The diameter of a spindle in a small motor is supposed to be 5 mm. If the spindle is either too small or too large, the motor will not work properly. The manufacturer measures the diameter in a sample of motors to determine whether the mean diameter has moved away from the target. What are the null and alternative hypotheses (H0 = null hypothesis and Ha = alternative hypothesis)?

(A) H0: Mean= 5 and Ha: Mean is not equal to 5
(B) H0: Mean = 5 and Ha: Mean <5
(C) H0: Mean < 5 and Ha: Mean > 5
(D) H0: Mean = 5 and Ha: Mean > 5

Answer: 66 ways

Step-by-step explanation:

Given;

Number of senior math club members = 6

Number of junior math club members = 4

Total number of members of the club = 10

To form a group of 5 members with at least 4 seniors.

N = Na + Nb

Na = number of possible ways of selecting 4 seniors and 1 junior

Nb = number of possible ways of selecting 5 seniors.

Since the selection is does not involve ranks(order is not important)

Na = 6C4 × 4C1 = 6!/4!2! × 4!/3!1! = 15 ×4 = 60

Nb = 6C5 = 6!/5!1! = 6

N = Na + Nb = 60+6

N = 66 ways

Using the combination formula, it is found that there are 66 ways to form the groups.

The order in which the students are selected is not important, hence, the combination formula is used to solve this question.

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, the possible groups are:

Hence:

[tex]T = C_{4,1}C_{6,4} + C_{6,5} = \frac{4!}{1!3!}\frac{6!}{4!2!} + \frac{6!}{5!1!} = 4(15) + 6 = 60 + 6 = 66[/tex]

There are 66 ways to form the groups.

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Answer:  P = 0.2

Step-by-step explanation:

let us define the expectations as;

Eнтн = no of flips to obtain HTH

Eн,тн = no of flips to obtain HTH where  H is flipped

Eнт,н =  no of flips to obtain HTH where HT is flipped

so let P and q represent the success and failure of probabilities

this gives;

Eнтн = 2 + P²Eн,тн + PqEнт,н + PqEн,тн + q²Eнтн

Eн,тн = 1 + PEн,тн + qEнт,н

     Eн,тн = (1 + qEнт,н) / q

Eнт,н =  1 + p*0 + qEнтн = 1 + qEнтн

from this expression we have that;

Eнтн = (2 +(P² +Pq) (1/q + 1) +Pq) / (1-qP² + 2Pq² + q²)

E(x) = 1/P + 1/P²q

      = 36.25

therefore the probability is P = 0.2

P = 0.2

Answer:

54,18

Step-by-step explanation:

Let one of the numbers be x. The other number be represented as 36-x

(x-(36-x ) = 36

open brackets

x - 36 + x  =  36

x + x = 36+ 36 = 72

2x = 72

x = 72/2 = 36

x =  36

The product can then be represented as y = x(36-x) or y=36x-x²

 The maximum or minimum is always on the axis of symmetry which has the formula x=-b/2a.

In our case, the axis of symmetry is -36/-2, so x=18.

If one number is 18 and the 2 numbers differentiate by 36, the other number is 18 + 36 = 54

So the 2 numbers are 18 and 54 and the minimum product is 972

The two numbers are 18 and 54 and the minimum product is 972

If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Let suppose one of the numbers be x and the other number be represented as 36-x

Then the equation would be;

(x-(36-x ) = 36

Now open brackets;

x - 36 + x  =  36

x + x = 36+ 36 = 72

2x = 72

x = 72/2 = 36

x =  36

Therefore, The product can be represented as y = x(36-x) or y=36x-x²

Here maximum or minimum is always on the axis of symmetry which is the formula x=-b/2a.

In our case; the axis of symmetry is -36/-2,

Then x=18.

If one number is 18 and the 2 numbers differentiate by 36 then the other number must be 18 + 36 = 54

Hence the two numbers are 18 and 54 and the minimum product is 972

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Answer:

See proof below.

Step-by-step explanation:

If we assume the following linear model:

[tex] y = \beta_o + \beta_1 X +\epsilon[/tex]

And if we have n sets of paired observations [tex] (x_i, y_i) , i =1,2,...,n[/tex] the model can be written like this:

[tex] y_i = \beta_o +\beta_1 x_i + \epsilon_i , i =1,2,...,n[/tex]

And using the least squares procedure gives to us the following least squares estimates [tex] b_o [/tex] for [tex]\beta_o[/tex] and [tex] b_1[/tex] for [tex]\beta_1[/tex]  :

[tex] b_o = \bar y - b_1 \bar x[/tex]

[tex] b_1 = \frac{s_{xy}}{s_xx}[/tex]

Where:

[tex] s_{xy} =\sum_{i=1}^n (x_i -\bar x) (y-\bar y)[/tex]

[tex] s_{xx} =\sum_{i=1}^n (x_i -\bar x)^2[/tex]

Then [tex] \beta_1[/tex] is a random variable and the estimated value is [tex]b_1[/tex]. We can express this estimator like this:

[tex] b_1 = \sum_{i=1}^n a_i y_i [/tex]

Where [tex] a_i =\frac{(x_i -\bar x)}{s_{xx}}[/tex] and if we see careful we notice that [tex] \sum_{i=1}^n a_i =0[/tex] and [tex]\sum_{i=1}^n a_i x_i =1[/tex]

So then when we find the expected value we got:

[tex] E(b_1) = \sum_{i=1}^n a_i E(y_i)[/tex]

[tex] E(b_1) = \sum_{i=1}^n a_i (\beta_o +\beta_1 x_i)[/tex]

[tex] E(b_1) = \sum_{i=1}^n a_i \beta_o + \beta_1 a_i x_i[/tex]

[tex] E(b_1) = \beta_1 \sum_{i=1}^n a_i x_i = \beta_1[/tex]

And as we can see [tex]b_1[/tex] is an unbiased estimator for [tex]\beta_1[/tex]

In order to find the variance for the estimator [tex]b_1[/tex] we have this:

[tex] Var(b_1) = \sum_{i=1}^n a_i^2 Var(y_i) +\sum_i \sum_{j \neq i} a_i a_j Cov (y_i, y_j) [/tex]

And we can assume that [tex] Cov(y_i,y_j) =0[/tex] since the observations are assumed independent, then we have this:

[tex] Var (b_1) =\sigma^2 \frac{\sum_{i=1}^n (x_i -\bar x)^2}{s^2_{xx}}[/tex]

And if we simplify we got:

[tex] Var(b_1) = \frac{\sigma^2 s_{xx}}{s^2_{xx}} = \frac{\sigma^2}{s_{xx}}[/tex]

And with this we complete the proof required.

Final answer:

β1^ in simple linear regression is normal because it is a ratio of linear combinations of normal variables Yi. It is unbiased as its expected value equals the true parameter β1. The variance of β1^ is σ2/sxx, derived from properties of variance and the independent nature of errors ϵi.

Explanation:

The student's question pertains to the properties of the estimated coefficient β1^ in a simple linear regression model. To show that β1^ is a normal random variable, we consider the linear combination of the normal random variables Yi, because a linear combination of normal random variables is also normally distributed. Since each Yi is normal and given by Yi=β0+β1xi+ϵi, and ϵi is N(0,σ2), the estimator β1^=sxy/sxx becomes a ratio of linear combinations of these normal variables and hence, normal.

Next, to prove that β1^ is an unbiased estimator, we take the expectation of β1^ and show that E[β1^]=β1. It's implied by calculating the expected value of the numerator sxy and denominator sxx separately and showing the ratio equals β1.

The variance of β1^, Var(β1^), can be shown to be σ2/sxx by leveraging properties of variance of the linear combinations of Yi and noting that ϵi's are independent random variables with variance σ2. The calculations involve squaring the deviations and utilizing expectations.

Answer:

Sample mean for U=21.5

Sample mean for F=8.4

Step-by-step explanation:

[tex]Sample mean of u=xbar_{u} =\frac{sum(xi)}{n}[/tex]

Where xi are the observations in the urban homes sample and n is the number of observations in the urban homes sample

[tex]sample mean of u=xbar_{u} =\frac{6+5+11+33+4+5+80+18+35+17+23}{11}[/tex]

[tex]sample mean of u=xbar_{u} =\frac{237}{11}=21.545[/tex]

Rounding it to one decimal places

[tex]sample mean of u=xbar_{u}=21.5[/tex]

Now for second sample

[tex]Sample mean of F=xbar_{F} \frac{sumxi}{n}[/tex]

Where xi are the observations in the farm homes sample and n is the number of observations in the farm homes sample

[tex]Sample mean of F=xbar_{F} =\frac{2+15+12+8+8+7+6+19+3+9.8+22+9.6+2+2+0.5)}{15}[/tex]

[tex]Sample mean of F=xbar_{F} =\frac{125.9}{15} =8.393[/tex]

Rounding it to one decimal places

[tex]Sample mean of F=xbar_{F} =8.4[/tex]

The sample mean for urban homes (U) is 21.54 and the sample mean for farm homes (F) is 7.73.

To find the sample mean, you sum up all the data points and then divide by the number of data points. For the Urban homes (U), we first add up all the data points: 6.0 + 5.0 + 11.0 + 33.0 + 4.0 + 5.0 + 80.0 + 18.0 + 35.0 + 17.0 + 23.0 = 237.0. The number of data points is 11, so the sample mean for U is 237.0 / 11 = 21.54 (rounded to two decimal places).

For Farm homes (F), add up all the data points: 2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0 + 3.0 + 9.8 + 22.0 + 9.6 + 2.0 + 2.0 + 0.5 = 115.9. The number of data points is 15, so the sample mean for F is 115.9 / 15 = 7.73 (rounded to two decimal places).

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Answer:

1) [tex] y =4x^2 +7[/tex]

2) [tex] y =7s^2 -3s^4 +181[/tex]

Step-by-step explanation:

Assuming that our function is [tex] y = f(x)[/tex] for the first case and [tex] y=f(s)[/tex] for the second case.

Part 1

We can rewrite the expression like this:

[tex] \frac{dy}{dx} =8x[/tex]

And we can reorder the terms like this:

[tex] dy = 8 x dx[/tex]

Now if we apply integral in both sides we got:

[tex] \int dy = 8 \int x dx[/tex]

And after do the integrals we got:

[tex] y = 4x^2 +c[/tex]

Now we can use the initial condition [tex] y(0) =7[/tex]

[tex] 7 = 4(0)^2 +c, c=7[/tex]

And the final solution would be:

[tex] y =4x^2 +7[/tex]

Part 2

We can rewrite the expression like this:

[tex] \frac{dy}{ds} =14s -12s^3[/tex]

And we can reorder the terms like this:

[tex] dy = 14s -12s^3 dx[/tex]

Now if we apply integral in both sides we got:

[tex] \int dy = \int 14s -12s^3 ds[/tex]

And after do the integrals we got:

[tex] y = 7s^2 -3s^4 +c[/tex]

Now we can use the initial condition [tex] y(3) =1[/tex]

[tex] 1 = 7(3)^2 -3(3)^4 +c, c=1-63+243=181[/tex]

And the final solution would be:

[tex] y =7s^2 -3s^4 +181[/tex]

Answer:

635599482kg

Step-by-step explanation:

A gourmet chocolate candy has 7.00 gg of dietary fat in each 22.7-gg piece

for the 22.7 piece , the gg is 7 x 22.7  = 158.9gg in total

1gg = 2204622.6218488 lbs

158.9gg in total  = 158.9 x 2204622.6218488 = 350314534.612

for the box of 4.00 lblb of candy

it is 350314534.612 x 4 = 1401258138.45lb

since our answer is needed in kilograms we convert 1401258138.45lb to kilograms

1lb = 0.453592kg

1401258138.45lb =  1401258138.45lb x 0.453592kg = 635599481.513 ≈ 635599482kg

Answer:

Red Marble is the color in the bag

Step-by-step explanation:

Total number of marbles ; 201

number of red marbles ; 101

number of blue marbles ; 101

first possibility ; From what is said as the condition, it can be inferred that the number of blue marbles is reduced by a factor of 1 , while the number of red marbles remains constant. this shows that the two marbles of the same color that was removed is most likely BLUE.

Second possibility ; Two red marbles are added (of the same color) and add a blue marbles, implies the the red marbles reduce by a factor of 2 while the blue marbles increase by a factor of 1. this is also possible.

Third possibility ; Two marbles of different color implies 1R and 1B , and then a red marble is added. this shows that the number of red marble still remains constant ( 1R removed, 1R added), while the number of blue reduce by 1 ( remove 1B, and not replaced back).

From the various possibilities, it can be inferred that the number of red marbles is either increasing by a factor of 2 or remains constant. this further shows that in all three , 1R marble will always remain in the bag no matter the possibility as such in all three scenerios, it further shows that if there is only a marble remaining, it would have been the RED MARBLE WHICH IS AUTOMATICALLY THE LAST MARBLE from the analysis done above.

Answer:

i) -28 ft/s

ii) -21.6 ft/s

iii) -20.8 ft/s

iv) -20.16 ft/s

b)  -20 ft/s

Step-by-step explanation:

if v represents vertical velocity then if the height y is

y = 44*t − 16*t²

and the instant velocity v is the derivative with respect to the time

v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*t

while the average velocity is va= (y-y₀)/(t-t₀)

where t₀ = 2 and y₀=y(t₀) =  44*2 − 16*2² = 24

then

va= (y-y₀)/(t-t₀) = (44*t − 16*t² -  24)/(t-2)

for

i) t= 0.5s + 2 s= 2.5 s

va = (44* 2.5  − 16* 2.5 ² -  24)/( 2.5 -2) = -28 ft/s

ii) t= 0.1s + 2 s= 2.1 s

va = (44* 2.1  − 16* 2.1 ² -  24)/( 2.1 -2) = -21.6 ft/s

iii) t= 0.05 s + 2 s= 2.05 s

va = (44* 2.05  − 16* 2.05 ² -  24)/( 2.05 -2) = -20.8 ft/s

iv) t= 0.01s + 2 s= 2.01 s

va = (44* 2.01  − 16* 2.01 ² -  24)/( 2.01 -2) = -20.16 ft/s

b) the instantaneous velocity when t=2

v (t=2) =  44- 32*(2) = -20 ft/s

Answer:

i) The first  -28 ft/s

ii) Then second -21.6 ft/s

iii) Thirds one is -20.8 ft/s

iv) Then  -20.16 ft/s

b)  -20 ft/s

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Answer:

The question is incomplete or missing some part, but I have answered it appropriately.

Step-by-step explanation:

The question is incomplete, but I will give a detailed description of where pie chart are appropriate to use. A pie chart as we known is a divided circle containing breakdown d and this are developed into a table from calculations and then a chart is developed. The circle in a pie chart are divided into sectors where the angle of each sector is directly proportional to the frequency of the item or data that is being represented.

The process is done by finding the sum of all the frequencies, the sum is then used to divide 360 degree which is the total degree in a complete circle. thereafter each of the frequencies is used to multiply this result to arrive at the sectoral angle for each item. for example, the breakdown of weekkly expenditures of a house wife. the expensiture are marked with the amount spent on each. this information can be represented or illustrated on pie chart.

It should be noted that a pie chart is only suitable for a single set of data and not more than one date as is the case of histogram where more than one grouped or un-grouped data maybe represented or ilustrated on a histogram. if this is the case, pie chart are best employed when making a description between a single part of a data and not the whole set.

Answer:

Option A) The null hypothesis should not be rejected.

Step-by-step explanation:

We are given the following in the question:

A one sample test for a proportion is being performed.

Critical Value = 2.33

Test statistic = 1.37

Since the critical value is positive and the calculated test statistic is positive, thus it is a right tailed test for a proportion.

The null and alternate hypothesis can be written as:

[tex]H_{0}: p \leq p_0\\H_A: p > p_0[/tex]

Since the calculated test statistic is less than the critical value, we fail to reject the null hypothesis.

We accept the null hypothesis.

Thus, the correct answer is

Option A) The null hypothesis should not be rejected.

Answer:

a. 241; b. 206.0; c. 272.0; d. 174 to 232; other answers (see below).

Step-by-step explanation:

First, at all, we need to organize the data from the smallest to the largest value:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

This step is important to find the median, the first quartile, and the third quartile. There are numerous methods to find the median and the other quartiles, but in this case, we are going to use a method described by Tukey, and it does not need calculators or software to estimate it.

In this case, we have 53 values (an odd number of values), the median is the value that has the same number of values below and above it, so what is the value that has 26 values below and above it? Well, in the organized data above this value is the 27th value, because it has 26 values above and below it:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

The median is 241.

For the first quartile, we need to calculate the median for the lower half of the values from the median previously obtained. Since we have an odd number of values (53), we have to include the median in this calculation. We have 27 values (including the median), so the "median" for these values is the value with 13 values below and above it.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241

Since we are asking to round the answer to one decimal place, the first quartile is 206.0

We use the same procedure used to find the first quartile, but in this case, using the upper half of the values.

241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

So, the third quartile is 272.0

The second quartile is the median and "50% of the data lies below this point" (Quartile (2020), in Wikipedia). Having this information into account, 50% of the weights lies below the median 241.

Thus, the middle 50% of the weights are from 174 to 232.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241

If the population were all professional football players, the right option is:

The above data would be a sample of weights because they represent a subset of the population of all football players.

It represents a sample. Supposing Football Team A are all professional, they can be considered a sample from all professional football players.

If the population were Football Team A, the right option is:

The data would be a population of weights because they represent all of the players on Football Team A.

The mean for the population of weights of Football Team A is the sum of all weights (12529 pounds) divided by the number of cases (53).

[tex] \\ \mu = \frac{12529}{53} = 236.39622641509433 [/tex]

[tex] \\ \mu = \frac{12529}{53} = 236.40 [/tex]

The standard deviation for the population is:

[tex] \\ \sigma = \sqrt(\frac{(x_{1}-\mu)^2 + (x_{2}-\mu)^2 + ... + (x_{n}-\mu)^2}{n}) [/tex]

In words, we need to take either value, subtract it from the population mean, square the resulting value, sum all the values for the 53 cases (in this case, the value is 74332.67924), divide the value by 53 (1402.50338) and take the square root of it (37.45001).

Then, the population standard deviation is 37.45.

The weight that is 3 standard deviations below the mean can be obtained using the following formula:

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

[tex] \\ \mu= 236.40\;and\;\sigma=37.45[/tex]

Then, in the case of three standard deviations below the mean, z = -3.

[tex] \\ -3 = \frac{x - 236.40}{37.45} [/tex]

[tex] \\ x = -3*37.45 + 236.40 [/tex]

[tex] \\ x = 124.05 [/tex]

For the player that weights 209 pounds:

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

[tex] \\ z = \frac{229 - 236.40}{37.45} [/tex]

[tex] \\ z = -0.19 [/tex]

For Team B, we have a mean and a standard deviation of:

[tex] \\ \mu=240.08\;and\;\sigma=44.38[/tex]

And Player B weighed 229 pounds.

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

[tex] \\ z = \frac{229 - 240.08}{44.38} [/tex]

[tex] \\ z = -0.70 [/tex]

This value says that Player B is lighter with respect to his team than Player A, because his weight is 0.70 below the mean of Football Team B, whereas Player A has a weight that is closer to the mean of his team. So, the answer is:

Player B, because he is more standard deviations away from his team's mean weight.

Answer:

a) No Unique solution

b) No Unique solution

c) No unique solution

Step-by-step explanation:

The step by step explanation is as shown in the attachment below

Answer:

P = 10/171

P = 0.058

The probability that both fish selected are Catfish is 0.0058

Step-by-step explanation:

Given:

The tank contains;

Algae eaters = 8

Catfish = 5

Tilapia fish = 6

Total number of fish = 8+5+6= 19

To determine the probability that both fish selected are Catfish P;

P = 5/19 × 4/18

( The number of the total number of fishes, and Catfish reduce by one after each selection because it is selected without replacement)

P = 10/171

P = 0.058

Answer:

a) 5 units/s

b) yes

c) counter-clockwise

d) yes

Step-by-step explanation:

part a

r(t) = cos (5t) i + sin (5t)j

v(t) = dr(t) / dt = -5sin(5t) i + 5cos(5t)j

[tex]mag( v(t)) = \sqrt{(-5sin(5t))^2 + (5cos(5t))^2} \\mag( v(t)) = \sqrt{25sin^2(5t) + 25cos^2(5t)} \\ \\mag( v(t)) = \sqrt{25*(sin(5t)^2 + cos(5t)^2)} \\\\mag( v(t)) = \sqrt{25} \\\\mag( v(t)) = 5 units/s[/tex]

Hence, the particle has a constant speed of 5 units/s

part b

a(t) = dv(t) / dt = -25cos(5t) i - 25sin(5t)j

To check orthogonality of two vectors their dot product must be zero

a(t) . v(t) = (-25cos(5t) i - 25sin(5t)j) . (-5sin(5t) i + 5cos(5t)j)

= 125cos(5t)*sin(5t) -125cos(5t)*sin(5t)

= 0

Yes, the particles velocity vector is always orthogonal to acceleration vector.

part c

Use any two values of t and compute results of r(t)

t = 0 , r(0) = 1 i

t = pi/2, r(0) = j

Hence we can see that the particle moves counter-clockwise

part d

Find the value r(t) at t=0

r(0) = cos (0) i + sin (0) j

r(0) = 1 i + 0 j

Yes, the particle starts at point ( 1, 0)

Answer:

[tex] T(t) = M -C_1 e^{-kt}[/tex]

And as we can see that represent the solution for the differential equation on this case.

Step-by-step explanation:

For this case we have the following differential equation:

[tex] \frac{dT}{dt}= k(M-T)[/tex]

We can rewrite this expression like this:

[tex] \frac{dT}{M-T} = k dt[/tex]

We can us the following susbtitution for the left part [tex] u = M-T[/tex] then [tex] du= -dt[/tex] and if we replace this we got:

[tex] \frac{-du}{u} = kdt[/tex]

We can multiply both sides by -1 and we got;

[tex] \frac{du}{u} =-k dt[/tex]

Now we can integrate both sides and we got:

[tex] ln |u| = -kt + C[/tex]

Where C is a contant. Now we can exponetiate both sides and we got:

[tex] u(t) =e^{-kt} *e^C = C_1 e^{-kt}[/tex]

Where [tex] C_1 = e^C[/tex] is a constant. And now we can replace u and we got this:

[tex] M-T = C_1 e^{-kt}[/tex]

And if we solve for T we got:

[tex] T(t) = M -C_1 e^{-kt}[/tex]

And as we can see that represent the solution for the differential equation on this case.

The solution to the differential equation representing Newton's law of cooling is calculated by rearranging the equation and then integrating both sides. The general solution for the temperature of an object over time is given by T = M - Ce^-kt where M is the ambient temperature, T is the temperature of the object, C is a constant, and k is the constant of proportionality.

To solve the given differential equation which represents Newton's law of cooling, we shall proceed with integrating both sides. This is in the form of a separable first order differential equation and can be written in the form as follows:

dT / (M - T) = k dt

By re-arranging and integrating both sides, we can find the solution. When we integrate both sides, we get:

- ln |M - T| = kt + C

Where C is the constant of integration. Simplifying further:

T = M - Ce-kt

This is the general solution to the temperature T of an object over time t according to Newton's law of cooling.

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Answer/Explanation

The complete question is:

Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]

Step-by-step explanation:

If we make the identification For ∅° (x) = 1 and  ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.

Therefore, in the first case, we have

(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx

This will therefore be equal to :

1/n sin nx lim(π) lim(-π) = 1/n  [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)

and in the second case, we have,,

(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx

This will therefore be equal to:

∫ lim(π) lim(-π) cos mx cos nx dx

Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)

1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ]  lim(π) lim(-π) = 0, m ≠ n

Now, to go ahead to find the norms in the given set intervals, we have,

for  ∅°(x) = 1 we have:

//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π

So therefore, //∅°(x)//² = √2π

For ∅°∨n(x)  = cos nx  , n > 0.

It then follows that,

//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π

Thus, for n > 0 , //∅°(x)// = √π

It is therefore ggod to note that,

Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be  normalized—that is, made into an orthonormal set by dividing each function by  its norm. It follows from the above equations that has been set.

Therefore,

{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.

Answer:

Qualitative variable

Step-by-step explanation:

The response variable is qualitative because it can be categorize as "Good time" or "Bad time". The time is also rated as "Good" or "Bad" for taking a lengthy vacation and so it involves ordinal scale of measurement.

We can say that our response variable is qualitative and ordinal. In short response variable  is of qualitative type of variable and the scale of measurement it involves is ordinal scale.

Answer:

First Part:

exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.

Second Part:

[tex]y=ce^{At}\\y'=Ay[/tex]

Third Part:

[tex]y=ce^t\\y'=ce^t\\y'=y[/tex]

Step-by-step explanation:

First Part:

In calculus exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.

Where:

t is independent variable.

Derivative is represented as:

[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]

Where:

c is any number.

Second Part:

Consider the constant A.

The function will become:

[tex]y=ce^{At}\\y'=\frac{d(ce^{At})}{dt} \\y'=cAe^{At}\\y'=Ay[/tex]

Third Part:

Derivative is represented as:

[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]

Where:

c is any number.

Final answer:

The differential equation is [tex]\( \frac{df}{dx} = Cf(x) \),[/tex] with the solution[tex]\( f(x) = Ce^x \), where \( C \)[/tex] is a constant.

Explanation:

The function that satisfies both conditions is [tex]\( f(x) = Ce^x \), where \( C \)[/tex] a constant. Its first derivative is[tex]\( f'(x) = Ce^x \),[/tex]which is a constant multiple of itself. This can be expressed as a first-order differential equation:

[tex]\[ \frac{df}{dx} = Cf(x) \][/tex]

with the solution:

[tex]\( f(x) = Ce^x \), where \( C \)[/tex]

This differential equation represents exponential growth or decay, depending on the sign of [tex]\( C \). For \( C > 0 \)[/tex], the function exponentially grows as [tex]\( x \) increases, while for \( C < 0 \)[/tex]it exponentially decays. The solution captures phenomena like population growth, radioactive decay, or compound interest, making it a fundamental concept in calculus and applied mathematics.

Hi, you haven't provided the concrete and hypothetical populations, I'll be using people with red hair in Texas as our population, but you can use a similar approach for the population you decide to use.

Answer:

Explanation:

As you can see from the questions, the main difference between statics and probability has to do with knowledge, what do we know about our population? For the first question (probability) we talked about a known group of people, and base on our knowledge we ask about the probability of selecting people with red hair. For the second question (statistics) we are asked to  make an inference about all the population based just on a sample (we don't have all the knowledge), also we say that the last question is Inferential because the data come from a large population selected randomly.

Final answer:

An example of a probability question could involve finding the likelihood of a statistics student using a product weekly based on a survey sample. Meanwhile, an inferential statistics question might use a t-test to infer differences in protein value between two lots of hay.

Explanation:

In the context of probability and inferential statistics, an example of a probability question for a concrete population might be: "If a random sample of 25 statistics students was surveyed, and six reported using a certain product within the past week, what is the probability that any given statistics student uses the product weekly?"

An example of an inferential statistics question using hypothetical population data could be: "Using a t-test to compare the mean protein value of two lots of alfalfa hay, can we infer that there is a significant difference between their protein contents?" The t-test determines if the observed difference in means is likely due to chance or if it is statistically significant.

These examples illustrate how probability is used to model uncertainty and random events, while inferential statistics are employed to make educated guesses about a population from sample data, such as calculating a confidence interval or testing a hypothesis.

Answer:

29C3 = 3654

Step-by-step explanation:

What is applied here is combination that n - combination of r which is a subscript and the n - is a superscript.

Attached is the detailed explanation.

Answer:

19.3795 m

Step-by-step explanation:

If sound travels at 343 m/s and it took Karen 0.113 s to hear the echo from the wall, the distance travelled by the sound is:

[tex]D=343\frac{m}{s} *0.113\ s\\D= 38.759\ m[/tex]

Note that the distance calculated above is the distance travelled from Karen to the wall and then back from the wall to Karen. Therefore, the distance between Karen and the wall is:

[tex]d=\frac{38.759}{2}\\d=19.3795\ m[/tex]

The wall is 19.3795 m away from Karen.

The wall is 19.39 meters away.

To find the distance to the wall based on the echo, we use the speed of sound and the time it takes for the echo to return. The time given includes both the journey to the wall and back, so we need to divide it by 2.

1.   Time for sound to travel to the wall and back: 0.113 s    

2.  Time for sound to travel one way:

0.113 s / 2 = 0.0565 s

3.  Speed of sound: 343 m/s

4.  Distance = Speed x Time

343 m/s * 0.0565 s = 19.39

Therefore, the wall is 19.39 meters away from Karen.

Answer:

Distance between the van and Morristown is changing at the rate of 13.22 miles per hour.

Step-by-step explanation:

From the figure attached,

Van starts from C (City of Morristown), reaches the point A 191 miles due North and then it travels with a speed of 25 miles per hour due East from A towards B.

We have to calculate the rate of change of distance BC, when the van reaches point B which is 119 miles away from A.

By Pythagoras theorem in the triangle ABC,

[tex]BC^{2}=AB^{2}+AC^{2}[/tex]

Distance AC is constant equal to 191 mi.

By differentiating the equation with respect to time 't'

[tex]2BC.\frac{d(BC)}{dt}=0+2AB.\frac{d(AB)}{dt}[/tex]

[tex]BC.\frac{d(BC)}{dt}=AB.\frac{d(AB)}{dt}[/tex]

Since BC² = (119)²+ (191)²

BC = √50642 = 225.04 miles

From the differential equation,

[tex](225.03).\frac{d(BC)}{dt}=119\times 25[/tex] [Since [tex]\frac{d(AB)}{dt}=25[/tex] miles per hour]

[tex]\frac{d(BC)}{dt}=13.22[/tex] miles per hour

Therefore, distance between the van and Morristown is changing at the rate of 13.22 miles per hour.

In the absence of all forces except gravity, the ball is approximately 0.28 feet below the ground when it crosses the home plate.

(a) In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of 130i +0j-3k ft/s (roughly 90mi/hr). How far above the ground is the ball when it cross the home plate.

To determine the vertical distance the ball is above the ground when it crosses home plate, we need to find the time it takes to reach the home plate using the vertical component of the initial velocity. The initial vertical velocity is -3 ft/s, and the vertical acceleration due to gravity is -32 ft/s2. Using the equation y = y0 + v0yt + 0.5at2, where y is the vertical distance above the ground, y0 is the initial vertical position, v0y is the initial vertical velocity, t is the time, and a is the acceleration due to gravity, we can solve for t. Plugging in the given values, we get:

y = 0 + (-3)t + 0.5(-32)t2

Simplifying the equation, we get 32t2 - 3t = 0. Factoring out t, we get t(32t - 3) = 0. So, either t = 0 (which is the initial time when the ball is released) or 32t - 3 = 0. Solving for t, we find that t = 3/32 seconds. Since the ball is in the air for this amount of time, we can substitute this value back into the equation to find the vertical distance above the ground when the ball crosses the home plate:

y = 0 + (-3)(3/32) + 0.5(-32)(3/32)2

Simplifying, we get y = -9/32 feet, which is approximately -0.28 feet above the ground. Therefore, the ball is 0.28 feet below the ground when it crosses the home plate.

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Answer:

[tex]\oint_C F \,dr = 0[/tex]

Step-by-step explanation:

In this problem, we have a vector field

                 [tex]F = \left \langle 2x-z^2,x^2+yz,-2xz +\frac{y^2}{2} \right \rangle[/tex].

We need to find the line integral

                                      [tex]\oint_C F \,dr[/tex]

where [tex]C[/tex] is a circle

                    [tex]r(t) = \left \langle 2 \cos t, 4\sin t, 5 \cos t \right \rangle, \quad 0 \leq t \leq 2 \pi[/tex].

As we can see, the vector filed [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] and its component functions have continuous partial derivatives.

First, we need to find the curl of the vector filed [tex]F[/tex].

[tex]\text{curl} F = \begin{vmatrix}i & j & k\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xy-z^2 & x^2 + yz & \frac{y^2}{2} - 2xz\\\end{vmatrix}[/tex]

Therefore,

[tex]\text{curl}F = i \left( \frac{\partial}{\partial y} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( x^2+yz\right)\right) - j \left( \frac{\partial}{\partial x} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( 2xy-z^2\right)\right)\\ \phantom{12356} +k \left( \frac{\partial}{\partial x} \left( x^2+yz\right) - \frac{\partial}{\partial y} \left( 2xy-z^2\right)\right)[/tex]

Now, we can easily calculate the needed partial derivatives and we obtain

[tex]\text{curl} F = i(y-y) - j(-2z + 2z) + k(2x-2x) = 0i + 0j+0k = \langle 0,0,0 \rangle[/tex]

So, the vector field [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] , its component functions have continuous partial derivatives and [tex]\text{curl} F = 0[/tex] .Therefore, by a well-known theorem, [tex]F[/tex] is a conservative field.

Since [tex]C[/tex] is a closed path, we obtain that

                                           [tex]\oint_C F \,dr = 0[/tex]

Answer:

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=13.8[/tex] represent the sample mea   n

[tex]s=0.3[/tex] represent the sample standard deviation

[tex]n=45[/tex] sample size  

[tex]\mu_o =68[/tex] represent the value that we want to test

[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is at least 14 oz, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 14[/tex]  

Alternative hypothesis:[tex]\mu < 14[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{13.8-14}{\frac{0.3}{\sqrt{45}}}=-4.47[/tex]    

Critical region

We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution

Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]

We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"

So then the correct answer for this case would be:

Reject H0 if test statistic < -2.05

And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.

The decision criterion to reject the null hypothesis that the mean weight is at least 14 oz at the 0.02 level of significance is to reject H0 if the test statistic is less than -2.33, as derived from the standard normal z-table.

To determine the criterion for rejecting the null hypothesis in a hypothesis test like the one presented - where a cereal company claims that the mean weight of the cereal in its packets is at least 14 oz - we use the level of significance (alpha). Since the level of significance in this problem is 0.02, we refer to the standard normal z-table to find the critical z-value that corresponds to this significance level for a one-tailed test, which in this case is a left-tailed test because we are checking if the mean weight is less than the claim (14 oz).

Using the z-table, we find that the critical z-value for an alpha of 0.02 in a one-tailed test is approximately -2.33. If our calculated test statistic is less than this critical value, we would reject the null hypothesis in favor of the alternative. Therefore, the decision criterion is: Reject H0 if test statistic < -2.33.

Since the given options are 'Reject H0 if test statistic < 2.05', 'Reject H0 if test statistic > -2.05', 'Reject H0 if test statistic < -2.05', and 'Reject H0 if test statistic < -2.33', the correct criterion for this test, considering the level of significance is 0.02, is 'Reject H0 if test statistic < -2.33'.

Answer:

67.84 km

Step-by-step explanation:

let her initial speed be 'u'

Time with speed u = 2 hours 50 minutes = 2.83 hours

Distance = speed × Time

therefore,

Distance = u × 2.83  ............(1)

when speed increase by 1 km/h

speed, v = u + 1

Time taken = 2 hours 50 minutes - 7 minutes

=  2 hours 43 minutes

= 2.7167 hours

Therefor,

Distance = ( u + 1 ) ×  2.7167 .............(2)

since, the distance in both the cases will be same

therefore,

from 1 and 2

u × 2.83 = ( u + 1 ) ×  2.7167

or

u × 1.0417 = u + 1

or

0.0417u = 1

or

u = 23.97 km/hr

Therefore,

Distance of loop = u × 2.83

= 23.97 × 2.83

= 67.84 km

The question involves calculating the distance of a loop that a bicyclist rides in two different conditions. Using speed, time and distance relation, we first set up two equations based on the given scenarios and solve them to find the distance. The distance turns out to be approximately 25.41 km.

This problem can be solved using the relationship between speed, distance and time. Let's denote the distance of the loop around Lake Washington as D km and her original average speed as S km/hr. Initially, she covers the distance in 2 hours and 50 minutes (which is equal to 2.83 hours when converted into hours).

So, we have the first equation as:  D = S x 2.83

Then, she increases her speed by 1 km/hr, and it reduces the total time by 7 minutes (which comes out to be 0.117 hours), resulting in a new time of 2.83 - 0.117 = 2.713 hours to cover the same distance. With her increased speed (S + 1 km/hr), this gives us the second equation as: D = (S + 1) x 2.713

Setting the equations equal to each other, we solve for S and subsequently find D. After solving, we get D approximately equal to 25.41 km.

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Final answer:

The differential equation governing the spread of flu among students on a college campus is modeled by the logistic equation dx/dt = kx(2000 - x), assuming a constant rate of interactions and no outside influences.

Explanation:

The differential equation governing the number of students x(t) who have contracted the flu on an isolated college campus, where the rate of disease spread is proportional to the number of interactions between infected and susceptible students, can be modeled using principles of epidemiology. The total number of students is 2000, and if we use k for the constant of proportionality, we can denote the number of susceptible students as 2000 - x, because x represents the number of infected students. Hence, the rate of change of x, given by dx/dt, is proportional to the product of the number of students who have the flu and the number who do not, which is kx(2000-x). The differential equation is thus:

dx/dt = kx(2000 - x)

This is a standard form of the logistic differential equation, often used in the SIR model in epidemiology to describe how a disease spreads in a population.

Answer:

Option B is right

Step-by-step explanation:

P(A or B)

In set theory if A and B are two sets we have either A occurs or B occurs or both as

P(A or B) = P(AUB)

This implies that A occurs or B occurs

Probability is calculated as P(A)+P(B)-P(AB)

P(AB) here represents both occurring and this is subtracted as this was added two times

Option A is wrong because this is AB

Option C is wrong because it is not necessary A occurs first

D is wrong because this is A'UB' where A' represents the complement of A

So correct answer out of four options is

B. the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur.

The notation P(A or B) in probability specifies the probability that in a single trial, either event A happens, event B happens, or they both happen.

The notation P(A or B) in probability theory denotes the probability that event A occurs, event B occurs, or both events occur in a single trial. If we look at the four options provided, option B fits this definition. Hence, P(A or B) indicates the probability that in a single trial, event A occurs, event B occurs, or they both occur. It does not imply that both events A and B have to occur concurrently in the same trial (option A), nor that event A has to occur in one trial followed by event B in another trial (option C), and certainly not the probability that event A or event B does not occur in a single trial (option D).

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Answer:

a) [tex]2.05[/tex]

b) [tex]z = 2.01[/tex]

c) No, we cannot conclude that a larger proportion of graduates have jobs than reported in the article.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 200

p = 0.333

Alpha, α = 0.02

Number of graduates had jobs , x = 80

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.333\\H_A: p > 0.333[/tex]

This is a one-tailed(right) test.  

b) Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{80}{200} = 0.4[/tex]

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

Putting the values, we get,

a) [tex]z = \displaystyle\frac{0.4-0.333}{\sqrt{\frac{0.333(1-0.333)}{200}}} = 2.01[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]

c) Since, the calculated z statistics less than the critical value, we fail to reject the null hypothesis and accept it.

Thus, same proportion of graduates have jobs as compared to previously reported.

Thus, we conclude that there is not enough evidence to support the claim that a larger proportion of graduates have jobs than previously reported.