac{25000}{x^2} = -2.5}[/tex]
Cross multiply
[tex]\mathbf{-2.5x^2 = -25000 }[/tex]
Make x^2 the subject
[tex]\mathbf{x^2 = 10000 }[/tex]
Take square roots of both sides
[tex]\mathbf{x = 100 }[/tex]
Hence, the average cost per unit will be minimized at 100 units of production level.
Read more about cost functions at:
https://brainly.com/question/13764254
A TV on ebay is described to be 35.7 inches wide and 20.1 inches
high. To the nearest whole number how many inches is it's diagonal?
(Enter your answer without units.)
Answer:
41 in.
Step-by-step explanation:
You have to use the Pythagorean Theorem. You have the values for the two sides (length and width). Now, you need to solve for the hypotenuse (diagonal).
a² + b² = c²
(35.7)² + (20.1)² = c²
1274.49 + 404.01 = c²
1678.5 = c²
√1678.5 = c
c = 40.97
c ≈ 41
The diagonal length is 41 in., to the nearest whole number.
Given the system:
2x – 4y = -34
-3x - y = 2
Solve for the variables that make up the coefficient
matrix:
[a b]
[c d]
a=
b =
c=
d=
Answer:
X= -3
Y= 7
a= -6
b = -28
C = 9
D= -7
Step-by-step explanation:
2x – 4y = -34
-3x - y = 2
2x – 4y = -34
-12x - 4y = 8
-14x = 42
X= 42/-14
X= -3
-3x - y = 2
-3(-3) -y = 2
9-y = 2
9-2= y
Y= 7
a = 2x
a= 2*-3
a= -6
b = -4y
b = -4*7
b = -28
C= -3x
C= -3*-3
C = 9
D= -y
D= -7
Answer:
A= 2 B=-4
C= -3 D= -1
Step-by-step explanation:
Took it on Edge
What is the domain of the following set of ordered pairs (-2,-5),(-3,8),(12,6),(8,3),(4,0),(-5,7)
Answer:
domain = {-5, -3, -2, 4, 8, 12}
Step-by-step explanation:
The domain is the set containing the x-coordinates of all ordered pairs.
domain = {-2, -3, 12, 8, 4, -5}
If you'd like, you can put the numbers in ascending order:
domain = {-5, -3, -2, 4, 8, 12}
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Answer:
97% Confidence interval = (12.62, 18.98)
Step-by-step explanation:
Complete Question
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Solution
We first compute the sample mean and standard deviation for this sample distribution
Sample mean = (Σx)/N = (158/10) = 15.8
Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 15.8
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 10 - 1 = 9
Significance level for 97% confidence interval
(100% - 97%)/2 = 1.5% = 0.015
t (0.015, 9) = 2.9982 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 3.36
n = sample size = 10
σₓ = (3.36/√10) = 1.0625
97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 15.8 ± (2.9982 × 1.0625)
CI = 15.8 ± 3.1809882246
97% CI = (12.6190117754, 18.9809882246)
97% Confidence interval = (12.62, 18.98)
Hope this Helps!!!
Any help would be great
A 12 ft ladder leans against the side of a house. The top of the ladder is 10ft off the ground. Find x, the angle of elevation of the ladder.
1. Remember to address each of the critical elements of the prompt:
Articulate your overall approach to solving this problem before tackling the details. In other words, think about what the question is actually asking, which pieces of information are relevant, and how you can use what you have learned to fill in the missing pieces.
2. Apply the mathematical process to solve the problem:
Interpret the word problem to identify any missing information.
Translate the word problem into an equation.
Appropriately use the order of operations and law of sines and cosines to determine the solution.
Check your work by ensuring that the known properties of triangles are met.
The image is missing, so i have attached it.
Answer:
x = 56.44°
Step-by-step explanation:
From the attached image, we can see that this is a right angle triangle which has opposite, adjacent and hypotenuse as sides. Since we want to find the angle x, thus, we can make use of trigonometric ratios.
From the attached image, the side opposite to angle x is 10ft and the hypotenuse is 12 ft.
From trigonometric ratios, we know that, sin x = opposite/hypotenuse
So, sin x = 10/12
x = sin^(-1) (10/12)
x = sin^(-1) 0.8333
x = 56.44°
A laptop computer is purchased for $2300. Each year, its value is 75% of its value the year before. After how many years will the laptop computer be worth $700 or less? (Use the calculator provided if necessary.) Write the smallest possible whole number answer.
Answer:
after the 1st year
Step-by-step explanation:
$2300 × 75% = $1725.00
$2300-$1725= $575
Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
Factor as the product of two binomials. x^2-8x+16
Answer:
(x-4) (x-4)
Step-by-step explanation:
x^2-8x+16
What 2 numbers multiply to 16 and add to -8
-4*-4 = 16
-4+-4 = -8
(x-4) (x-4)
Answer:
correct ^
Step-by-step explanation: