Final answer:
There are approximately 132489 prime numbers between 2³¹and 2³², with a ratio of primes to all numbers being approximately 0.1156.
Explanation:
To find the number of primes between 2³¹and 2³², we can use the Sieve of Eratosthenes algorithm. With this algorithm, we can mark all the multiples of each prime number, and the remaining unmarked numbers will be prime.
Using this method, we can calculate that there are approximately 132489 primes between 2³¹ and 2³². The ratio of primes to all the numbers between 2³¹and 2³²is approximately 0.1156.
A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.04 with 90% confidence if (a) she uses a previous estimate of 0.38? (b) she does not use any prior estimates?
Answer:a-396
b-420
Step-by-step explanation:
[tex]\alpha[/tex] =0.1
Margin of Error=0.04
Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]
Previous estimate[tex]\left ( p\right ) =0.38[/tex]
sample size is given by:
n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]
n=[tex]\frac{1.64}{0.04}^{2}0.38\left ( 1-0.38\right )=396.0436\approx 396[/tex]
[tex]\left ( b\right )[/tex]Does not use prior estimate
Assume
[tex]\alpha [/tex]=0.1
Margin of Error=0.04
Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]
Population proportion[tex]\left ( p\right )[/tex]=0.5
n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]
n=[tex]\frac{1.64}{0.04}^{2}0.5\left ( 1-0.5\right )[/tex]
n=420.25[tex]\approx 420[/tex]
Real estate ads suggest that 58 % of homes for sale have garages, 39 % have swimming pools, and 6 % have both features. What is the probability that a home for sale has a) a pool or a garage? b) neither a pool nor a garage? c) a pool but no garage?
Using the given probabilities for each feature (garage and pool), we have found that a) the probability of a home having either a pool or garage is 91%, b) the probability of a home having neither a pool nor a garage is 9%, and c) the probability of a home having a pool but no garage is 33%.
Explanation:The question is asking about the probability of certain features in homes for sale, namely garages and swimming pools. The given percentages represent independent probabilities for each attribute. Let's denote garage as 'G' and pool as 'P'. Then the probabilities given are P(G)=0.58, P(P)=0.39, and P(G and P)=0.06.
a) The probability a home has a pool or a garage: This is determined using the formula for the union of two events: P(G U P) = P(G) + P(P) - P(G and P) = 0.58+0.39-0.06 = 0.91 or 91% of homes for sale.
b) The probability a home has neither a pool nor a garage: This is the complement of the event in part a. So, P(Neither G nor P) = 1 - P(G U P) = 1 - 0.91 = 0.09 or 9% of homes for sale.
c) The probability a home has a pool but not a garage: This is determined using the formula for the difference of two events: P(P - G) = P(P) - P(G and P) = 0.39 - 0.06 = 0.33 or 33% of homes for sale.
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The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. Based on a large number of observations, the distribution of trout weights is normally distributed with a mean of 402.7 grams and a standard deviation of 8.8 grams. What is the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams?
Answer: 0.0222
Step-by-step explanation:
Given : The distribution of trout weights is normally distributed with
Mean : [tex]\mu=402.7\text{ grams}[/tex]
Standard deviation : [tex]\sigma=8.8\text{ grams}[/tex]
Sample size : [tex]n=40[/tex]
The formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Let x be the weight of randomly selected trout.
Then for x = 405.5 , we have
[tex]z=\dfrac{405.5 -402.7}{\dfrac{8.8}{\sqrt{40}}}\approx2.01[/tex]
The p-value : [tex]P(405.5<x)=P(2.01<z)[/tex]
[tex]1-P(2.01)=1-0.9777844=0.0222156\approx0.0222[/tex]
Thus,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.
The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams is 0.0222.
The distribution of trout weights is normally distributed with
We have given that
Mean=402.7 grams
Standard deviation =8.8 grams
Sample size (n)=40
We have to calculate
The probability that the mean weight for a sample of 40 trout exceeds 405.5 grams
What is the to calculate the z-score?Te formula of Z score is given by,
[tex]z=\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]
n= the ample size
x=mean
sigma=standard deviation
So by using the formula we have,
Let x is the weight of randomly selected trout.
Then for x = 405.5
[tex]z=\frac{405.5-\402.7}{\frac{\8.8 }{\sqrt{40}}}\\\\\z=2.01[/tex]
we have
The p-value :(405.5<x)
(1-2.01)=1-0.97778
=0.02221
=0.0222
Therefore,the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams= 0.0222.
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Four marbles are to be selected at random with replacement from a jar that contains 10.0 red marbles, 9.0 blue marbles, 6.0 green marbles, and 7.0 yellow marbles. Find the probability of getting exactly 1.0 yellow marbles.
Answer:
7/32
Step-by-step explanation:
.Given: F(x) = 3x2+ 1, G(x) = 2x - 3, H(x) = x
G-1(x) =
a. -2x + 3
b. (x + 3)/2
c. 2(x + 3)
F(x) + G(x) =
a. 3x^2 + 2x - 2
b. 5x^3 - 2
c. 3x^2 + 2x + 4
F(-2) =
a. -11
b. 13
c. 37
F(3) + G(4) - 2H(5) =
a. 13
b. 23
c. 33
For this case we have the following functions:
[tex]F (x) = 3x ^ 2 +1\\G (x) = 2x-3\\H (x) = x[/tex]
We have to:
[tex]G (x) * - 1[/tex] is given by:
[tex](2x-3) * - 1 = -2x +3[/tex]
Thus, the correct option is the option is A.
On the other hand,
[tex]F (x) +G (x) = 3x ^ 2 +1 +(2x - 3) = 3x ^ 2+ 1+ 2x-3 = 3x ^ 2 +2x-2[/tex]
Thus, the correct option is the option is A.
We also have:
[tex]F (-2) = 3 (-2) ^ 2+ 1 = 3 (4)+ 1 = 12+ 1 = 13[/tex]
Thus, the correct option is the option is B.
Last we have:
[tex]F (3)+G (4) -2H (5) = (3 (3) ^ 2+ 1)+ (2 (4) -3) -2 (5) = (3 (9)+ 1) - (8-3) 10 = 28+ 5-10 = 33[/tex]
Thus, the correct option is the option is C.
ANswer:
Option A, A, B, C
Answer:
G-1(x)= -2x+3
F(x)+G(x)=3x^2+2x-2
F(-2)=13
F(3)+G(4)-2H(5)=33
A,A,B,C are the answers to the equations
Find the? inverse, if it? exists, for the given matrix.
@MATX{{3;3;-1};{-12;-12;4};{2;6;0}}
Answer:
Step-by-step explanation:
The Inverse of the matrix doesn't exist because the determinant is equal to 0.
Answer:
The inverse of given matrix is not exist, since determinant is 0.
Step-by-step explanation:
The inverse of a square matrix [tex]A[/tex] is [tex]A^{-1}[/tex] such that
[tex]A A^{-1}=I[/tex] where I is the identity matrix.
Consider, [tex]A = \left[\begin{array}{ccc}3&3&-1\\-12&-12&4\\2&6&0\end{array}\right][/tex]
[tex]\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}[/tex]
[tex]\det \begin{pmatrix}3&3&-1\\ -12&-12&4\\ 2&6&0\end{pmatrix}\ne 0[/tex]
[tex]\det \begin{pmatrix}3&3&-1\\ -12&-12&4\\ 2&6&0\end{pmatrix}[/tex]
[tex]\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \:[/tex]
[tex]\det \begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}=a\cdot \det \begin{pmatrix}e&f\\ h&i\end{pmatrix}-b\cdot \det \begin{pmatrix}d&f\\ g&i\end{pmatrix}+c\cdot \det \begin{pmatrix}d&e\\ g&h\end{pmatrix}[/tex]
[tex]=3\cdot \det \begin{pmatrix}-12&4\\ 6&0\end{pmatrix}-3\cdot \det \begin{pmatrix}-12&4\\ 2&0\end{pmatrix}-1\cdot \det \begin{pmatrix}-12&-12\\ 2&6\end{pmatrix}[/tex]
[tex]=3\left(-24\right)-3\left(-8\right)-1\cdot \left(-48\right)[/tex]
[tex]3\left(-24\right)-3\left(-8\right)-1\cdot \left(-48\right)=0[/tex]
Therefore, the inverse of given matrix is not exist, since determinant is 0.
Which director made the Beatles films, A Hard Day's Night and Help? a. Blake Edwards b. Stanley Kubrick c. Richard Lester d. Mike Nichols
Answer: Option(c) Richard Lester is correct.
Step-by-step explanation:
Both the films were directed by Richard Lester.
A Hard Day's night was a scripted comic farce and its main focus on Beatlemania and the band's hectic touring lifestyle. It is a black and white movie.
Help! film also directed by Richard Lester. And this film was shot in various exotic locations. Help! was the first Beatles film that is filmed in colour.
Calculate the mean, median, and mode for each of the following populations of numbers: (a) 17, 23, 19, 20, 25, 18, 22, 15, 21, 20 N (Population) Mean Median Mode (b) 505, 497, 501, 500, 507, 510, 501 N (Population) Mean Median Mode
Answer: i dont now
Step-by-step explanation:
u have to add them togther i guess
Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. (Assume that n begins with 1.){1/2,1/4,1/6,1/8,1/10,...}a_n = ?
Answer:
[tex]a_{n}=\frac{1}{2n}[/tex] [Where a ≥ 1 ]
Step-by-step explanation:
The pattern of the given sequence is {[tex]\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},\frac{1}{10},......[/tex]
We have to find a formula for the general term [tex]a_{n}[/tex] of the given sequence.
We can rewrite the terms of the sequence as
[tex]\frac{1}{2}=\frac{1}{(2)(1)}[/tex]
[tex]\frac{1}{4}=\frac{1}{(2)(2)}[/tex]
[tex]\frac{1}{6}=\frac{1}{(2)(3)}[/tex]
[tex]\frac{1}{8}=\frac{1}{(2)(4)}[/tex]
[tex]\frac{1}{10}=\frac{1}{(2)(5)}[/tex]
Now we can write the term [tex]a_{n}[/tex] as
[tex]a_{n}=\frac{1}{2n}[/tex]
Where n = 1, 2, 3, 4, 5......
Show that the differential equation (on the left) is a solution of the function (on the right)
d^2u/dt^2 = a^2 * (d^2u/dx^2) u(x,t) = f(x-at) + g(x+at)
We have to show that
[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]
for [tex]\frac{\partial ^{2}u}{\partial t^{2}}[/tex] we have
[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]
[tex]\frac{\partial ^{2}u}{\partial t^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial t^{2}}[/tex]
[tex]=\frac{\partial }{\partial t}[\frac{\partial[f(x-at)+g(x+at)] }{\partial t}][/tex]
[tex]\frac{\partial }{\partial t}[-a\cdot f'(x-at)+a\cdot g'(x+at)][/tex]
[tex]=a^{2}f''(x-at)+a^{2}g''(x+at)[/tex]
[tex]=a^{2}[f''(x-at)+g''(x+at)].............(i)[/tex]
similarly,
[tex]\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial x^{2}}[/tex]
[tex]=\frac{\partial }{\partial x}[\frac{\partial[f(x-at)+g(x+at)] }{\partial x}][/tex]
[tex]=\frac{\partial }{\partial x}[f'(x-at)+g'(x+at)][/tex]
[tex]=f''(x-at)+g''(x+at).......(ii)[/tex]
Comparing i and ii we get
[tex]a^{2}\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}u}{\partial t^{2}}[/tex]
Hence proved
A chef is going to use a mixture of two brands of Italian dressing. The first brand contains 7% vinegar, and the second brand contains 12% vinegar. The chef wants to make 270 milliliters of a dressing that is 9% vinegar. How much of each brand should she use?
Answer: There is 162 ml of first brand and 108 ml of second brand.
Step-by-step explanation:
Since we have given that
Percentage of vinegar that the first brand contains = 7%
Percentage of vinegar that the second brand contains = 12%
Percentage of vinegar in mixture = 9%
Total amount of dressing = 270 ml
We will use "Mixture and Allegation":
First brand Second brand
7% 12%
9%
--------------------------------------------------------
12%-9% : 9%-7%
3% : 2%
So, ratio of first brand to second brand in a mixture is 3:2.
So, Amount of first brand she should use is given by
[tex]\dfrac{3}{5}\times 270\\\\=162\ ml[/tex]
Amount of second brand she should use is given by
[tex]\dfrac{2}{5}\times 270\\\\=108\ ml[/tex]
Hence, there is 162 ml of first brand and 108 ml of second brand.
Pediatricians work an average of 48 h per week. The standard deviation is 12 hours. What percentage of pediatricians work more than 72 h per week
Approximately 2.28% of pediatricians work more than 72 hours per week according to the table.
Explanation:To find the percentage of pediatricians who work more than 72 hours per week, we need to calculate the z-score for this value and then use a standard normal distribution table to find the corresponding percentage.
Calculate the z-score using the formula:
[tex]z = (x - u) / \alpha[/tex]
where x is the value (72 hours), u is the mean (48 hours), and a is the standard deviation (12 hours).
Substitute the values into the formula: z = (72 - 48) / 12 = 2.
Using a standard normal distribution table, find the percentage of values that are greater than 2.
Based on the table, approximately 2.28% of pediatricians work more than 72 hours per week.
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A pollster wants to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better. A Gallup poll taken in July 2010 estimates this proportion to be 0.33. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.052 ?
To construct a 95% confidence interval with a margin of error of 0.052, a sample size of 300 is needed.
Explanation:To determine the sample size needed for the 95% confidence interval with a margin of error of 0.052, we can use the formula:
n = (Z^2 * p * (1 - p)) / (E^2)
where n is the sample size, Z is the Z-score corresponding to the desired confidence level (in this case, 1.96), p is the estimated proportion (0.33), and E is the margin of error (0.052).
Substituting the given values into the formula:
n = (1.96^2 * 0.33 * (1 - 0.33)) / (0.052^2)
Simplifying the equation:
n = 299.5554
Rounding up to the nearest whole number, the sample size needed is 300.
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To construct a 95% confidence interval for the proportion of adults who believe economic conditions are getting better with a margin of error of 0.052, the required sample size is approximately 577 participants.
To calculate the sample size needed to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better, with a Gallup poll estimate of 0.33 and a desired margin of error of 0.052, the formula for determining the sample size (n) is:
n = (Z^2*p*(1-p))/E^2,
where Z is the Z-score corresponding to the 95% confidence level, p is the estimated proportion (0.33) and E is the margin of error (0.052). The Z-score for a 95% confidence level is 1.96. Plugging in the values gives:
n = (1.96^2*0.33*(1-0.33))/(0.052^2),
Solving this, we find the sample size required:
n= 576.7.
Since we cannot have a fraction of a person, we round up to the next whole number. Therefore, the sample size needed is 577 participants.
The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 590 peas with 139 of them having yellow pods. Find the value of the test statistic.
Final answer:
The test statistic for the proportion of yellow pea pods being 25% is calculated using the sample proportion, hypothesized proportion, and sample size. Using the given data, the test statistic (Z-score) comes out to approximately -1.412.
Explanation:
To find the value of the test statistic for the claim that the proportion of peas with yellow pods is equal to 0.25 (25%), we use the sample statistics provided from the experiment. You mentioned that there were 590 peas in total, with 139 having yellow pods. First, we check if the conditions for the binomial distribution are met, which in this case they are as we are dealing with two outcomes (yellow pods and not yellow pods), a fixed number of trials (590 peas), and each pea is independent of the others.
The test statistic for a proportion is calculated using the formula:
Z = (p' - p) / (sqrt(p(1 - p) / n))
Where:
p' is the sample proportion (139 / 590 = 0.2356)p is the hypothesized population proportion (0.25)n is the sample size (590)Now, we calculate the test statistic:
Z = (0.2356 - 0.25) / (sqrt(0.25 × (1 - 0.25) / 590))
Z ≈ -1.412
This Z-score tells us how many standard deviations the observed sample proportion (0.2356) is from the hypothesized proportion (0.25).
The formula P = 0.672x^2 - 0.046x+ 3 models the approximate population P, in thousands, for a species of frogs in a particular rain forest, x years after 1999. During what year will the population reach 182 frogs? a) 2015 b) 2018 c) 2017 d) 2016 e) none
Answer:
The correct option is d.
Step-by-step explanation:
The approximate population P, in thousands, for a species of frogs in a particular rain forest, x years after 1999 is given by the formula
[tex]P=0.672x^2-0.046x+3[/tex]
We need to find the year it which the population reach 182 frogs.
Substitute P=182 in the given formula.
[tex]182=0.672x^2-0.046x+3[/tex]
Subtract 182 from both the sides.
[tex]0=0.672x^2-0.046x+3-182[/tex]
[tex]0=0.672x^2-0.046x-179[/tex]
Multiply both sides by 1000 to remove decimals.
[tex]0=672x^2-46x-179000[/tex]
Quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substitute a=672, b=-46 and c=-179000 in the quadratic formula.
[tex]x=\frac{-\left(-46\right)\pm\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}[/tex]
On simplification we get
[tex]x=\frac{-\left(-46\right)+\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx 16.355[/tex]
[tex]x=\frac{-\left(-46\right)-\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx -16.287[/tex]
The value of x can not be negative because x is number of years after 1999.
x=16.35 in means is 17th year after 1999 the population reach 182 frogs.
[tex]1999+17=2016[/tex]
The population reach 182 frogs in 2016. Therefore the correct option is d.
If possible, find a matrix B such that AB = A2 + 2A.
Answer:
[tex]\large\boxed{B=A+2I}[/tex]
Step-by-step explanation:
It's possible if dimensions of a matrix A and matrix B are n × n
[tex]AB=A^2+2A\qquad\text{multiply both sides on the left by}\ A^{-1}\\\\A^{-1}AB=A^{-1}A^2+A^{-1}(2A)\qquad\text{we know}\ A^{-1}A=I\\\\IB=A^{-1}A\cdot A+2A^{-1}A\\\\IB=IA+2I\qquad\text{we know}\ IA=A\\\\B=A+2I[/tex]
Matrices is an array of numbers, usually 2 dimensional, but can be single dimensional too.
A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as
[tex]B = A + 2I = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]
(Assuming A is left invertible and [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])
When can we cancel out matrix multiplied on both sides of an equation?Suppose that there is an equation
[tex]AB = AC[/tex]
We cannot always say that [tex]B = C[/tex]
If we assume that A is left invertible, then only we can surely say that we have got [tex]B = C[/tex]
Similarly, for [tex]BA = CA[/tex] to imply [tex]B = C[/tex], we need A to be right invertible.
Assuming that we have A as a left invertible matrix, say
[tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex]
and [tex]L_A[/tex] be its left inverse, then [tex]L_A A = I[/tex] ([tex]I[/tex] is identity matrix)
Then,
[tex]AB = A^2 + 2A\\A(B) = A(A + 2I_2)\\\\\Multiplying L_{A}\text{ on left side of both terms,}\\\\L_{A} AB = L_{A}A(A + 2I_2)\\B = A + 2I_2\\\\B = \left[\begin{array}{cc}2&0\\0&2\end{array}\right] + \left[\begin{array}{cc}2&0\\0&2\end{array}\right] = \left[\begin{array}{cc}4&0\\0&4\end{array}\right] = 4I_2\\\\B = 4I_2[/tex]
Thus, i
A matrix B such that [tex]AB = A^2 + 2A[/tex] is given as
[tex]B = \left[\begin{array}{cc}4&0\\0&4\end{array}\right][/tex]
(Assuming A is left invertible and [tex]A = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex])
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Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant between the circles with center the origin and radii 2 and 3
Set
[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta[/tex]
The region [tex]R[/tex] is given in polar coordinates by the set
[tex]R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}[/tex]
So we have
[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}[/tex]
The Cartesian coordinates are converted into polar coordinates so that the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ. But this specific integral can't be solved analytically, yet using polar coordinates can simplify other integration issues related to circular regions or distances from the origin.
Explanation:To evaluate the given integral using polar coordinates, one must firstly translate the Cartesian coordinates (x,y) into polar coordinates (r,θ), so that x is replaced with rcosθ and y with rsinθ. Consequently, the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ, with r varying from 2 to 3, and θ from 0 to π/2 (since we are only dealing with the first quadrant).
However, this integral becomes very complex and is not feasible to solve analytically. You would need to use a numeric method to get an approximate answer. In the context of other problems, switching to polar coordinates can simplify the integration, especially when dealing with circular regions or equations related to distances from the origin.
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solve the system of equation by guess jordan method
2x1-6x2-2x3=14, 3x1+4x2-7x3= 16, 3x1-6x2+9x3=21
Answer with explanation:
The System of equations which we have to solve by Gauss Jordan Method:
[tex]1.\rightarrow 2x_{1}-6x_{2}-2x_{3}=14, 2.\rightarrow 3x_{1}+4x_{2}-7x_{3}= 16, 3.\rightarrow 3x_{1}-6x_{2}+9x_{3}=21[/tex]
Writing it in the form of Augmented Matrix=3 Rows and 4 Columns:
[tex]\left[\begin{array}{cccc}2&-6&-2&14\\3&4&-7&16\\3&-6&9&21\end{array}\right]\\\\R_{1}=\frac{R_{1}}{2},R_{3}=\frac{R_{3}}{3}\\\\ \left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\1&-2&3&7\end{array}\right]\\\\R_{3}\rightarrow R_{3}-R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\0&1&4&0\end{array}\right]\\\\R_{2}\rightarrow R_{2}-3R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&1&4&0\end{array}\right][/tex]
[tex]R_{3}\rightarrow R_{2}+R_{3}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&14&0&-5\end{array}\right]\\\\\rightarrow14 x_{2}= -5\\\\x_{2}=\frac{-5}{14}\\\\\rightarrow 13 x_{2}-4x_{3}=-5\\\\ \frac{-65}{14}-4 x_{3}=-5\\\\-4x_{3}=-5+\frac{65}{14}\\\\x_{3}=\frac{5}{56}\\\\x_{1}-3x_{2}-x_{3}=7\\\\x_{1}+\frac{15}{14}-\frac{5}{56}=7\\\\x_{1}+\frac{55}{56}=7\\\\x_{1}=7-\frac{55}{56}\\\\x_{1}=\frac{337}{56}[/tex]
Solution set
[tex]=(\frac{337}{56},\frac{-5}{14},\frac{5}{56})[/tex]
. CAR WASH Shea and Tucker are washing their father's car. Shea can wash it by herself in 20 minutes. Tucker can wash it by himself in 30 minutes. 19 How long does it take them to wash the car if they work together?
Answer:
12 minutes
Step-by-step explanation:
First we figure out how much of a car each person can wash in 1 minute.
Shea can wash the car by herself in 20 minutes
Therefore, she can wash a car in 1 minute = [tex]\frac{1}{20}[/tex]
Tucker can wash a car in 30 minutes.
Tucker can wash the car in 1 minute = [tex]\frac{1}{30}[/tex]
Thus, in one minute together they wash a car
[tex]\frac{1}{20}[/tex] + [tex]\frac{1}{30}[/tex] = [tex]\frac{50}{600}[/tex] = [tex]\frac{1}{12}[/tex]
In one minute together they can wash = [tex]\frac{1}{12}[/tex] of a car
Time needed to wash entire car together = 12 minutes.
It takes them 12 minutes to wash the car.
Final answer:
Shea and Tucker can wash their father's car in 12 minutes if they work together.
Explanation:
To find out how long it takes Shea and Tucker to wash the car together, we can use the concept of work rates.
Shea can wash the car by herself in 20 minutes, so her work rate is 1/20 of the car per minute.
Tucker can wash the car by himself in 30 minutes, so his work rate is 1/30 of the car per minute.
When two people work together, their work rates are additive. So, if Shea and Tucker work together, their combined work rate is 1/20 + 1/30 = 1/12 of the car per minute.
Since the work rate is the reciprocal of the time taken, the time it takes them to wash the car together is 12 minutes. Therefore, it takes Shea and Tucker 12 minutes to wash their father's car if they work together.
M1Q7.) Construct a box plot from the data below
There are 16 numbers.
The median is 92.5 ( find the middle two values and divide by 2).
Minumum is 81
Maximum is 109
First quartile is 88.25 (Find median of the lower half of numbers).
Third quartile is 97.75 (Find median of the upper half of numbers.)
The interquartile range is 9.5 ( Difference between the first and third quartile).
Plotting that data in a box plot, the correct one looks like #1
Find the median.
92 and 93 are both middle numbers so add and divide by two.
92 + 93 = 185
185 / 2 = 92.5
Minimum (smallest number): 81
Maximum (largest number): 109
Find the median of the lower values behind the median.
88.25
Find the mean of the higher values ahead of the median.
97.75
Subtract to find the interquartile range.
97.75 - 88.25 = 9.5
The only option with these characteristics is Option A.
Best of Luck!
1000 mL of D5W is ordered to be infused over 5 hours. The drop factor is 10 gtt/mL. How many gtt/min should be given to infuse the 1000 mL over 5 hours?
Answer:
flow rate of to infuse the 1000 mL over 5 hours is 33.33 gtt/min
Step-by-step explanation:
Given data
volume = 1000 mL
time = 5 hours
drop factor = 10 gtt/min
to find out
flow rate of to infuse the 1000 mL over 5 hours
Solution
we know flow rate formula i.e.
flow rate = ( volume × drop factor ) / time .................1
here time will be in min so time = 5 hours = 5 × 60 = 300 min
put volume, drop factor and time value in equation 1 and we get flow rate
flow rate = ( volume × drop factor ) / time
flow rate = ( 1000 × 10 ) / 300
flow rate = 33.33 gtt/min
flow rate of to infuse the 1000 mL over 5 hours is 33.33 gtt/min
Buses headed to Longmont arrive in downtown Boulder every 30 minutes starting at 8:37am, whereas buses heading to Denver arrive 15 minutes starting at 8:31am. (a) If a passenger arrives at the station uniformly between 8:30am and 9:30am and then gets onto the first bus that arrives, what is the probability she goes to Longmont? (b) What is the passenger arrives uniformly between 8:45am and 9:45am?
Answer:
(a) 20% to Longmont; 80% to Denver
(b) 20% to Longmont; 80% to Denver
Step-by-step explanation:
(a) The bus to Longmont is the first bus to arrive, only between 8:31 and 8:37, and again between 9:01 and 9:07. That is, for a total of 12 minutes every hour, the Longmont bus is the first to arrive. The probability of going to Longmont is 12/60 = 1/5 = 20%.
__
(b) Same as for (a). As long as passenger arrival times are uniform within an hour, the probability is the same.
what is the solution if the inequality shown below? a-1>11
Answer:
a > 12
Step-by-step explanation:
Isolate the variable a. Treat the > sign like the = sign. What you do to one side, you do to the other. Add 1 to both sides:
a - 1 (+1) > 11 (+1)
a > 11 + 1
a > 12
a > 12 is your answer.
~
Answer:
[tex]\huge \boxed{a>12}[/tex]
Step-by-step explanation:
Add by 1 from both sides.
[tex]\displaystyle a-1+1>11+1[/tex]
Simplify, to find the answer.
[tex]\displaystyle 11+1=12[/tex]
[tex]\displaystyle a>12[/tex], which is our answer.
Find the least squares approximation of the the data (0, 1), (1, 2), (2, 1/2) (3, 3) using the quadratic function p(x) = a_0 + a_1 x + a_2 x^2. Plot p(x) along with the data to compare.
Answer:
The required function is [tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex].
Step-by-step explanation:
The given data points are (0, 1), (1, 2), (2, 1/2) and (3, 3).
Let the quadratic function is defined as
[tex]p(x)=a_0+a_1x+a_2x^2[/tex] .... (1)
Using graphing calculator, we get
[tex]a_0=1.325[/tex]
[tex]a_1=-0.675[/tex]
[tex]a_2=0.375[/tex]
Substitute [tex]a_0=1.325[/tex], [tex]a_1=-0.675[/tex] and [tex]a_2=0.375[/tex] in function (1), to find the quadratic function.
[tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex]
Therefore the required function is [tex]p\left(x\right)=1.325-0.675x+0.375x^2[/tex].
The graph of data points and quadratic function is shown below.
Consider two functions f and g on [1, 8] such that integral^8_1 f(x) dx = 9, integral^8_1 g(x) dx = 5, integral^8_5 f(x) dx = 4, and integral^5_1 g (x) dx = 3. Evaluate the following integrals. a. integral^5_1 2f(x) dx = (Simplify your answer.) b. integral^8_1 (f(x) - g (x)) dx = (Simplify your answer.) c. integral^5_1 (f (x) - g (x)) dx = (Simplify your answer.) d. integral^8_5 (g(x) - f(x)) dx = (Simplify your answer.) e. integral^8_5 7g(x) dx = (Simplify your answer.) f. integral^1_5 3f(x) dx = (Simplify your answer.)
I'll abbreviate the definite integral with the notation,
[tex]I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx[/tex]
We're given
[tex]I(f,1,8)=9[/tex][tex]I(g,1,8)=5[/tex][tex]I(f,5,8)=4[/tex][tex]I(g,1,5)=3[/tex]Recall that the definite integral is additive on the interval [tex][a,b][/tex], meaning for some [tex]c\in[a,b][/tex] we have
[tex]I(f,a,b)=I(f,a,c)+I(f,c,b)[/tex]
The definite integral is also linear in the sense that
[tex]I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)[/tex]
for some constant scalars [tex]k,\ell[/tex].
Also, if [tex]a\ge b[/tex], then
[tex]I(f,a,b)=-I(f,b,a)[/tex]
a. [tex]I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}[/tex]
b. [tex]I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}[/tex]
c. [tex]I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}[/tex]
d. [tex]I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}[/tex]
e. [tex]I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}[/tex]
f. [tex]I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}[/tex]
In this integral calculus problem, we leverage properties of definite integrals to compute the values of various expressions. Key steps usually involve substituting given integral values and multiplying by constant factors when required
Explanation:To solve the problem, we first need to consider the properties of integral calculus, specifically those of definite integrals. A fundamental rule that is applicable here is that the product of a constant and an integral is the constant times the value of the integral.
So for problem a, integral^5_1 2f(x) dx = 2* integral^5_1 f(x) dx = 2 * 5 = 10.
Similarly, for problem b, integral^8_1 (f(x) - g (x)) dx = integral^8_1 f(x) dx - integral^8_1 g(x) dx = 9 - 5 = 4.
Following through similar steps of substitutions, we obtain the following solutions:
c. integral^5_1 (f (x) - g (x)) dx = 1d. integral^8_5 (g(x) - f(x)) dx = 1e. integral^8_5 7g(x) dx = 21 f. integral^1_5 3f(x) dx = 15Learn more about Integral Calculus here:https://brainly.com/question/34730103
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A medical laboratory tested 8 samples of human blood for acidity on the pH scale, with the results below. 7.1 7.5 7.6 7.4 7.3 7.3 7.3 7.5 a. Find the mean and standard deviation. b. What percentage of the data is within 2 standard deviations of the mean?
Answer:
Mean = 7.38
SD = 0.148
b. 95%
Step-by-step explanation:
Given data is:
7.1 7.5 7.6 7.4 7.3 7.3 7.3 7.5
Mean:
Mean = Sum/No. of values
= (7.1+7.5+7.6+7.4+7.3+7.3+7.3+7.5)/8
=59/8
=7.38
Standard Deviation:
x x-x' (x-x')^2
7.1 -0.28 0.0784
7.5 0.12 0.0144
7.6 0.22 0.0484
7.4 0.02 0.0004
7.3 -0.08 0.0064
7.3 -0.08 0.0064
7.3 -0.08 0.0064
7.5 0.12 0.0144
Total : 0.1752
Variance = Sum of squares/No of items
= 0.1752/8 = 0.0219
SD =√0.0219 = 0.148
b. What percentage of the data is within 2 standard deviations of the mean?
95% of data is within two standard deviations of mean in a standard normal distribution ..
The mean of the pH test results is 7.375, and the standard deviation is approximately 0.086.
Explanation:
To find the mean and standard deviation of the pH test results, we can use the following formulas:
Mean: Add up all the pH values and divide by the total number of samples (in this case, 8). So, (7.1 + 7.5 + 7.6 + 7.4 + 7.3 + 7.3 + 7.3 + 7.5) / 8 = 7.375.
Standard Deviation: Calculate the difference between each pH value and the mean, square each difference, calculate the mean of those squared differences, and then take the square root. Let's break it down into steps: Subtract the mean from each pH value: (7.1 - 7.375), (7.5 - 7.375), (7.6 - 7.375), (7.4 - 7.375), (7.3 - 7.375), (7.3 - 7.375), (7.3 - 7.375), (7.5 - 7.375). Square each difference: (0.0425)^2, (0.125)^2, (0.225)^2, (0.025)^2, (-0.075)^2, (-0.075)^2, (-0.075)^2, (0.125)^2. Calculate the mean of the squared differences: (0.0018 + 0.0156 + 0.0506 + 0.000625 + 0.005625 + 0.005625 + 0.005625 + 0.0156) / 8 = 0.0074. Take the square root of the mean: √0.0074 ≈ 0.086.
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The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs. Find the probability that the weight of a randomly selected steer is greater than 920lbs. Round your answer to four decimal places.
Answer: 0.7257
Step-by-step explanation:
Given : The weights of steers in a herd are distributed normally.
[tex]\mu= 1100\text{ lbs }[/tex]
Standard deviation : [tex]\sigma=300 \text{ lbs }[/tex]
Let x be the weight of the randomly selected steer .
Z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]
[tex]z=\dfrac{920-1100}{300}=-0.6[/tex]
The the probability that the weight of a randomly selected steer is greater than 920 lbs using standardized normal distribution table :
[tex]P(x>920)=P(z>-0.6)=1-P(z<-0.6)\\\\=1-0.2742531=0.7257469\approx0.7257[/tex]
Hence, the probability that the weight of a randomly selected steer is greater than 920lbs =0.7257
Dana leaves Las Vegas for LA at 2 p.m. driving at 55 mph. At 4 p.m. Lance leaves LA for Las Vegas driving at 45 mph along the same route. If the cities are 260 miles, what time do they meet?
Answer: They meet after 1 hour 42 minutes.
Step-by-step explanation:
Since we have given that
Dana leaves Las Vegas for LA at 2 p.m. driving at 55 mph.
Let the time taken by Dana be 't'.
Distance traveled by Dana would be 55t.
At 4 p.m. Lance leaves LA for Las Vegas driving at 45 mph along the same route.
It means after 2 hours Lance leave for LA.
So, time taken by Lance be 't-2'.
Distance traveled by Lance would be 45(t-2)
Total distance = 260 miles
According to question, it becomes,
[tex]55t+45(t-2)=260\\\\55t+45t-90=260\\\\100t=260-90\\\\100t=170\\\\t=1.7\ hours=1\dfrac{7}{10}=1\ hour\ and\ \dfrac{7\times 60}{10}\ minutes=1\ hour\ 42\ minutes[/tex]
Hence, they meet after 1 hour 42 minutes.
A 20% TIP ON A MEAL THAT COSTS $29.17. CHOOSE THE CORRECT ESTIMATE BELOW. A.$ 58.00 B.$ 5.80 C.$ 0.58 D. $ 8.70
Answer:
$5.80 Option B.
Step-by-step explanation:
It is given that a 20% tip on a meal that costs $29.17.
The cost of the meal = $29.17
Tip on a meal = 20%
Therefore, 20% of $29.17
= [tex]\frac{20}{100}[/tex] × 29.17
= 0.20 × 29.17
= 5.834
= $5.80
The correct estimate would be $5.80 Option B.
Find X.
A.124
B.138
C.282
D.69
To find x, find the difference between the outer angle and middle angle, then divide by two.
X = (210 - 72) / 2
x = 138 / 2
x = 69
The answer is D.
Answer
subtract 210 with 72 and then divide that by 2 and you get 138. so x=138.