Express the sum of the polynomial 5x^2+6x−17 and the square of the binomial (x+6) as a polynomial in standard form.

We divide [0, 3] into [tex]n[/tex] subintervals,

[tex]\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\left[\dfrac6n,\dfrac9n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right][/tex]

so that the right endpoint of each subinterval is given according to the arithmetic sequence,

[tex]r_k=\dfrac{3k}n[/tex]

for [tex]1\le k\le n[/tex].

The Riemann sum is then

[tex]\displaystyle\sum_{k=1}^nf(r_k)\Delta x_k[/tex]

where

[tex]\Delta x_k=r_k-r_{k-1}=\dfrac{3k}n-\dfrac{3(k-1)}n=\dfrac3n[/tex]

With [tex]f(x)=2x^2[/tex], we have

[tex]\displaystyle\frac3n\sum_{k=1}^n2\left(\frac{3k}n\right)^2=\frac{54}{n^3}\sum_{k=1}^nk^2[/tex]

Recall that

[tex]\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]

The area under the curve [tex]f(x)[/tex] over the interval [0, 3] is then

[tex]\displaystyle\int_0^32x^2\,\mathrm dx=\lim_{n\to\infty}\frac{54n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}[/tex]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]

(a) Exactly two will be drunken drivers.

[tex]P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}[/tex]

[tex]P(x=2)=66(0.2)^{2}(0.8)^{10}[/tex]

[tex]P(x=2)=\approx 0.28347[/tex]

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

[tex]P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)[/tex]

[tex]P(x=3\text{ or }x=4)=P(x=3)+P(x=4)[/tex]

Using binomial we get

[tex]P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}[/tex]

[tex]P(x=3\text{ or }x=4)=0.236223+0.132876[/tex]

[tex]P(x=3\text{ or }x=4)\approx 0.369099[/tex]

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

[tex]P(x\geq 7)=1-P(x<7)[/tex]

[tex]P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)][/tex]

[tex]P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155][/tex]

[tex]P(x\leq 7)=1-[0.9961][/tex]

[tex]P(x\leq 7)=0.0039[/tex]

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

[tex]P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)[/tex]

[tex]P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315[/tex]

[tex]P(x\leq 5)=0.9806[/tex]

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

a) Exactly two will be drivers: 0.2835. b) Three or four will be drivers: 1.5622; c) At least 7 will be drivers: 32.5669 (rounded to four decimal places). d) At most 5 will be drivers: 0.8749

a) Exactly two will be drivers:

For this case, we'll use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Where:

n is the number of trials (in this case, the number of drivers in the sample), which is 12.

k is the number of successful trials (in this case, the number of drivers), which is 2.

p is the probability of success in a single trial (in this case, the probability of a driver being), which is 0.20.

(n choose k) is the number of combinations of n items taken k at a time.

Calculating:

P(X = 2) = (12 choose 2) * (0.20)^2 * (0.80)^10

= 66 * 0.04 * 0.1073741824

= 0.2834678413

b) Three or four will be drivers:

For this case, we'll find P(X = 3) and P(X = 4) and then add them together.

For P(X = 3):

P(X = 3) = (12 choose 3) * (0.20)^3 * (0.80)^9

= 220 * 0.008 * 0.134456

= 0.23757696

For P(X = 4):

P(X = 4) = (12 choose 4) * (0.20)^4 * (0.80)^8

= 495 * 0.016 * 0.16777216

= 1.3245924272

Adding them together:

P(X = 3 or 4) = 0.23757696 + 1.3245924272

= 1.5621693872

c) At least 7 will be drivers:

To find this probability, we need to calculate P(X = 7) + P(X = 8) + ... + P(X = 12).

For P(X = 7):

P(X = 7) = (12 choose 7) * (0.20)^7 * (0.80)^5

= 792 * 0.128 * 0.32768

= 32.5669376

Similarly, find P(X = 8), P(X = 9), P(X = 10), P(X = 11), and P(X = 12) using the same method.

Finally, add all these probabilities together.

d) At most 5 will be drivers:

To find this probability, we need to calculate P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).

For P(X = 0):

P(X = 0) = (12 choose 0) * (0.20)^0 * (0.80)^12

= 1 * 1 * 0.0687194767

= 0.0687194767

Similarly, find P(X = 1), P(X = 2), P(X = 3), P(X = 4), and P(X = 5) using the same method.

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Answer:

Step-by-step explanation:

The F statistic, calculated through one-way ANOVA, for this problem is 1.07, which aims to examine the differences in the averages of multiple groups. However, without details such as the significance level and degrees of freedom, this task cannot determine if there's a significant difference between the average gas mileages of the four car models.

The average gas mileage comparison across four car models represented by the car dealer is an example of a problem solved by the One-Way ANOVA statistical approach. This test aims to determine if there is a statistically significant difference between the means of multiple groups, in this case, the average mileage of four different car models.

To find the test statistic, we consider the between-group mean square (MS Between) and the within-group mean square (MS Within). In ANOVA, the F statistic is used which performs the test of two variances, and is calculated as the ratio of MS Between to MS Within. In this case, MS Between is represented by 'MS Rows' (47.5500) and MS Within by 'MS Error' (44.5167). So, the F statistic = MS Between / MS Within = 47.5500 / 44.5167 = 1.07 (rounded to two decimal places).

However, the value of the F statistic alone is not enough to conclude the test. The conclusion depends on the significance level, degrees of freedom, and the value from the F-distribution table. Without these details, we cannot conclude whether there's a significant difference in the average gas mileages of the four car models.

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Answer:

  8

Step-by-step explanation:

You can skip directly to the formula for the sum of an infinite sequence with first term a₁ and common ratio r:

  S = a₁/(1-r)

Your values of the variables in this formula are a₁ = 6 and r = 2/8. Putting these into the formula gives ...

  S = 6/(1 -2/8) = 6/(6/8) = 8

The sum of the infinite geometric sequence is 8.

_____

The above formula is the degenerate form of the formula for the sum of a finite sequence:

  S = a₁((rⁿ -1)/(r -1))

When the common ratio r has a magnitude less than 1, the term rⁿ tends to zero as n gets very large. When that term is zero, the sum of the infinite sequence is ...

  S = a₁(-1/(r-1)) = a₁/(1-r)

Answer: height of kite is 147.042 feets

Step-by-step explanation:

The diagram of the kite is shown in the attached photo

Triangle ABC is formed and it is a right angle triangle.

The kite string made an angle of 33 degrees with the ground. The string used was 90 yards We will convert the 90 yards to feets.

I yard = 3 feets

90 yards would become

90×3 = 270 feets

This 270 feets form the hypotenuse of the triangle.

To determine the height of the kite h, we will use trigonometric ratio

Sin# = opposite / hypotenuse

Where

# = 33 degrees

Hypotenuse = 270 feets

Opposite = h feets

Sin 33 = h/270

h = 270sin33

h = 270 × 0.5446 = 147.042 feets

Answer:

  7 ft 1 1/4 in

Step-by-step explanation:

The total width of 3 paintings and 2 spaces is ...

  (10.5 in) + (3 in) + (3 ft 6 in) + (3 in) + (2 ft 2 3/4 in)

  = (10.5 +3 +6 +3 +2.75) in + (3 +2) ft

  = 25.25 in + 5 ft = 1.25 in + 24 in + 5 ft

  = 7 ft 1 1/4 in

The total width of wall space needed for the paintings is 7 feet 1 1/4 inches.

_____

If two more 3-inch spaces are added, one on each end, then the total width is 7 feet 7 1/4 inches. The problem isn't clear about that, saying only that there are spaces between the paintings.

Answer:

m= -0.1

b= -1.2

Step-by-step explanation:

The equation of the given line is

y=10x, slope of the given line is 10.

As the line is perpendicular to this line, the slope  is [tex]-\frac{1}{10}[/tex]

m= -0.1

( product of slope of two perpendicular lines is -1)

thus the equation is

[tex]y = (-\frac{1}{10})x + b[/tex]

[tex]y = (-0.1)x + b[/tex]

now to find the value of b, input the point coordinates (8, -2) in the above equation.

[tex]-2 = (-0.1)(8) + b[/tex]

[tex]b = -2 + (0.8)[/tex]

b = -1.2

Answer:3

Step-by-step explanation:plug in the numbers in the algebraic expression. Should look like this: 6(9+3)/(3+9)2. After you set it up like this you have to get rid of the parentheses by multiplying the outside number to the numbers in the parentheses. So do: 6×9=54, then 6×3=18 so now your problem should look like this 54+18/ (3+9)2 do the same with the other side so: 2×3=6, then 9×2=18 now your problem should look like this: 54+18/6+18 now you add each side up 54+18=72 and 6+18=24 . Then divide those 2 answers which looks like this 72/24 = 3.

By Green's theorem,

[tex]\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy[/tex]

where [tex]C[/tex] is the circle [tex]x^2+y^2=9[/tex] and [tex]D[/tex] is the interior of [tex]C[/tex], or the disk [tex]x^2+y^2\le1[/tex].

Convert to polar coordinates, taking

[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta[/tex]

Then the work done by [tex]\vec F[/tex] on the particle is

[tex]\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0[/tex]

The work done in a force field on a particle moving around a circle is found by calculating and integrating the line integral of the force field over the path defined by the circle. It involves substituting the parametric representation of the circle into the force field equation and incorporating the directional aspect of the line integral.

This problem is related to calculating work done in a force field and involves concepts from vector calculus. The work done is calculated based on the line integral of the force field F over a path C, defined by the parametric representation of the circle. More specifically, we'll need to find the line integral of F along the path C, and then integrate that from 0 to 2π (since the particle moves around the circle once).

The parametric representation of the circle x² + y² = 9 is x = 3cosθ, y = 3sinθ, where -π ≤ θ ≤ π. Substitute these into the force field equation you'll get F(3cosθ, 3sinθ) = 9cos²θ i + 9cosθsinθ j.

To find the work done, we'll compute the line integral of F over the path C, which in this case is the circle. Since the movement is in the clockwise direction and when it comes to line integrals, the direction matters, we'll need to use -θ instead of θ to represent our parameter. So we're integrating F along C from 0 to 2π.

The exact calculation of the integral might require a bit of time and effort, but you should end up with the work done by the force field on the particle that moves once around the circle oriented in the clockwise direction.

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Answer:

$ 725

Step-by-step explanation:

Price of call option = 7.25

buyer have to pay for one call option contract. Assume each contract is for 100 = 100 * 7.25 =  $ 725

Answer:

32.9 mpg

Step-by-step explanation:

Population mean (μ) = 25.5 mpg

Standard deviation (σ) = 4.5 mpg

Assuming a normal distribution for gasoline use, the manufacturer wants his car to be at the 95-th percentile of the distribution. The 95-th percentile has a corresponding z-score of 1.645. The expression for the z-score for a given gasoline use rate 'X' is:

[tex]z=\frac{X-\mu}{\sigma} \\1.645=\frac{X-25.5}{4.5} \\X=32.9\ mpg[/tex]

The gasoline use rate for the new car must be at least 32.9 mpg

To outperform 95% of the current compacts in fuel economy, the gasoline use rate for the new car must be at least 33.7775 miles per gallon.

To develop a compact car that outperforms 95% of the current compacts in fuel economy, the gasoline use rate for the new car must be at least as good as the top 5% of the current compacts. To find this, we use the z-score formula: z = (x - mean) / standard deviation. Since we want to find the value for x that corresponds to the top 5% (or 0.05) of the distribution, we can find the z-score by using the inverse normal distribution table. Once we have the z-score, we can use the formula z = (x - mean) / standard deviation to solve for x.

Using the inverse normal distribution table, we find that the z-score corresponding to the top 5% is approximately 1.645. Plugging this value into the formula and rearranging to solve for x, we get:

x = mean + (z * standard deviation)

Substituting in the given values, we have:

x = 25.5 + (1.645 * 4.5) = 33.7775

Therefore, the gasoline use rate for the new car must be at least 33.7775 miles per gallon to outperform 95% of the current compacts in fuel economy.

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Answer:

test statistic is 0.176

Step-by-step explanation:

Given Data

p>33

a=0.05

n=50

x=33.3

d(population deviation)=12

Test statistics=?

Solution

Test statistic z=(p-x)\(d/sqrt(50))

z=(33.3-30)\(12\sqrt(50))

z=0.176

Answer:

The lengths of the sides of a right triangle are

Longer leg = 0.4 units.

Shorter leg = 0.3 units.

Step-by-step explanation:

Given:

Hypotenuse = 0.5 units

Let the length of shorter leg of right triangle be x units then

According to the given condition,

length of longer leg will be (0.1 + x) units

Now,we know for a right triangle,by Pythagoras theorem we have

[tex](\textrm{Hypotenuse})^{2} = (\textrm{Longer leg})^{2}+(\textrm{Shorter leg})^{2}[/tex]

substituting the values we get

[tex]0.5^{2}= (x+0.1)^{2}+ x^{2}[/tex]

Applying [tex](a+b)^{2}= a^{2}+2ab+b^{2}[/tex]  we get

[tex]0.25= x^{2} +2\times 0.1\times x+ 0.1^{2} + x^{2} \\2x^{2} +0.2x+0.01-0.25=0\\2x^{2} +0.2x-0.24=0\\[/tex]

which is a quadratic equation

dividing the equation throughout by two we get

[tex]x^{2} +0.1x-0.12=0\\\textrm{on factorizing we get}\\x^{2} +0.4x-0.3x-0.12=0\\(x+0.4)(x-0.3)=0[/tex]

[tex]\therefore (x-0.3)= 0\\\therefore x=0.3[/tex]

Since x cannot be negative we  take

x = 0.3 units

∴ Longer leg = x + 0.1

                     = 0.3+0.1

                     =0.4 units

So, the lengths of the sides of a right triangle are

Longer leg = 0.4 units.

Shorter leg = 0.3 units.

Final answer:

To find the lengths of the sides of the right triangle with a hypotenuse of 0.5 units and one leg being 0.1 units longer than the other, you can use the Pythagorean theorem to set up an equation and solve for the lengths of the legs.

Explanation:

To find the lengths of the sides of the triangle, let's assume the length of the shorter leg is x units. Then, the length of the longer leg would be x + 0.1 units. Applying the Pythagorean theorem, where a and b are the legs and c is the hypotenuse:

a² + b² = c²

(x)² + (x + 0.1)² = (0.5)²

Simplifying the equation and solving for x, we get x ≈ 0.226 units for the shorter leg, and the longer leg would be x + 0.1 ≈ 0.326 units.

In probability, the sampling distribution of the sample proportions can be approximated by a normal probability distribution when D. Both a and b are true.

The sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever np is greater than or equal to 5 and when n(1-p) is greater than or equal to 5.

When the continuous random variable is uniformly distributed between a and b, the probability density function between a and b is 1/(b - a).

The probability that the sample mean will be between 80.54 and 88.9 will be:

= P[Z = 88.9 - 84)/(12/✓36)] - P(Z = 80.54 - 84)/(12/✓36)]

= P(Z = 2.45) - P(Z = -1.73)

= 0.9929 - 0.0418

= 0.9511

Therefore, the probability is 0.9511.

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Answer: The upper confidence limit for the 90% confidence interval would be 0.415.

Step-by-step explanation:

Since we have given that

n = 400

x = 150

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{150}{400}=0.375[/tex]

At 90% confidence interval, z = 1.645

So, margin of error would be

[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.375\times 0.625}{400}}\\\\=0.0398[/tex]

So, the upper limit would be

[tex]\hat{p}+0.0398\\\\=0.375+0.0398\\\\=0.415[/tex]

Hence, the upper confidence limit for the 90% confidence interval would be 0.415.

Answer:

P(The grade of the boy is A| He has played video games)

is, [tex]\simeq 0.54[/tex]

P(The grade of the boy is B| He has played video games)

is [tex]\simeq 0.33[/tex]

P(The grade of the boy is C| He has played video games)

is [tex]\simeq 0.14[/tex]

Step-by-step explanation:

The total no. of boys who have played video games,

= (730 + 444 + 190)

=1360

Now, from the given data,

P(The grade of the boy is A| He has played video games)

= [tex]\frac {730}{1360}[/tex]

[tex]\simeq 0.54[/tex]

P(The grade of the boy is B| He has played video games)

= [tex]\frac {444}{1360}[/tex]

[tex]\simeq 0.33[/tex]

P(The grade of the boy is C| He has played video games)

= [tex]\frac {190}{1360}[/tex]

[tex]\simeq 0.14[/tex]

The conditional distribution of grades for boys who have played games shows that 53% received A's and B's, 33% received C's, and 14% received D's and F's. This is calculated by dividing each grade category by the total number of boys who played games and rounding to two decimal places.

To find the conditional distribution of grades for boys who have played games, we need to calculate the proportion of each grade category relative to the total number who played.

Steps to Calculate Conditional Distribution

Therefore, the conditional distribution of grades for boys who have played games is 0.53 for A's and B's, 0.33 for C's, and 0.14 for D's and F's.

Answer: time it will take him to 17.5 minutes to complete the test from start to finish

Step-by-step explanation:

During the standardized test, Paul answered the first 22 questions in 5 minutes.

If he answers 22 questions 5 minutes

He would answer one question in 5/22 = 0.227 minutes

He continues to answer questions at the same rate. This means that his unit rate of answering 1 question in 0.227 minutes is constant throughout the test.

Total number of questions on the test is 77. The time it will take him to complete the test from start to finish will be

Unit rate of answering questions × total number of questions

= 0.227 × 77 = 17.5 minutes

Answer:

As the sample size increases, the variability decreases.

Step-by-step explanation:

Variability is the measure of actual entries from mean.  The less the deviations the less would be the variance.

For a sample  of size n, we have by central limit theorem the mean of sample follows a normal distribution for random samples of large size.

X bar will have std deviation as [tex]\frac{s}{\sqrt{n} }[/tex]

where s is the square root of variance of sample

Thus we find the variability denoted by std deviation is inversely proportion of square root of sample size.

Hence as sample size increases, std error decreases.

As the sample size increases, the variability decreases.

In statistics, as the sample size increases, the variability typically decreases because more data points allow a closer approximation of the true population mean. Therefore, larger sample sizes provide a narrower confidence interval, leading to less variability.

The variability of a distribution is a measure of the differences from the mean that occur in the data points. The sample size refers to the number of data points collected in your sample from a population. In statistics, as the sample size increases, the variability or scatter of your dataset normally decreases, because a larger number of data points give a more accurate representation of the population you are studying.

Variability is affected by sample size in the following way: Increasing the sample size leads to a decrease in the error bound and makes a narrower confidence interval. This is because more data points enable a closer estimation of the true population mean. Thus, as your sample size grows larger, the variability decreases, and your data forms a tighter grouping around the mean.

For example, if you are conducting a survey, and you take four different samples of 50 people each from the same population, you might see differing outcomes due to sample variability. However, if you were to increase your sample size to perhaps 500 people, the results are likely to have less sample variability.

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Answer:

34% of lightbulb replacement requests numbering between 39 and 50.

Step-by-step explanation:

The 68-95-99.7 rule states that, for a normally distributed random variable:

68% are within 1 standard deviation of the mean(34% between one standard deviation below and the mean, 34% between the mean and one standard deviation above the mean).

95% are within 2 standard deviations of the mean.

99.7% are within 3 standard deviations of the mean.

In this problem, we have that:

The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11.

Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?

50 is the mean

39 is one standard deviation below the mean.

This means that 34% of lightbulb replacement requests numbering between 39 and 50.

Answer:

a. Sample proportion ^p= 0.12

b. It is appropiate.

c. [0.0447;0.1953]

d. [^p ± d]

Step-by-step explanation:

Hello!

Given the information I'll assume that the variable of study has a binomial distribution:

X~Bi(n;ρ)

The sample data:

n= 50

"Success" x= 6

Sample proportion ^p= x/n = 6/50 = 0.12

Now, your study variable has a binomial distribution, but remember that the Central Limit Theorem states that given a big enough sample size (usually n≥ 30) you can approximate the sample proportion distribution to normal.

Since the sample is 50 you can apply the approximation, your sample proportion will have the following distribution:

^p≈ N( p; [p(1 - p)]/n)

With E(^p)= p and V(^p)= [p(1 - p)]/n.

This allows you to estimate the population proportion per Confidence Interval using the Z-distribution:

[^p±[tex]Z_{1-\alpha /2}[/tex]*√(^p(1 - ^p)/n)]

Since you are estimating the value of p, you'll use the estimated standard deviation (i.e. with the sample proportion instead of the population proportion)

to calculate the interval.

At level 90% the interval is:

[0.12±1.64*√([0.12(1 - 0.12)]/50)]

[0.0447;0.1953]

The margin of error (d) of an interval is half its amplitude (a)

if a= Upper bond - Low bond

then d= (Upper bond - Low bond)/2

d= (0.1953-0.0447)/2

d= 0.0753

And since the interval structure is "estimator" -/+ "margin of error" you can write it as:

[^p ± d]

I hope you have a SUPER day!

By Green's theorem,

[tex]\displaystyle\int_C(7(x^2-y)\,\vec\imath+3(y^2+x)\,\vec\jmath)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial3(y^2+x)}{\partial x}-\frac{\partial7(x^2-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]

[tex]\displaystyle=10\iint_D\mathrm dx\,\mathrm dy[/tex]

where [tex]D[/tex] is the region bounded by the closed curve [tex]C[/tex]. The remaining integral is 10 times the area of [tex]D[/tex].

Since [tex]D[/tex] is a circle in both cases, and we're given the equations for them right away, it's just a matter of determining the radius of each one and plugging it into the well-known formula for the area of a circle with radius [tex]r[/tex], [tex]\pi r^2[/tex].

(a) [tex]C[/tex] is a circle with radius 3, so the line integral is [tex]10\pi(3^2)=\boxed{90\pi}[/tex].

(b) [tex]C[/tex] is a circle with radius [tex]R[/tex], so the line integral is [tex]\boxed{10\pi R^2}[/tex].

Let ,

First calculate the value for given equation by using green's theorem,

Since, the formula for greens theorem is:

The given equation is,

[tex]\int\ C [7(x^2-y)i +3(y^2+x)j]dr[/tex]

Here,

[tex]F=[7(x^2-y)i +3(y^2+x)j][/tex]

Now to calculate the value of [tex]Curl F[/tex],

[tex]Curl F=\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial k} \\7(x^2-y)&(y^2+x)&0\end{array}\right] \\\\CurlF=[\frac{\partial}{ \partial y} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (y^2+x)-\frac{\partial}{\partial y} (7x^2-y)]\\\\Curl F=0+0+10k\\\\CurlF=10k[/tex]

Substitute in equation (1),

[tex]\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10k *kdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10dA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=10\int\ \int\ dA[/tex]

The remaining integral is [tex]10[/tex] times the area of region .

The general equation is,

[tex]x^2+y^2=r^2[/tex]

The area of a circle is [tex]\pi r^2[/tex]  .

Hence, area of region of circle is [tex]10\pi r^2[/tex].

Now,

(a) The given equation is [tex](x-2)^2+(y-3)^2=3^2[/tex],

  C is a circle with radius 3, so the line integral is

         [tex]10\pi (3)^2=90\pi[/tex] .

(b) The given equation is [tex](x-a)^2+(y-b)^2=R^2[/tex],

    C is a circle with radius [tex]R[/tex] , so the line integral is,

         [tex]10\pi (R)^2=10\pi R^2[/tex].

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Answer:

a) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.166[/tex]

b) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]

Step-by-step explanation:

Part a

So then the random variable is given by this table

X     | 1      | 2    | 3    | 4     |

P(X) | 0.2 | 0.2 | 0.2 | 0.4 |

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=1*0.2 +2*0.2 +3*0.2 +4*0.4=2.8[/tex]

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=(1^2 *0.2)+(2^2 *0.2)+(3^2 *0.2)+(4^2 *0.4)=9.2[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =9.2-(2.8)^2 =1.36[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.17[/tex]

Part b

So then the random variable is given by this table

X     | -20  | -10 | 0   | 10  |20  |

P(X) | 0.1 | 0.2 | 0.4 | 0.1 | 0.2  |

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=(-20*0.1) +(-10*0.2) +(0*0.4) +(10*0.1)+(20*0.2)=1[/tex]

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=((-20)^2 *0.1)+((-10)^2 *0.2)+(0^2 *0.4)+(10^2 *0.1)+(20^2 *0.2)=150[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =150-(1)^2 =149[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]

The standard deviation of the first probability distribution is 1.10 and of the second distribution is 15.23.

The standard deviation σ of a probability distribution is calculated by first finding its mean μ, then using the formula:

σ = √[Σ(x-μ)^2 * P(X = x)]

For the first distribution, the mean μ = (1*0.2) + (2*0.2) + (3*0.2) + (4*0.4) = 3.2. The standard deviation σ = √[(1-3.2)^2 * 0.2 + (2-3.2)^2 * 0.2 + (3-3.2)^2 * 0.2 + (4-3.2)^2 * 0.4] = 1.10.

For the second distribution, the mean μ = (-20*0.1) + (-10*0.2) + (0*0.4) + (10*0.1) + (20*0) + (30*0.2) = -2. The standard deviation σ = √[(-20+2)^2 * 0.1 + (-10+2)^2 * 0.2 + (0+2)^2 * 0.4 + (10+2)^2 * 0.1 + (20+2)^2 * 0 + (30 + 2)^2 * 0.2] = 15.23.

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Answer:

a) 7,858,539,612

b) 2,080,201,662

c) 346,700,277

d) 7,511,839,335

e) 410,040

Step-by-step explanation:

a. How many ways are there to extend the 6 offers to 6 of the 136 candidates?

Combinations of 136 (candidates) taken 6 (offers) at a time without repetition:

[tex]\large \binom{136}{6}=\frac{136!}{6!(136-6)!}=\frac{136!}{6!130!}=7,858,539,612[/tex]

b. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer, but we do not know which?

There are 6 ways Computer Joe can get an offer. Now there are left 5 offers and 135 candidates. So there are  

6 times combinations of 135 taken 5 at a time without repetition:

[tex]\large 6*\binom{135}{5}=6*\frac{135!}{5!(135-5)!}=6*\frac{135!}{5!130!}=2,080,201,662[/tex]

c. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer for job number 2?

Now, we only have 5 offers and 135 candidates. So there are combinations of 135 taken 5 at a time without repetition:

[tex]\large \binom{135}{5}=\frac{135!}{5!(135-5)!}=\frac{135!}{5!130!}=346,700,277[/tex]

d. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is not getting any offers?

Here we have 6 offers and 135 candidates, given that  Computer Joe is out. So there are combinations of 135 taken 6 at a time without repetition:

[tex]\large \binom{135}{6}=\frac{135!}{6!(135-6)!}=\frac{135!}{6!129!}=7,511,839,335[/tex]

e. How many ways are there for 3 interviewers to select 3 resumes (one resume for each interviewer) from the pile of 136 resumes for the first interview round?

There are combinations of 136 taken 3 at a time without repetition:

[tex]\large \binom{136}{3}=\frac{136!}{3!(136-3)!}=\frac{136!}{3!133!}=410,040[/tex]

Answer:

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)

Step-by-step explanation:

Given that y(t) satisfies the differential equation

[tex]\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)[/tex]

Separate the variables to have

[tex]\frac{dy}{y^2(y-1)(y-5)} =dt[/tex]

Left side we can resolve into partial fractions

Let [tex]\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}[/tex]

Taking LCD we get

[tex]1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 =  -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B\\[/tex]

By equating coeff of y^3 we have

A+C+D=0

[tex]C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}[/tex]

Hence left side =

[tex]\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C[/tex]

b) y is increasing whenever dy/dt>0

dy/dt =0 at points y =0, 1 and 5

dy/dt >0 in (-\infty,0) U (1,5)

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)

Answer:

a) y = 0 , 5,1

b) y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Step-by-step explanation:

Given data:

differential equation is given as

[tex]\frac{dy}[dt} = y^4 -6y^3+ 5y^2[/tex]

a) constant solution

[tex] y^4 -6y^3+ 5y^2 = 0 [/tex]

taking y^2 from all part

[tex]y^2(y^2 - 6y -5) = 0[/tex]

solution of above equation is

y = 0 , 5,1

b) for which value y is increasing

[tex]\frac{dy}{dt}  > 0[/tex]

y^2(y - 5) (y -1) > 0

y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Answer:

150.51 dB

Step-by-step explanation:

Data provided in the question:

decibel level of sound at 161 km distance = 180 dB

d₁ = 161 km

d₂ = 4800 km

I₁ = 180 db

The formula for intensity of sound is given as:

I = [tex]10\log(\frac{I_2}{I_1})[/tex]

and the relation between intensity and distance is given as:

I ∝ [tex]\frac{1}{d^2}[/tex]

or

Id² = constant

thus,

I₁d₁² = I₂d₂²

or

[tex]\frac{I_2}{I_1}=\frac{d_1}{d_2}[/tex]

therefore,

I = [tex]10\log(\frac{d_1}{d_2})^2[/tex]

or

I = [tex]10\times2\times\log(\frac{161}{4,800})[/tex]

or

I = 20 × (-1.474)

or

I = -29.49

Therefore,

the decibel level on Rodriguez Island, 4,800 km away

= 180 - 29.49

= 150.51 dB

To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, you can use the inverse square law and the formula dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, you can calculate the decibel level on Rodriguez Island to be approximately 201.94 dB.

To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, we can use the inverse square law. According to the problem, the decibel level at a distance of 161 km is 180 dB. To find the decibel level at 4,800 km, we can use the formula: dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, we get: dB2 = dB1 - 20log10(d1/d2). Substituting dB1 = 180 dB, d1 = 161 km, and d2 = 4800 km, we can calculate the decibel level on Rodriguez Island.

dB2 = 180 - 20log10(161/4800) ≈ 180 - 20(-1.097) ≈ 180 + 21.94 ≈ 201.94 dB

Therefore, the decibel level on Rodriguez Island, 4,800 km away, is approximately 201.94 dB.

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Answer: y = - 466.375

Step-by-step explanation:

y varies jointly as x and z.

This means that y varies directly as x and also varies directly as z.

In order to remove the proportionality symbol, we will introduce a constant of proportionality, k. Therefore,

y = kxz

The next step is to determine the value of k

if y = 205 when x = –5 and z = –8.

we will substitute these values into the equation to determine k.

205 = k × -5 × -8

205 = 40k

k = 205/40 = 5.125

Therefore, the equation becomes

y = 5.125xz

We want to determine y when x = - 13 and z = 7

y = 5.125 × - 13 × 7

y = 5.125 × - 91

y = - 466.375

Answer:

A. Direct, strong

Step-by-step explanation:

If the Pearson's correlation coefficient is positive the relationship coefficient is direct. If it is negative, it is inverse.

If is considerate to be strong if it is larger than 0.7

In this problem, we have that:

The Pearson's correlation coefficient was found to be .78. It is positive, and larger than 0.7. So the correct answer is:

A. Direct, strong

Answer:

Step-by-step explanation:

a) What is the size of each cache?

Direct mapped cache= 2^index * size of cache line= 2^10 * 1B lines = 1KB.

2-way set associative cache= 2^index * size of cache line * 2 ways=2^7 * 4 words *2ways= 128 4B lines * 2 ways = 1KB

Fully associative cache= number of cache lines* size of each line= 256 * 4B lines = 1KB

b) How much space does each cache need to store tags?

Direct mapped cache= 1024 * 6-bit tags = 6Kb

2-way set associative cache= 256 * 7-bit tags = 1792 bits

Fully associative cache= 256 * 14-bit tags = 3584 bits

c) Which   cache   design   has   the   most   conflict   misses?   Which   has   the   least?    

Direct mapped cache has likely the most conflict misses, because it is direct mapped. Fully associative cache has the least since it is fully associative so it can never have conflict misses.

d) The   following   information   is   given   to   you: hit   rate   for   the   3   caches   is   50%,   70%   and   90%  but   did   not   tell   you   which   hit   rate   corresponds   to   which   cache,   which   cache   would   you   guess  corresponded   to   which   hit   rate?   Why?    

Since the size of all three caches is same size and as we said in the previous answer that direct mapped cache has more conflict misses and fully associative has the least so direct mapped will have 50%, 2-way set associative 70%, and Fully associative will have 90% hit rate.

e) Assuming   the   miss   time   for   each   is   20   cycles,   what   is   the   average   service   time   for   each? (Service   Time   =   (hit   rate)*(hit   time)   +   (miss   rate)*(miss   time)

We are given hit rates and miss rates. Also miss time=2o cycles for each cache and hit time= 1, 2, 5 for direct mapped, 2-way set associative and fully associative cache respectively.

Direct mapped= 0.5*1 + 0.5*20 = 10.5 cycles

2-way set associative= 0.7*2 + 0.3*20 = 7.4 cycles

Fully associative cache= 0.9*5 + 0.1*20 = 6.5 cycles.

The size of each cache, the space needed to store tags, the cache design with the most and least conflict misses, guessing which cache corresponds to each hit rate, and calculating the average service time for each cache.

Each cache design has different levels of conflict misses. The direct-mapped cache (D1) has the most conflict misses because multiple memory locations map to the same cache line. The 2-way set associative cache (D2) has fewer conflict misses because each set can hold two cache lines, reducing the chance of multiple memory locations mapping to the same set. The fully associative cache (D3) has the least conflict misses because any memory location can be stored in any cache line, reducing conflicts.

Answer:

The correct option is C. either a t test or an analysis of variance can be performed.

Step-by-step explanation:

Consider the provided information.

Hence, Either a t test or an analysis of variance can be performed to determine whether the means of two population are equal.

Therefore, the correct option is C. either a t test or an analysis of variance can be performed.

To determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.

The t-test is a statistical test that is used to compare the means of two groups, to determine if there is any significant difference between the two groups or populations.

The ANOVA, like the t-test is also used to compare means, however, it is used when the groups or populations involved are more than two.

Therefore, to determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.

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Answer:

The hypotheses used in this situation

[tex]H_0:\mu = 14.6[/tex]

[tex]H_a:\mu \neq 14.6[/tex]

Step-by-step explanation:

We are given that  Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours.

Mean = [tex]\mu = 14.6[/tex]

Claim : The amount UMSL students study is different from this 14.6 hour benchmark.

The hypotheses used in this situation

[tex]H_0:\mu = 14.6[/tex]

[tex]H_a:\mu \neq 14.6[/tex]