An article in Fortune (September 21, 1992) claimed that nearly one-half of all engineers continue academic studies beyond the B.S. degree, ultimately receiving either an M.S. or a Ph.D. degree. Data from an article in Engineering Horizons (Spring 1990) indicated that 117 of 484 new engineering graduates were planning graduate study. Are the data from Engineering Horizons consistent with the claim reported by Fortune? Use a = 0.10 in reaching your conclusions. Find the P-value. Give your answer. The true proportion of engineering students planning graduate studies significantly different from 0.5 at a = 0.10. The P-value is less than (choose the least possible).

Answer:

add the decimals thats all

Step-by-step explanation:

Answer:

a)  13/52    and the second  12/51

b) Two solutions:

b.1 if we did not picked up an eight in the first two cards   4/50

b.2 there is an eight i the two previous    3/59

c ) (48/52)*(47/51)*(46/50)

Step-by-step explanation:

Condition: Cards are taken out without replacement

a) Probability of first card is heart

There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways

we have  13 heart cards and 52 cards  then probability of one heart card is  13/52   = 0.25

or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25

Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51

b) There are 4 eight (one for each suit )   P =  4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50

c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is

(48/52)*(47/51)*(46/50)

Answer:

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.

Step-by-step explanation:

1) Notation and definitions

[tex]X=34[/tex] number of people living at USA with a particular disease.

[tex]n=527[/tex] random sample taken

[tex]\hat p=\frac{34}{527}=0.0645[/tex] estimated proportion of people living at USA with a particular disease

[tex]p[/tex] true population proportion of people living at USA with a particular disease.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435[/tex]

[tex]0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855[/tex]

The 95% confidence interval would be given by (0.0435;0.0855)

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.

Answer:

Option D is right

Step-by-step explanation:

given that a two sample test for mean of independent samples to be done.

We create hypotheses as:

[tex]H_0: \ bar x = \bar y[/tex] vs alternate suitably right or left or two tailed according to the needs.

The conditions needed for conducting this test would be

I. Both populations are approximately normally distributed

II. Both sample sizes greater than 30

III. Population of differences is approximately normally distributed

i.e. either i, ii or III

Option D is right.

Answer:

a) t =  2.7638

b) t = - 2.6245

c) t = 3.1058      on the right side    and

   t  = -3.1058    on the left

Step-by-step explanation:

a)Determine critical value for a right-tail test  for α = 0.01 level of significance and 10 degrees of fredom

From t-student table we find:

gl  =  10   and α = 0.01       ⇒  t =  2.7638

b)Determine   critical value for a left-tail test  for α = 0.01 level of significance  and sample size n = 15

From t-student table we find:

gl  =  14   and α = 0.01           gl  =  n - 1      gl  = 15 - 1    gl = 14

t = - 2.6245

c) Determine critical value for a two tails-test  for α = 0.01 level of  significance the  α/2   =   0.005  and sample size  n = 12

Then

gl  =  11   and α = 0.005

t = 3.1058      on the right side of the curve and by symmetry

t = - 3.1058  

From t-student table we find:

x is replaced with Y, so it is a reflection across the line y = x

Answer:

y=x

Step-by-step explanation:

Answer: 95% confidence interval would be (0.344,0.456).

Step-by-step explanation:

Since we have given that

n = 295

x = 118

so, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{118}{295}=0.4[/tex]

At 95% confidence, z = 1.96

So, margin of error would be

[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.4\times 0.6}{295}}\\\\=0.056[/tex]

so, 95% confidence interval would be

[tex]\hat{p}\pm \text{margin of error}\\\\=0.4\pm 0.056\\\\=(0.4-0.056,0.4+0.056)\\\\ =(0.344,0.456)[/tex]

Hence, 95% confidence interval would be (0.344,0.456).

Answer:

Step-by-step explanation:

Answer:

10 ft x 10 ft

Area = 100 ft^2

Step-by-step explanation:

Let 'S' be the length of the southern boundary fence and 'W' the length of the eastern and western sides of the fence.

The total area is given by:

[tex]A=S*W[/tex]

The cost function is given by:

[tex]\$ 120 = \$3*2W+\$6*S\\20 = W+S\\W = 20-S[/tex]

Replacing that relationship into the Area function and finding its derivate, we can find the value of 'S' for which the area is maximized when the derivate equals zero:

[tex]A=S*(20-S)\\A=20S-S^2\\\frac{dA}{dS} = \frac{d(20S-S^2)}{dS}\\0= 20-2S\\S=10[/tex]

If S=10 then W =20 -10 = 10

Therefore, the largest area enclosed by the fence is:

[tex]A=S*W\\A=10*10 = 100\ ft^2[/tex]

Answer: [tex]60,540< \mu<70,060[/tex]

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex]\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\sigma[/tex] = Population standard deviation.

n= sample size

[tex]\overline{x}[/tex] = Sample mean

z* = Critical z-value .

Given :  [tex]\sigma=\$17,000[/tex]

n= 49

[tex]\overline{x}= \$65,300[/tex]

Two-tailed critical value for 95% confidence interval = [tex]z^*=1.960[/tex]

Then, the 95% confidence interval would be :-

[tex]65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}[/tex]

[tex]=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}[/tex]

[tex]=65,300-4760< \mu<65,300+4760[/tex]

[tex]=60,540< \mu<70,060[/tex]

Hence, the 95​% confidence interval for estimating the population mean [tex](\mu)[/tex] :

[tex]60,540< \mu<70,060[/tex]

Answer:

Option b) Increase the sample mean

Step-by-step explanation:

Given that a researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6.

This is a paired test with test statistic

=mean diff/std error

Mean difference would increase if sample mean increases.

This would increase the test statistic

Or otherwise decrease in variance will increase the test statistic

Or Increase in sample size would also increase test statistic

Of all these the II option is definite in increasing the  likelihood of rejecting the null hypothesis because this would definitely increase the chances of rejecting H0.

Others may also have effect but not as much direct as sample mean difference.

Because variance and sample size have influence only upto square root of the difference.

Answer:

Step-by-step explanation:

Descartes' rule of signs tells you this function, with its signs {- - - + +}, having one sign change, will have one positive real root.

When odd-degree terms have their signs changed, the signs {- + - - +} have three changes, so there will be 1 or 3 negative real roots.

The constant term (10) tells us the y-intercept is positive. The sum of coefficients is -11 -5 -9 +12 +10 = -3, so f(1) < 0 and there is a root between 0 and 1.

When odd-degree coefficients change sign, the sum becomes -11 +5 -9 -12+10 = -17, so there is a root between -1 and 0.

__

Synthetic division by (x+1) gives a quotient of -11x^3 +6x^2 -15x +27 -17/(x+1), which has alternating signs, indicating -1 is a lower bound on real roots.

Real roots are located in the intervals [-1, 0] and [0, 1].

_____

The remaining roots are complex. All roots are irrational.

The attached graph confirms that roots are in the intervals listed here. Newton's method iteration is used to refine these to calculator precision. Dividing them from f(x) gives a quadratic with irrational coefficients and complex roots.

Answer:

a)[tex]H_0 :\mu = 4\\ H_1 : \mu \neq 4[/tex] , n = 15 , X=3.4 , S=1.5 , α = .05

Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}[/tex]

[tex]t =-1.549[/tex]

p- value = 0.607(using calculator)

α = .05

p- value > α

So, we failed to reject null hypothesis

b)[tex]H_0 :\mu = 21\\ H_1 : \mu < 21[/tex] , n =75 , X=20.12 , S=2.1 , α = .10

Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}[/tex]

[tex]t =-3.6290[/tex]

p- value = 0.000412(using calculator)

α = .1

p- value< α

So, we reject null hypothesis

(c) [tex]H_0 :\mu = 10\\ H_1 : \mu \neq 10[/tex], n = 36, p-value = 0.061.

Assume α = .05

p-value = 0.061.

p- value > α

So, we failed to reject null hypothesis

The decision to reject the null hypothesis in each scenario depends on comparing calculated test statistics (or given p-value) to critical values associated with the significance level. In (a) and (c), we may not reject H0, and in (b) we would reject H0 if our test statistic is larger than the critical value.

In hypothesis testing, we compare the test statistic, often a t-value, to the critical value determined by our chosen significance level, α. If the test statistic is greater, we reject the null hypothesis (H0).

(a) In the first scenario, we must calculate the test statistic: t = (X - µ) / (S/√n) = (3.4 - 4) / (1.5/√15); if this absolute value is less than our critical value associated with α = .05 and degrees of freedom = 14, we do not reject H0.

(b) For the second, the test statistic would be (20.12 - 21) / (2.1/√75); compare its value to the critical value associated with α = .10 and degrees of freedom = 74. If it is larger, we reject H0 due to the lower tail test in this scenario.

(c) In the third scenario, the p-value is given. Since our p-value = 0.061 is larger than our α = .05, we would not reject H0.

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Answer:

± 0.0736

Step-by-step explanation:

Data provided in the question:

randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4

Confidence level = 90%

sample size, n = 120

Now,

For 90% confidence level , z-value = 1.645

Width of the confidence interval = ± Margin of error

= ± [tex]z\times\sqrt\frac{p\times(1-p)}{n}[/tex]

= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]

= ± 0.07356 ≈ ± 0.0736

Hence,

The correct answer is option  ± 0.0736

The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice

To find the width of a confidence interval for a proportion, we can use the formula:

[tex]\[ \text{Width} = Z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]

where:

[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,

[tex]\( p \)[/tex] is the sample proportion,

[tex]\( n \)[/tex] is the sample size.

Given:

[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],

[tex]Sample\ size \( n = 120 \)[/tex]

[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]

Substitute the values into the formula:

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]

[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]

[tex]\[ \text{Width} = 0.0736 \][/tex]

Answer:

0.6284,0.4335,0.1953.0.9786

Step-by-step explanation:

Given that X the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk is following a poisson distribution with parameter 4

a) [tex]P(X\leq 4)=0.6284\\P(X<4)=0.4335[/tex]

b) [tex]P(4\leq x<5)\\=P(4)=0.1953\\[/tex]

c) P([tex]5\leq x)[/tex]=0.8046

d) the probability that the number of anomalies does not exceed the mean value by more than one standard deviation

=[tex]P(0\leq X\leq 8)\\=F(8)-F(0)\\=0.9786[/tex]

Answer:

sin(t)j - cos(t)i

Step-by-step explanation:

Let's start with A:

Position vector = cos(t)i

Velocity vector = -sin(t)i (differentiating the position vector)

acceleration vector = -cos(t)i   (differentiating the velocity vector)

Then we go to B:

Position vector = sin(t)j

Velocity vector = cos(t)j

acceleration vector = -sin(t)j

Relative acceleration = -cos(t)i - (-sin(t)j) = sin(t)j - cos(t)i

Answer:

y = 2x^2

Step-by-step explanation:

The the coordinates (2,8) are expressed in terms of (x,y). Substitute the x with 2 and y with 8 and plug them into the formulas. The one that has the two sides equal to each other is the function that has (2,8) on its line.

Final answer:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. We will define the null and alternative hypotheses, calculate the Z-test statistic, compare the p-value to the significance level, and make a conclusion based on the results.

Explanation:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. Let's define our hypotheses:

Null hypothesis (H0): The average price of homes sold in the U.S. is not significantly different from $220,000.

Alternative hypothesis (Ha): The average price of homes sold in the U.S. is significantly less than $220,000.

Next, we need to calculate the test statistic. Since the population standard deviation is known, we can use a Z-test. The formula for the Z-test statistic is:

Z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

In this case, the sample mean is $210,000, the population mean is $220,000, the population standard deviation is $36,000, and the sample size is 81. Plugging these values into the formula:

Z = (210,000 - 220,000) / (36,000 / sqrt(81))

Calculating this gives us a Z-statistic of -1.6875.

The final step is to compare the p-value of the test statistic to the significance level. Since the significance level is 1%, the critical value is 0.01. We can use a Z-table or calculator to find the p-value corresponding to a Z-statistic of -1.6875. The p-value is approximately 0.0466.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. This means that there is sufficient evidence to conclude that there has been a significant decrease in the average price of homes sold in the U.S.

Answer: D) 0.438 < p < 0.505

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size

[tex]\hat{p}[/tex] = Sample proportion.

z* = critical value.

Given : A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.

i.e. n= 865

[tex]\hat{p}=\dfrac{408}{865}\approx0.4717[/tex]

Two-tailed critical avlue for 95% confidence interval : z* = 1.96

Then, the 95 % confidence interval for the true proportion of all voters in the state who favor approval will be :-

[tex]0.4717\pm (1.96)\sqrt{\dfrac{0.4717(1-0.4717)}{865}}\\\\\approx0.4717\pm 0.03327\\\\=(0.4717-0.03327,\ 0.4717+0.03327)=(0.43843,\ 0.50497)\approx(0.438,\ 0.505)[/tex]

Thus, the required 95% confidence interval : (0.438, 0.505)

Hence, the correct answer is D) 0.438 < p < 0.505

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

[tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]

We conclude that the line integral is 1/2

To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field and the area enclosed by the square. The line integral of the vector field along the square is equal to the double integral of the curl over the region enclosed by the square. Using this method, we can find the value of the line integral to be 1/3.

To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field. In this case, the vector field is F(x, y) = 7y^2x i + 8x^2y j. Taking the partial derivatives of its components with respect to x and y, we get curl(F) = (8x^2 - 14xy^2) k.

Next, we need to find the area enclosed by the square C, which is 1 unit^2. Using Green's Theorem, the line integral of F along C is equal to the double integral of curl(F) over the region D enclosed by C. Integrating curl(F) with respect to y, we get -7xy^2 + 6x^2y. Integrating this with respect to x over the given limits, we find the value of the line integral to be 1/3.

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Answer:

209 points

Step-by-step explanation:

Mean points scored (μ) = 182 points  

Standard deviation (σ) = 14 points

The z-score for any given game score 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

At, the 97th percentile of a normal distribution, the z-score, according to a z-score table, is 1.881.

Therefore, the minimum score, X, needed for a bowler to receive a trophy is:

[tex]1.881=\frac{X-182}{14}\\X=208.334[/tex]

Since only whole point scores are possible, X=209 points.

To find the minimum score at the 97th percentile for bowlers in a league, one must calculate the z-score for the 97th percentile and then apply the formula Score = μ + (z * σ) using the league's mean and standard deviation.

To find the minimum score needed for a bowler to receive a trophy (which is at or above the 97th percentile), we need to use the normal distribution properties. With a mean (μ) of 182 points and a standard deviation (σ) of 14 points, we can find the z-score corresponding to the 97th percentile using a z-table or a calculator with normal distribution functions. Once we have the z-score, we can use the formula:

Score = μ + (z * σ)

to calculate the score that corresponds to the 97th percentile.

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Answer:

a) Nina decreases the confidence level to 90%?  (Decrease)

b) Nina decreases the sample size to 34 locations? (Increase)

c) Nina increases the sample size to 70 locations?   (Decrease)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n=48 represent the original sample size  

Confidence =95% or 0.95

ME=4.28 represent the margin of error.

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

And the margin of error is given by the following expression:

[tex]ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (4)

Based on the formula (4) we can answer all the questions involved:

a) Nina decreases the confidence level to 90%?

On this case the value for [tex]t_{\alpha/2}[/tex] will also decrease so the margin of error would decrease.

b) Nina decreases the sample size to 34 locations?

If we analyze the original sample size of 48 we see that if we reduce the value of n to 34, the margin of error would increase, because n is on the denominator of the margin of error.

c) Nina increases the sample size to 70 locations?

If we analyze the original sample size of 48 we see that if we increase the value of n to 70, the margin of error would decrease, because n is on the denominator of the margin of error.

Answer:

53¢

Step-by-step explanation:

First, I'll put these in order.

20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;  

Then, I'll combine like terms.

30+30=60

40+40+40+40(or 40 x 4)=160

55+55=110

65+65=130

60+160+110+130+20+75+50+$1.50=$7.55/14=53¢

PLZ correct me if i'm wrong :-D

Answer:

absolute max= (4.243,18)

absolute min =(-1,-5.916)

absolute max=(pi/6, 2.598)

absolute min = (pi/2,0)

Step-by-step explanation:

a) [tex]f(t) = t\sqrt{36-t^2} \\[/tex]

To find max and minima in the given interval let us take log and differentiate

[tex]log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)[/tex]

It is sufficient to find max or min of Y

[tex]y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243[/tex]

In the given interval only 4.243 lies

And we find this is maximum hence maximum at  (4.243,18)

Minimum value is only when x = -1 i.e. -5.916

b) [tex]f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\[/tex]

Equate I derivative to 0

-2sint +1-2sin^2 t=0

sint = 1/2 only satisfies I quadrant.

So when t = pi/6 we have maximum

Minimum is absolute mini in the interval i.e. (pi/2,0)

Answer:

B

Step-by-step explanation:

Answer:

The answer is a 100,000-choose-99 (a combination.)

[tex]P(100000, 99) = \displaystyle \left( \begin{array}{c}100000\cr 99\end{array}\right)[/tex].

Step-by-step explanation:

A combination [tex]C(n, r)[/tex] or equivalently [tex]\displaystyle \left(\begin{array}{c}n \cr r\end{array}\right)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements.

A permutation [tex]P(n, r)[/tex] also gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements. On top of that, it accounts for the order of the elements. Two elements in different order counts twice in a permutation, but only once in a combination.

The question emphasize that the council members are "co-equal." That implies that the order of the members don't really matter. Hence a combination with [tex]n = 100000[/tex] and [tex]r = 99[/tex] would be a more suitable choice.

Answer:

f(2)=8

Step-by-step explanation:

f(2)=(-5)*f(1)-7=

(-5)*(-3)-7=8

f(2)=8

[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]

is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

[tex]\dfrac{\partial f}{\partial z}=3[/tex]

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

[tex]f(x,y,z)=xy+g(y,z)[/tex]

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]

[tex]f(x,y,z)=xy+h(z)[/tex]

[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]

[tex]f(x,y,z)=xy+3z+C[/tex]

so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]

[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]

Answer:

hi

Step-by-step explanation:

6n - 3(2n-5)

mutiply the bracket by -3

(-3)(2n)= -6n

(-3)(-5)= 15

6n-6n+15

0+15

answer:

15