True or false. If a is any odd integer, then a^2 + a is even. Explain this.

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

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%.

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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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:

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.

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......

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.

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]

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

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.

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.

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

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

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:

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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

To construct a 95% confidence interval with a margin of error of 0.052, a sample size of 300 is needed.

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.

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.

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:

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).

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.

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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Answer: i dont now

Step-by-step explanation:

u have to add them togther i guess

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]

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.

[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]

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[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]

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

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.

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.

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

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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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.

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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Answer:

7/32

Step-by-step explanation:

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.

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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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.

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!

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])

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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