new vintage 1965 convertible requires an ol change every 400 miles and replacement of all fBuids every 13,000 mles If these services have pust been performed by the dealer, how many mles from now will both be due at the same tme? The services will both be due at the same ime agan in mdes (Type a whole rumber)

Check the picture below.

let's recall that a kite is a quadrilateral, and thus is a polygon with 4 sides

sum of all interior angles in a polygon

180(n - 2)      n = number of sides

so for a quadrilateral that'd be 180( 4 - 2 ) = 360, thus

[tex]\bf 3b+70+50+3b=360\implies 6b+120=360\implies 6b=240 \\\\\\ b=\cfrac{240}{6}\implies b=40 \\\\[-0.35em] ~\dotfill\\\\ \overline{XY}=\overline{YZ}\implies 3a-5=a+11\implies 2a-5=11 \\\\\\ 2a=16\implies a=\cfrac{16}{2}\implies a=8[/tex]

Answer:

D.)a = 8; b = 40

Step-by-step explanation:

it is on edgeinuity

a. All points on the board are equally likely to be hit with a probability of 1/(area of board), or

[tex]f_{X,Y}(x,y)=\begin{cases}\dfrac1{4\pi}&\text{for }x^2+y^2\le4\\\\0&\text{otherwise}\end{cases}[/tex]

b. To find the marginal distribution of [tex]X[/tex], integrate the joint distribution with respect to [tex]y[/tex], and vice versa. We can take advantage of symmetry here to compute the integral:

[tex]\displaystyle\int_y f_{X,Y}(x,y)\,\mathrm dy=2\int_0^{\sqrt{4-x^2}}\frac{\mathrm dy}{4\pi}=\frac{\sqrt{4-x^2}}{2\pi}[/tex]

[tex]f_X(x)=\begin{cases}\dfrac{\sqrt{4-x^2}}{2\pi}&\text{for }-2\le x\le2\\\\0&\text{otherwise}\end{cases}[/tex]

and by the same computation you would find that

[tex]f_Y(y)=\begin{cases}\dfrac{\sqrt{4-y^2}}{2\pi}&\text{for }-2\le y\le2\\\\0&\text{otherwise}\end{cases}[/tex]

c. We get the conditional distributions by dividing the joint distributions by the respective marginal distributions:

[tex]f_{X\mid Y=y}(x)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}[/tex]

[tex]f_{X\mid Y=y}(x)=\begin{cases}\dfrac1{2\sqrt{4-y^2}}&\text{for }-2\le y\le2\text{ and }x^2\le4-y^2\\\\0&\text{otherwise}\end{cases}[/tex]

and similarly,

[tex]f_{Y\mid X=x}(y)=\begin{cases}\dfrac1{2\sqrt{4-x^2}}&\text{for }-2\le x\le2\text{ and }y^2\le4-x^2\\\\0&\text{otherwise}\end{cases}[/tex]

d. You can compute this probability by integrating the joint distribution over a part of the circle (call it "B" for bullseye):

[tex]\displaystyle\iint_Bf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^{0.25}\frac r{4\pi}\,\mathrm dr\,\mathrm d\theta=\frac1{64}[/tex]

(using polar coordinates) The easier method would be to compute the area of a circle with radius 0.25 instead, then divide that by the total area of the dartboard.

[tex]\dfrac{\pi\left(\frac14\right)^2}{\pi\cdot2^2}=\dfrac1{64}[/tex]

e. The event that [tex]X>1[/tex] is complementary to the event that [tex]X\le1[/tex], so

[tex]P(X>1)=1-P(X\le1)=1-F_X(1)[/tex]

where [tex]F_X(x)[/tex] is the marginal CDF for [tex]X[/tex]. We can compute this by integrate the marginal PDF for [tex]X[/tex]:

[tex]F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<-2\\\\\dfrac12+\dfrac1\pi\sin^{-1}\dfrac x2+\dfrac{x\sqrt{4-x^2}}{4\pi}&\text{for }-2\le x<2\\\\1&\text{for }x\ge2\end{cases}[/tex]

Then

[tex]P(X>1)=1-F_X(1)=\dfrac13-\dfrac{\sqrt3}{4\pi}\approx0.1955[/tex]

f. We found that either random variable conditioned on the other is a uniform distribution. In particular,

[tex]f_{Y\mid X=1}(y)=\begin{cases}\dfrac1{2\sqrt3}&\text{for }y^2\le3\\\\0&\text{otherwise}\end{cases}[/tex]

Then

[tex]P(Y>0.3\mid X=1)=1-P(Y\le0.3\mid X=1)=1-F_{Y\mid X=1}(0.3)[/tex]

where [tex]F_{Y\mid X=x}(y)[/tex] is the CDF of [tex]Y[/tex] conditioned on [tex]X=x[/tex]. This is easy to compute:

[tex]F_{Y\mid X=1}(y)=\displaystyle\int_{-\infty}^yf_{Y\mid X=1}(t)\,\mathrm dt=\begin{cases}0&\text{for }y<-\sqrt3\\\\\dfrac{y+\sqrt3}{2\sqrt3}&\text{for }-\sqrt3\le y<\sqrt3\\\\1&\text{for }y\ge\sqrt3\end{cases}[/tex]

and we end up with

[tex]P(Y>0.3\mid X=1)=\dfrac{10-\sqrt3}{20}\approx0.4134[/tex]

Final answer:

The joint PDF for a uniform distribution on a circular dart board is constant within the dart board and zero outside. Marginal and conditional PDFs are derived from the joint PDF. To compute probabilities such as the bulls eye hit or X > 1, integrate the corresponding PDF over the relevant range.

Explanation:

Finding the Joint and Marginal PDFs, and Conditional Probabilities

The joint probability density function (PDF) of X and Y for a uniform distribution over a circular dart board with radius 2 inches is constant within the circle and zero outside. First, we calculate the area of the circle, A = πr² = π(2)² = 4π square inches. The joint PDF f(x, y) will be 1/A for all points inside the dart board, and 0 otherwise.

The marginal PDFs are derived by integrating the joint PDF over the other variable. For instance, fX(x) is found by integrating f(x, y) over y, and fY(y) is found by integrating f(x, y) over x.

The conditional PDFs fX|Y and fY|X are derived from the joint PDF divided by the marginal PDF of the conditioned variable.

To find the probability of hitting the bulls eye, a circle of radius 0.25 inches, you'd integrate the joint PDF over the area of the bulls eye or simply calculate the area ratio of the bulls eye to that of the entire dart board.

The probability that X > 1 is found by integrating fX(x) from 1 to 2. If you know that X = 1, the conditional PDF of Y is fY|X(y|X=1). The probability that Y > 0.3 given X = 1 is calculated by integrating this conditional PDF from 0.3 to the upper limit set by the circle's boundary.

Answer: Last option.

Step-by-step explanation:

Given the inequality [tex]|x| > 5[/tex] you need to set up two posibilities. These posibilities are the following:

- FIRST POSIBILITY :

[tex]x>5[/tex]

- SECOND POSIBILTY:

 [tex]x<-5[/tex]

Then you get that:

 [tex]x<-5\ or\ x>5 [/tex]  

Therefore, through this procedure, you get that the solution set of the inequality  [tex]|x| > 5[/tex]  is this:

{[tex]x|x<-5\ or\ x>5[/tex]}

Answer:

C. {x|x < -5 or x > 5}

Step-by-step explanation:

Answer:

The value of [tex]f^{-1}(2)=8[/tex]

Step-by-step explanation:

* Lets revise how to find the inverse function

- At first write the function as y = f(x)

- Then switch x and y

- Then solve for y

- The domain of f(x) will be the range of f^-1(x)

- The range of f(x) will be the domain of f^-1(x)

* Now lets solve the problem

- The inverse of the logarithmic function is an exponential function

∵ [tex]f(x)=log_{4}(x + 8)[/tex]

- Write the function as y = f(x)

∴ [tex]y=log_{4}(x+8)[/tex]

- Switch x and y

∴ [tex]x=log_{4}(y+8)[/tex]

- Lets solve it to find y

# Remember: [tex]log_{a}b=n=====a^{n}=b[/tex]

- Use this rule to find y

∴ [tex]4^{x}=(y + 8)[/tex]

- Subtract 8 from both sides

∴ [tex]4^{x}-8=y[/tex]

∴ [tex]f^{-1}(x)=4^{x}-8[/tex]

- Lets substitute x by 2

∴ [tex]f^{-1}(2)=4^{2}-8[/tex]

- The value of 4² = 16

∴ [tex]f^{-1}(x)=16-8=8[/tex]

* The value of [tex]f^{-1}(2)=8[/tex]

To find the inverse of the function f(x) = log base 4 of (x+8) when evaluated at 2, we solve the equation log base 4 of (x+8) = 2 to find x. This gives us x = 8, so f^-1(2) = 8.

To find the value of f-1(2) for the function f(x) = log4(x+8), we first need to understand that f-1(x) denotes the inverse function of f(x). This means that if f(a) = b, then f-1(b) = a. To solve for f-1(2), we set f(x) equal to 2 and solve for x:

log4(x+8) = 2

42 = x+8 (using the property that if logb(a) = c then bc = a)

16 = x+8

x = 16 - 8

x = 8

Therefore, f-1(2) = 8.

Answer:

30 cubic feet

Step-by-step explanation:

Here we are given the volume of rectangular pyramid as 90 cubic feet as we are required to find the volume of rectangular prism.

For that we need to use the theorem which says that

the volume prism is always one third of the volume of the pyramid . Whether it is rectangular of triangular base. Hence in this case also the volume of the rectangular prism will be one third of the volume of the rectangular pyramid.

Volume of Rectangular prism = [tex]\frac{1}{3}[/tex] * Volume of rectangular pyramid

=  [tex]\frac{1}{3}[/tex] * 90

= 30

Answer:

270 is the answer just took test.

Step-by-step explanation:

For this case we have the following expression:

[tex]\frac {b ^ 7} {b ^ 6}[/tex]

By definition of division of powers of the same base, we have to place the same base and subtract the exponents, that is:

[tex]\frac {a ^ m} {a ^ n} = a ^ {n-m}[/tex]

So:

[tex]\frac {b ^ 7} {b ^ 6} = b ^ {7-6} = b ^ 1 = b[/tex]

Answer:

b

Answer: [tex]b[/tex]

Step-by-step explanation:

 You need to remember a property called "Quotient of powers property". This property states the following:

[tex]\frac{a^m}{a^n}=a^{(m-n)}[/tex]

You can observe that the bases of the expression [tex]\frac{b^7}{b^6}[/tex] are equal, then you can apply the property mentioned before.

Therefore, you can make the division indicated in the exercise.

Then you get this result:

[tex]\frac{b^7}{b^6}=b^{(7-6)}=b[/tex]

Answer:

The mean  of a sampling distribution of sample means is 87

The standard deviation of a sampling distribution of sample = 4

Step-by-step explanation:

* Lets revise some definition to solve the problem  

- The mean of the distribution of sample means is called μx

- It is equal to the population mean μ

- The standard deviation of the distribution of sample means is

 called  σx

- The rule of σx = σ/√n , where σ is the standard  deviation and n

  is the size of the sample

* lets solve the problem  

- A population has a mean (μ) is 87

∴ μ = 87

- A standard deviation of 24

∴ σ = 24

- A sampling distribution of sample means with sample size n = 36

∴ n = 36

∵ The mean of the distribution of sample means μx = μ

∵ μ = 87

∴ μx = 87

* The mean  of a sampling distribution of sample means is 87

∵ The standard deviation of a sampling distribution of sample

   means σx  = σ/√n

∵ σ = 24 and n = 36

∴ σx = 24/√36 = 24/6 = 4

* The standard deviation of a sampling distribution of sample = 4

The mean of the sampling distribution of sample means is 87, identical to the population mean. The standard deviation of this sampling distribution is 4, which is the population standard deviation (24) divided by the square root of the sample size (36).

When working with populations and sampling distributions, the Central Limit Theorem is critical for understanding the behavior of sample means. Given a population with mean (μ) and standard deviation (σ), the mean of the sampling distribution of sample means will be the same as the population mean, and the standard deviation of the sampling distribution (σx) is equal to the population standard deviation divided by the square root of the sample size (n).

In this instance, the population has a mean (μ) of 87 and a standard deviation (σ) of 24. For a sample size (n) of 36, the mean of the sampling distribution of sample means (μx) is equal to the population mean:

μx = μ = 87

The standard deviation of the sampling distribution of sample means (σx) is calculated as follows:

σx = σ / √ n = 24 / √ 36 = 24 / 6 = 4

Therefore, the mean of the sampling distribution is 87, and standard deviation is 4.

Final answer:

To assess if the frame can support a 20 kN load given the factor of safety for buckling of member AB is 3, the critical buckling loads for both axes, considering the end conditions, should be calculated and compared after adjusting for the safety factor.

Explanation:

To determine if the frame can support a load of P = 20 kN with a factor of safety with respect to buckling of member AB being F.S. = 3, we need to calculate the critical buckling load for member AB. The member's critical buckling load depends on its end conditions, which are pinned for the x-x axis and fixed for the y-y axis buckling. The Euler Buckling Formula, which is Pcr = (π2EI) / (KL2), where E is the modulus of elasticity, I is the moment of inertia, K is the column effective length factor, and L is the actual length of the column, can be applied separately for each axis. For pin-ended columns, K=1, and for fixed-ended columns, K=0.5.

To ensure safety, the critical load calculated for both axes should be divided by the factor of safety, F.S. = 3. If the adjusted critical load for either axis is greater than the applied load (P = 20 kN), the frame can support the load. However, specific values for E, I, and L are required to perform these calculations, which are not provided in the question. Generally, for steel, E=200 GPa is a typical value, and the values for I and L need to be determined based on the cross-sectional and length dimensions of member AB. Without these specifics, an exact answer cannot be provided.

[tex]\bf 5y^4+11y^2+2\implies 5(y^2)^2+11y^2+2\implies (5y^2+1)(y^2+2)[/tex]

For this case we must factor the following polynomial:

[tex]5y ^ 4 + 11y ^ 2 + 2[/tex]

We rewrite [tex]y ^ 4[/tex]as [tex](y^ 2) ^ 2[/tex]:

[tex]5 (y ^ 2) ^ 2 + 11y ^ 2 + 2[/tex]

We make a change of variable:

[tex]u = y ^ 2[/tex]

We replace:

[tex]5u ^ + 11u + 2[/tex]

we rewrite the middle term as a sum of two terms whose product of 5 * 2 = 10 and the sum of 11.

So:

[tex]5u ^ 2 + (1 + 10) u + 2[/tex]

We apply distributive property:

[tex]5u ^ 2 + u + 10u + 2[/tex]

We factor the highest common denominator of each group.

[tex](5u ^ 2 + u) + (10u + 2)\\u (5u + 1) +2 (5u + 1)[/tex]

We factor again:

[tex](u + 2) (5u + 1)[/tex]

Returning the change:

[tex](y ^ 2 + 2) (5y ^ 2 + 1)[/tex]

ANswer:

[tex](y ^ 2 + 2) (5y ^ 2 + 1)[/tex]

Answer:   [tex]\bold{\dfrac{h^9}{g^9}}[/tex]

Step-by-step explanation:

The power rule is "multiply the exponents".

You must understand that the exponent of both h and g is 1.

Multiply 1 times 9 for both variables.

[tex]\bigg(\dfrac{h}{g}\bigg)^9=\bigg(\dfrac{h^1}{g^1}\bigg)^9=\dfrac{h^{1\times 9}}{g^{1\times 9}}=\large\boxed{\dfrac{h^9}{g^9}}[/tex]

Answer:

D

Step-by-step explanation:

It's a repeating pattern

Answer:

D

Reasoning:

the pattern keeps the previous color and alternates between blue and red.  The pattern is also increasing in number by 1.  Therefore there would be no reason for the fourth step to change previous color or numbers.  That eliminates A and B.  For C, since the pattern is adding a different color clockwise in each step, it is more likely that the pattern will continue to alternate, therefore D is most likely.

Answer:

  r ≈ 13.01

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that ...

  Cos = Adjacent/Hypotenuse

  cos(67.4°) = 5.00/r . . . . . . filling in the given values

Solving for r gives ...

  r = 5.00/cos(67.4°) ≈ 13.01

_____

Check your requirements for rounding. We rounded to 2 decimal places because the x-coordinate, 5.00, was expressed using 2 decimal places.

Answer:

[tex]\large\boxed{43^o}[/tex]

Step-by-step explanation:

[tex]\text{Faucets}\to12\%\\\\p\%=\dfrac{p}{100}\to12\%=\dfrac{12}{100}=0.12\\\\12\%\ \text{of}\ 360^o\to0.12\cdot360^o=43.2^o\approx43^o[/tex]

Answer: [tex]43^{\circ}[/tex]

Step-by-step explanation:

From the given pie-chart, the percentage for faucets = 12 %

We know that every circle has angle of [tex]360^{\circ}[/tex].

Now, the central angle for faucets is given by :-

[tex]\text{Central angle}=\dfrac{\text{Percent of Faucets}}{\text{100}}\times360^{\circ}\\\\\Rightarrow\ \text{Central angle}=\dfrac{12}{100}\times360^{\circ}=43.2^{\circ}\approx43^{\circ}[/tex]

Hence, the measure of the central angle for faucets [tex]\approx43^{\circ}[/tex]

Answer: There are 6045 millions world population during the period of 1950 to 2000.

Step-by-step explanation:

Since we have given that

The world population in the second half of the 20 the century by the equation:

[tex]P(t)=2560e^{0.017185t}[/tex]

We need to find the average world population during the period of 1950 to 2000.

So, there are 50 years between 1950 to 2000.

So, t = 50 years.

Therefore, the  average population would be

[tex]P(50)=2560e^{0.017185\times 50}\\\\P(50)=6045.15\\\\P(50)\approx 6045\ millions[/tex]

Hence, there are 6045 millions world population during the period of 1950 to 2000.

Answer:

D. 4

Step-by-step explanation:

Percent of drivers who wear seat belts = 68

Percent of drivers who do not wear seat belts = 100 - 68 = 32

Now, we know that for every 100 pull overs, 32 drivers will not be wearing belt.

32 drivers without seat-belt = 100 pullovers

1 driver without seat-belt = 100/32 pullovers

1 driver without seat-belt = 3.125 pullovers (4 pullovers)

So, a police officer should expect 4 pullovers until she finds a driver not wearing a seat-belt.

The police officer should expect to pull over approximately 4 drivers until finding one not wearing a safety belt.

To determine how many people a police officer should expect to pull over until she finds a driver not wearing a safety belt, given that only 68% of drivers wear their safety belt, follow these steps:

Step 1:

Calculate the probability of a driver not wearing a safety belt.

Since 68% of drivers wear their safety belt, the probability of a driver not wearing a safety belt is [tex]\( 1 - 0.68 = 0.32 \).[/tex]

Step 2:

Calculate the expected number of people to pull over.

The expected number of people to pull over is the reciprocal of the probability of a driver not wearing a safety belt. So, it's  [tex]\( \frac{1}{0.32} \).[/tex]

Step 3:

Perform the calculation.

[tex]\[ \frac{1}{0.32} \approx 3.125 \][/tex]

Step 4:

Interpret the result.

Since we can't pull over a fraction of a person, rounding up to the nearest whole number, the police officer should expect to pull over 4 drivers until finding one not wearing a safety belt.

So, the correct answer is option C: 4.

Answer:

(a)  99%

(b)  50%

Step-by-step explanation:

(a) All factorials after 1! have 2 as a factor, so are even. Thus 99 of the 100 factorials are even, for a probability of 99%.

__

(b) The first two sums are odd; the next two sums are even. The pattern repeats every four sums. There are 25 repeats of that pattern in 100 sums, so 2/4 = 50% of sums are even.

Answer:

  [2, 15]

Step-by-step explanation:

The graph shows the n/4 algorithm to be better (smaller run time) for n in the range 2 to 15.

We're given an inner product defined by

[tex]\langle p,q\rangle=p(-2)q(-2)+p(0)q(0)+p(2)q(2)[/tex]

That is, we multiply the values of [tex]p(x)[/tex] and [tex]q(x)[/tex] at [tex]x=-2,0,2[/tex] and add those products together.

[tex]p(x)=2x^2+6x+1[/tex]

[tex]q(x)=3x^2-5x-6[/tex]

The inner product is

[tex]\langle p,q\rangle=-3\cdot16+1\cdot(-6)+21\cdot(-4)=-138[/tex]

To find the norms [tex]\|p\|[/tex] and [tex]\|q\|[/tex], recall that the dot product of a vector with itself is equal to the square of that vector's norm:

[tex]\langle p,p\rangle=\|p\|^2[/tex]

So we have

[tex]\|p\|=\sqrt{\langle p,p\rangle}=\sqrt{(-3)^2+1^2+21^2}=\sqrt{451}[/tex]

[tex]\|q\|=\sqrt{\langle q,q\rangle}=\sqrt{16^2+(-6)^2+(-4)^2}=2\sqrt{77}[/tex]

Finally, the angle [tex]\theta[/tex] between [tex]p[/tex] and [tex]q[/tex] can be found using the relation

[tex]\langle p,q\rangle=\|p\|\|q\|\cos\theta[/tex]

[tex]\implies\cos\theta=\dfrac{-138}{22\sqrt{287}}\implies\theta\approx1.95\,\mathrm{rad}\approx111.73^\circ[/tex]

The question is about calculating an inner product, norms, and the angle between two polynomials in a two-dimensional space, represented by their coefficients. We use the provided polynomials and input values from the question to calculate these values.

The mathematics problem given refers to calculating the inner product and magnitude of two polynomials, as well as the angle between them, if they are represented in a two-dimensional space.

To find the values asked in this problem, we will use the given polynomials p(x) = 2x^2+6x+1 and q(x) = 3x^2-5x-6. The inner product < p,q > is calculated as follows: p(-2)q(-2) + p(0)q(0) + p(2)q(2).

To find the magnitudes, or norms, ||p|| and ||q||, we need to solve the expression for < p,p > and < q,q > respectively, and then take the square root of the result. The angle between the vectors, denoted as Theta (θ), can be computed using the formula cos θ = < p,q > / (||p||*||q||). Because cos θ is the cosine of the angle, θ = arccos(< p,q > / (||p||*||q||)).

Note that the exact value of the inner product, norms, and angle depends on the input values of -2, 0, and 2 for each polynomial.

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Answer:

  f(x) = f(-x) = x^2 -1 is an even function

Step-by-step explanation:

When f(x) = f(-x), the function is symmetrical about the y-axis. That is the definition of an even function.

___

An odd function is symmetrical about the origin: f(x) = -f(-x).

Answer:

(a) shortest ladder length ≈ 35.7 ft (rounded to tenth)

(b) L = (d/x +1)√(16+x²) . . . . where 16 is fence height squared

Step-by-step explanation:

It works well to solve the second part of the problem first, then put in the specific numbers.

We have not been asked anything about "b", so we can basically ignore it. Using the Pythagorean theorem, we find the length GH in the attached drawing to be ...

  GH = √(4²+x²) = √(16+x²)

Then using similar triangles, we can find the ladder length L to be that which satisfies ...

  L/(d+x) = GH/x

  L = (d +x)/x·√(16 +x²)

The derivative with respect to x, L', is ...

  L' = (d+x)/√(16+x²) +√(16+x²)/x - (d+x)√(16+x²)/x²

Simplifying gives ...

  L' = (x³ -16d)/(x²√(16+x²))

Our objective is to minimize L by making L' zero. (Of course, only the numerator needs to be considered.)

___

(a) For d=24, we want ...

  0 = x³ -24·16

  x = 4·cuberoot(6) ≈ 7.268 . . . . . feet

Then L is

  L = (24 +7.268)/7.268·√(16 +7.268²) ≈ 35.691 . . . feet

__

(b) The objective function is the length of the ladder, L. We want to minimize it.

  L = (d/x +1)√(16+x²)

The objective function for the length of the ladder in terms of x is L = sqrt((x+b)^2 + d^2).

To find the length of the shortest ladder that extends from the ground to the house without touching the fence, we can use the concepts of similar triangles. Let L be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house, and b be the distance from the ground to the point the ladder touches the house. In terms of x, the objective function for the length of the ladder is:

L = sqrt((x+b)^2 + d^2)

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Answer:

a) False. Because you can live in California AND watch American Idol at the same time

b) True. Because the number of patients is a whole number, like 1, 2 or 3. There is no 1.5 patient

c) False. The actual probability should become closer to the theoretical probablity

d) True

Answer:

True

Step-by-step explanation:

6m (2m + 10)

= (6m)(2m) + (6m)(10)

= 12m² + 60m

= 4(3m² + 15m)

Because we factored out a "4" in the expression, by observation, we can see that regardless of what value of integer m is chosen, the entire expression is divisible by the "4" which has been factored out of the parentheses. Hence it is True.

What is the divisibility test of 4?

If a number's last two digits are divisible by 4, the number is a multiple of 4 and totally divisible by 4.

6m(2m+10) = 6m cannot be said to be always divisible by 4 but is divisible by 2 as 6m = 2*3m

(2m+10) cannot say if it is divisible by but can say is divisible by 2 as we can take 2 commons from both the integer and rewrite it as 2*(m+5).

Hence multiplying both the numbers we get = 2*3m*2(m+5) and now we group 2's together we get 4*3m*(m+5), as the whole number is multiplied by 4 it is at least divisible by 4 once.

Hence proved that 6m(2m+10) is divisible by 4 and the answer is true.

learn more about Divisibility here

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Answer:

f(-3)=5x^2+5x*3

-f*3=5x^2+5*3x

-3f=5x^2+15x

f=-5x(x+3)/3

f(-9)=5x^2+5x*3

-f*9=5x^2+5*3x

-9f=5x^2+15x

f=-5x(x+3)/9

Step-by-step explanation:

hope it helps you?

Answer:

True

Step-by-step explanation:

If two events X and Y are mutually exclusive,

Then,

P(X∪Y) = P(X) + P(Y)

Let A represents the event of a diamond card and B represent the event of a heart card,

We know that,

In a deck of 52 cards there are 4 suit ( 13 Club cards, 13 heart cards, 13 diamond cards and 13 Spade cards )

That is, those cards which are heart can not be diamond card,

⇒ A ∩ B = ∅

⇒ P(A∩B) = 0

Since, P(A∪B) = P(A) + P(B) - P(A∩B)

⇒ P(A∪B) = P(A) + P(B)

By the above statement,

Events A and B are mutually exclusive,

Hence, the probability of selecting a 4 of diamonds or a 4 of hearts is an example of a mutually exclusive event is a true statement.

Answer:

[tex]N(L)=\frac{k}{L^2}[/tex]

Step-by-step explanation:

Here, N represents the number of animal species and L represents a certain body length,

According to the question,

[tex]N\propto \frac{1}{L^2}[/tex]

[tex]\implies N=\frac{k}{L^2}[/tex]

Where, k is the constant of proportionality,

Since, with increasing the value of L the value of N is decreasing,

So, we can say that, N is dependent on L, or we can write N(L) in the place of N,

Hence, the required function formula is,

[tex]N(L)=\frac{k}{L^2}[/tex]

Certainly! When we say that the number of animal species \( N \) is inversely proportional to the square of the body length \( L \), what we mean mathematically is that as the body length increases, the number of species decreases at a rate that is the square of the increase in length. This can be represented by the following formula:

\[ N = \frac{k}{L^2} \]

Here \( N \) is the number of species, \( L \) is the body length, and \( k \) is the constant of proportionality. This constant \( k \) represents the number of species at the unit body length (when \( L = 1 \)). The constant of proportionality is determined by the specific biological context, based on empirical data or theoretical considerations.

In this formula, \( L^2 \) denotes the body length squared, and the fraction represents the inverse relationship.

In summary, to find the number of species \( N \) for a given body length \( L \), we use the inverse square relationship with the constant of proportionality \( k \).

Answer:

True

Step-by-step explanation:

The sample space consists of any possible outcome

The total number of ways are:

                       462

When we are asked to select r items from a set of n items that the rule that is used to solve the problem is:

Method of combination.

Here the total number of bills of different values are: 7

i.e. n=7

(  $1, $2, $5, $10, $20, $50, and $100 )

and there are atleast five of each type of bill.

Also, we have to choose 5 bills i.e. r=5

The repetition is  allowed while choosing bills.

Hence, the formula is given by:

[tex]C(n+r-1,r)[/tex]

Hence, we get:

[tex]C(7+5-1,5)\\\\i.e.\\\\C(11,5)=\dfrac{11!}{5!\times (11-5)!}\\\\C(11,5)=\dfrac{11!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}\\\\\\C(11,5)=462[/tex]

           Hence, the answer is:

                  462

There are 462 different possible ways to choose five bills from a set of seven types with repetition allowed.

To determine the number of different ways to choose five bills from a set of bills with values $1, $2, $5, $10, $20, $50, and $100, we can use the combinatorial concept known as "combinations with repetition."

Given:

- Bill values:  [tex]\( \{1, 2, 5, 10, 20, 50, 100\} \)[/tex]

- Total number of types of bills: ( n = 7 )

- Number of bills to choose: ( r = 5 )

The formula for combinations with repetition (also known as "stars and bars") is:

\[

\binom{n + r - 1}{r}

\]

Substituting the values ( n = 7 ) and ( r = 5 ):

[tex]\[\binom{7 + 5 - 1}{5} = \binom{11}{5}\][/tex]

Now, calculate  [tex]\( \binom{11}{5} \):[/tex]

[tex]\[\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!}\][/tex]

First, compute the factorials:

[tex]\[11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\][/tex]

Next, compute [tex]\( \frac{11!}{6!} \):[/tex]

[tex]\[\frac{11!}{6!} = 11 \times 10 \times 9 \times 8 \times 7\]\[= 11 \times 10 = 110\]\[110 \times 9 = 990\]\[990 \times 8 = 7920\]\[7920 \times 7 = 55440\][/tex]

Now, compute:

[tex]\[\frac{55440}{120} = 462\][/tex]

Therefore, the number of different possible ways to choose the five bills is:

[tex]\[\boxed{462}\][/tex]

Answer:  0.1037

Step-by-step explanation:

Given : Mean : [tex]\mu=80\text{ miles per day}[/tex]

Standard deviation : [tex]\sigma = 30\text{ miles per day}[/tex]

The formula to calculate the z-score is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 97 miles per day ,

[tex]z=\dfrac{97-80}{30}\approx0.57[/tex]

For x = 107 miles per day ,

[tex]z=\dfrac{97-80}{30}=0.9[/tex]

The P-value =[tex]P(0.57<z<0.9)=P(z<0.9)-P(z<0.57)[/tex]

[tex]=0.8159398-0.7122603=0.1036795\approx0.1037[/tex]

Hence, the probability that a truck drives between 97 and 107 miles in a day = 0.1037

The probability that a truck drives between 97 and 107 miles in a day is approximately 0.101. We calculated this using the Z-scores for 97 and 107 miles, the standard normal distribution, and the properties of the normal distribution.

To find the probability that a truck drives between 97 and 107 miles in a day, we'll use the normal distribution. This probability is the equivalent of finding the area under the curve of the normal distribution between 97 and 107 miles.

First, we calculate the Z-scores for 97 and 107, where Z = (X - μ)/σ. Here, μ is the mean (80 miles), and σ is the standard deviation (30 miles).

Now, we look up these Z-scores in the standard normal distribution table or use a calculator with this function. Suppose the table gives us P(Z < 0.567) = 0.715 and P(Z < 0.9) = 0.816.

Then, the probability that a truck drives between 97 to 107 miles is P(0.567 < Z < 0.9) = P(Z < 0.9) - P(Z < 0.567) = 0.816 - 0.715 = 0.101.

Please note that results can vary slightly depending on the Z-table or calculator you use. All of the numbers after the decimal point are rounded to 4 decimal places.

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Answer:

[tex]\large\boxed{5y^4+11y^2+2=(y^2+2)(5y^2+1)}[/tex]

Step-by-step explanation:

[tex]5y^4+11y^2+2=5y^4+10y^2+y^2+2=5y^2(y^2+2)+1(y^2+2)\\\\=(y^2+2)(5y^2+1)[/tex]