Answer:
Step-by-step explanation:
Given that [tex]r = 0.952[/tex]
We have coefficient of determination
[tex]r^2 =0.952^2\\=0.906304[/tex]
=90.63%
This implies that nearly 91% of variation in change in dependent variable is due to the change in x.
The coefficient of determination is the square of the correlation (r) between predicted y scores and actual y scores; thus, it ranges from 0 to 1.
An R2 of 0 means that the dependent variable cannot be predicted from the independent variable.
An R2 of 1 means the dependent variable can be predicted without error from the independent variable.
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.
Use a substitution method to solve both of the following DEs, stating the general solution clearly, and showing all work clearly. a. dy/dx - y = e^xy^2 (Solve explicitly for y.) b. dy/dx = x + y/x - y (You can leave the General Solution in implicit form.)
[tex]1.\rightarrow \frac{dy}{dx}-y=e^x y^2\\\\\rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=e^x\\\\ \text{put},\frac{-1}{y}=z\\\\ \frac{dy}{y^2} =d z\\\\ \frac{dy}{dx} \times \frac{1}{y^2}=\frac{dz}{dx}\\\\\frac{dz}{dx} +z=e^x\\\\ \text{Integrating factor}=e^{\int {1} \, dx}\\\\=e^x \\\\ \text{Multiplying both sides by }e^x\\\\e^x(\frac{dz}{dx} +z)=e^{2x}\\\\ \text{Integrating both sides}\\\\z e^x=\frac{e^{2x}}{2}+C\\\\ \frac{-e^x}{y}=\frac{e^{2x}}{2}+C[/tex]
-----------------------------------------------------------------------------------------------------------------
[tex]\rightarrow \frac{dy}{dx}=x+\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}-x=\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}+y(1-\frac{1}{x})=x\\\\\text{Integrating factor}=e^{\int{1-\frac{1}{x}}\,dx}\\\\=e^{x-\log x}\\\\ \text{Multiplying both sides by} e^{x-\log x}\\\\e^{x-\log x}\times[\frac{dy}{dx}+y(1-\frac{1}{x})]=x \times e^{x-\log x}\\\\y\times e^{x-\log x} =\int x \times e^{x-\log x} \, dx}\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{e^{\log x}}\,dx\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{x} \, dx\\\\y\times e^{x-\log x}=e^x+K[/tex]
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!
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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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.
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).
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.
d) neither one-to-one nor onto. 15. Determine whether each of these functions is a bijection from R to R. a) f(x)=2x+1 b) f(x)=x2+1 c) f(x) r3 d) f(x) (x2 +1)/(r2 +2) a function f(x)-ex from the set of real
Answer:
The only bijection is f(x)=2x+1.
I took r to be a constant.
Step-by-step explanation:
Bijections are both onto and one-to-one.
Onto means every element of the codomain gets hit. Here the codomain is the set of real numbers. So you want every y to get hit.
One-to-one means you don't want any y to get hit more than once.
f(x)=2x+1 is a linear function. It is diagonal line so every element of the codomain will get hit and hit only once so this function is onto and one-to-one.
f(x)=x^2+1 is a quadratic function. It is parabola so not every element of our codomain will get not get hit and of those that do get hit they get hit more than once. So this is neither onto or one-to-one.
f(x)=r^3 is a constant function. It is a horizontal line so not every y will get hit so it isn't onto. The same y is being hit multiple times so it isn't one-to-one.
f(x)=(x^2+1)/(r^2+2) is a quadratic. It is a parabola. Quadratic functions are not onto or one-to-one.
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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Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. To be 99% sure that we will not be late in completing the project, we should request a completion time of _______ work-days.
Answer:
233 days.
Step-by-step explanation:
Problems of normally distributed samples can be solved using 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 problem, we have that:
[tex]\mu = 200, \sigma = 10[/tex]
To be 99% sure that we will not be late in completing the project, we should request a completion time of ...
This is the value of X when Z has a pvalue of 0.99. So X when Z = 2.325.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.325 = \frac{X - 200}{10}[/tex]
[tex]X - 200 = 10*2.325[/tex]
[tex]X = 232.5[/tex]
So the correct answer is 233 days.
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......
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.
Give the largest interval I over which the general solution is defined. PLEASE EXPLAIN HOW!!!
(x^2-1)dy/dx+2y=(x+1)^2
Divide both sides by [tex]x^2-1[/tex] to get a linear ODE,
[tex]\dfrac{\mathrm dy}{\mathrm dx}+\dfrac2{x^2-1}y=\dfrac{x+1}{x-1}[/tex]
In order for this operation to be valid in the first place, we require that [tex]x\neq\pm1[/tex] (since that would make [tex]\dfrac1{x^2-1}[/tex] undefined, which we don't want to happen). Then we are forcing any solution to the ODE to exist on any of the three intervals, [tex](-\infty,-1)[/tex], [tex](-1, 1)[/tex], or [tex](1,\infty)[/tex], and either the first or third of these can be chosen as the largest interval.
In case you also need to solve the ODE: Multiply both sides by [tex]\dfrac{1-x}{1+x}[/tex], so that
[tex]\dfrac{1-x}{1+x}\dfrac{\mathrm dy}{\mathrm dx}-\dfrac2{(1+x)^2}y=-1[/tex]
Then the left side can be condensed as the derivative of a product, since
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}\right]=-\dfrac2{(1+x)^2}[/tex]
and we have
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}y\right]=-1[/tex]
Integrate both sides:
[tex]\displaystyle\int\frac{\mathrm d}{\mathrm dx}\left[\frac{1-x}{1+x}y\right]\,\mathrm dx=-\int\mathrm dx[/tex]
[tex]\dfrac{1-x}{1+x}y=-x+C[/tex]
[tex]\implies\boxed{y=\dfrac{(-x+C)(1+x)}{1-x}}[/tex]
The largest interval over which the general solution is defined for the given differential equation is [-1, ∞).
Here's how:
Rewrite the differential equation in proper form.Analyze the coefficients to determine the interval of definition.In this case, the interval is determined by the denominator of the coefficient of dy/dx.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
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.
. 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.
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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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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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.
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
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 new drug on the market is known to cure 20% of patients with breast cancer. If a group of 20 patients is randomly
selected, what is the probability of observing, at most, one patient who will be cured of breast cancer?
A• (20)/1 (0-20)^1 (0.80)^19
B. 1-(20)/1 (.20)'(0.230)
C. (20)/0( .80)^20+(20/1)(.20)^1(.80)^19
D • (20)/0 (.80)^20
E. 1-(20/0)(.80)^20
01-(20)10.2010
Answer:
It's actually C
Step-by-step explanation:
don't forget about the probability of 0 too
you have to add the two probability formulas
The probability of observing, at most, one patient who will be cured of breast Cancer will be P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹). Then the correct option is C.
What is probability?Its basic premise is that something will almost certainly happen. The percentage of favorable events to the total number of occurrences.
A new drug on the market is known to cure 20% of patients with breast cancer.
p = 0.20
q = 1 – 0.20
q = 0.80
If a group of 20 patients is randomly selected.
The probability of observing, at most, one patient who will be cured of breast Cancer will be
P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹)
Then the correct option is C.
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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:
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
.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
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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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.
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.
A civil service exam yields scores which are normally distributed with a mean of 81 and a standard deviation of 5.5. If the civil service wishes to set a cut-off score on the exam so that 15% of the test takers fail the exam, what should the cut-off score be? Remember to round your z-value to 2 decimal places.
Answer:
The cutoff score should be 75.8 marks
Step-by-step explanation:
The cut-off should be set as the value corresponding to an area 15% in the normal distribution diagram.
Using the standard distribution tables we have value of standard normal deviate (Z) corresponding to area of 15% = -1.04
Thus we have
[tex]-1.04=\frac{X-\bar{X}}{\sigma }\\\\X=-1.04\times 5.5+81\\X=75.8[/tex]
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]