Lucy is using a one-sample t ‑test based on a simple random sample of size n = 22 to test the null hypothesis H 0 : μ = 16.000 cm against the alternative H 1 : μ < 16.000 cm. The sample has mean ¯¯¯ x = 16.218 cm and standard deviation is s = 0.764 cm. Determine the value of the t ‑statistic for this test. Give your answer to three decimal places.

Answer:

2.03 × 10¹⁴ different possibilities.

Step-by-step explanation:

We want to know the different ways 1000 people can come out to watch a movie at 6 different times of the day.

This is a permutation and combination problem.

Dividing 1000 people amongst 6 movie showings = 1005C5 = 1005!/(1005-5)!5!

= (1005×1004×1003×1002×1001×1000!)/(1000!5!) = (1005×1004×1003×1002×1001)/5! = 2.03 × 10¹⁴ different possibilities.

Answer:

a) [tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

b) [tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]

[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]

[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]

[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]

[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]

And adding we got:

[tex] P(X<4) =0.000864[/tex]

c) [tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]

[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]

[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]

[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]

[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]

[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]

[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

And replacing we got:

[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]

d) [tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]

[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]

[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]

And replacing we got:

[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]

e) [tex] E(X) = n*p = 10*0.8 =8[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=10, p=0.8)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

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

Part a

We want to find this probability:

[tex] P(X=5)[/tex]

And using the pmf we got:

[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

Part b

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

[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]

[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]

[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]

[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]

And adding we got:

[tex] P(X<4) =0.000864[/tex]

Part c

For this case we can use the complement rule like this:

[tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]

[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]

[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]

[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]

[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]

[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]

[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

And replacing we got:

[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]

Part d

[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]

[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]

[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]

[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]

And replacing we got:

[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]

Part e

The expected number is given by:

[tex] E(X) = n*p = 10*0.8 =8[/tex]

This answer provides step-by-step explanations for finding probabilities of professors with doctoral degrees and the expected number of professors with doctoral degrees.

Expected number of college professors with a doctoral degree:

Find the probability that five have a doctoral degree: Using the binomial probability formula, [tex]P(X = 5) = 10C5 * 0.8^5 * 0.2^5[/tex]

Find the probability that fewer than 4 have a doctoral degree: Calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).

Find the probability that at least 6 have a doctoral degree: Calculate P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)).

Find the probability that between 5 and 7 inclusive have a doctoral degree: Calculate P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7).

Expected number of professors with a doctoral degree: Multiply the total number of professors by the probability (0.8).

Answer:

The proof is shown in the explanation below.

Step-by-step explanation:

Analysis:

The proof by induction focuses on n. In this case, let n = 1, and [tex]L^{1}[/tex] will be a linear operator since [tex]L^{1} = L[/tex]

The exercise will show that [tex]L^{n}[/tex] is a linear operator on V and that [tex]L^{n+1}[/tex] is also a linear operator on V. This, follows that:

[tex]L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})[/tex]

Answer/Step-by-step explanation:

For the mathematical induction,

We show that the equation

L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) is true for

L = 1,

Assume it is true for L = n and show that it is true for L = n + 1.

If L = 1, the equation become

(α1v1 + α2v2 +· · ·+αnvn)= α1(v1) + α2 (v2)+· · ·+αn (vn). Therefore, the Right Hand side(RHS) = Left Hand side(LHS)

When L = n, we assume the following is true

(α1nv1 + α2nv2 +· · ·+αnvn)= α1n(v1) + α2n (v2)+· · ·+αn (vn)

Then, when L = n + 1,

n +1 (α1v1 + α2v2 +· · ·+αvn)= α1(n +1) (v1) + α2(n +1) (v2)+· · ·+αn(n + 1)(vn).

Open the bracket,

n(α1v1 + α2v2 +· · ·+αvn) + α1v1 + α2v2 +· · ·+αnvn = α1n (v1) + α2v2 +· · ·+αvn ) + α1(v1) + α2v2+· · ·+αn(vn)

Since we assume the the equation is true for L = n, and eliminating some terms, then

L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn)

Answer:

Reliability

Step-by-step explanation:

Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack reliability.

Reliability can be defined as the extent to which an  experiment or test provides the same result on  repeated trials.

The huge difference in Dylan's IQ test scores in a short period of time suggests that the Whaddayanno IQ test is not reliable.  

Answer:

Part (a) : 0.2297

Part (b) :  0.1112

Part (c) : 0.1469

Part (d) : 77,171

Step-by-step explanation:

Given info on Health Sciences:

Mean = $51,541

Standard Deviation = $11,000

Given info on Business:

Mean = $53,901

Standard Deviation = $15,000

Part (a)

Let X represents the new college graduate in business,

P (X ≥ 65,000) = 1 - P (X < 65,000)

= 1 - P ( z < [tex]\frac{65,000 - 53,901}{15,000}[/tex] )

= 1 - P ( z < 0.74)

= 1 - 0.77035

= 0.2297

Part (b)

Let Y represents the new college graduate in Health Sciences,

P (Y ≥ 65,000) = 1 - P (Y < 65,000)

= 1 - P ( z < [tex]\frac{65,000 - 51,541}{11,000}[/tex] )

= 1 - P ( z < 1.22)

= 1 - 0.88877

= 0.1112

Part (c)

Let Y represents the new college graduate in Health Sciences,

P (Y  < 40,000) = P (Y < [tex]\frac{40,000-51,541}{11,000}[/tex])

= P ( z < -1.05 )

= 0.1469

Part (d)

To have a starting salary higher than 99%, the z-score = 2.33. Let A represents the salary of a new college graduate in health sciences higher than 99% of all starting salaries.

[tex]2.33 = \frac{A - 51,541}{11,000}[/tex]

[tex]A = 77,171[/tex]

77,171 new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in health sciences.

In factual tests, an invalid speculation recommends that there is no importance between factors in a bunch of noticed information. In the given decisions, the assertion shows no distinction in the pace of ability improvement between school gymnasts who practice contemplation and the people who don't address the invalid speculation.

In factual theory testing, an invalid speculation is an assertion recommending that no measurable relationship and importance exists in a bunch of noticed information between factors. It expects that any noticed contrasts are because of possibility. So in the given choices, the invalid speculation is: 'There will be no distinction in the pace of expertise improvement between school gymnasts who practice contemplation and the people who don't.' This assertion proposes no effect of the variable (reflection) on the result (ability improvement), which is what an invalid theory addresses in a trial of importance.

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Answer:

Face validity.

Step-by-step explanation:

Face validity is considered the weakest form of validity as it does a superficial, subjective assessment which does not involve objective approach, revealing the deeper intent of the test being used. It involved a process similar to skimming the surface of an item or a book to make opinions. Though it is a weak form of validity, it is the easiest to apply to research. Face validity is an informal approach to identifying how suitable the content of a test is to the research.

It generally asks the question:

Is the content of the test suitable to its aims?

Answer:

ht j5yh bgmn gmjdt ky

Step-by-step explanation:

w245358697809908

The cross sections of the given mathematical functions provide a range of shapes, from parabolas to circles or lines, depending on whether x, y or z are fixed.

Interpreting and sketching the three-dimensional functions in question gives us an understanding of the cross-sections. For the first function i) z = x2 + y2 + 6, when x is fixed, the cross section gives us a parabola facing upwards in yz-plane. Moving ahead, when y is fixed, it gives us a similar parabola in the xz-plane. When z is fixed, we end up with a circle in xy-plane.

Going to the second function ii) z = 3x2, cross sections with x fixed results in a vertical line in yz-plane as z is not a function of y here. For y being fixed, it will again give us a vertical line but in the xz-plane because z and y are again unrelated. If z is fixed, we obtain a parabola opening upwards (or downwards depending on the sign) parallel to xy-plane.

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Answer:

Step-by-step explanation:

Given that an ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations.

Hence total observations are 30*4 =120

No of groups = 3

Hence numerator df = 3-1 =2

Now total degrees of freedom = 120-1 =119

So denominator degrees of freedom = 119-2 = 117

Thus F statistic will have numerator as 2 degrees of freedom and denominator as 117 degrees of freedom.

The test statistic for the given hypothesis test for a proportion is calculated to be 0.356.

Given that:

The claim is that the proportion of people who are right-handed is greater than 70%.

So, p > 0.7 is the alternative hypothesis.

From the setup of the hypothesis test, the test is a one-tailed test since the alternative hypothesis has a sign ">".

Sample size, n = 600

Number of people right-handed = 424

So, the observed sample proportion is:

[tex]\hat{\text{p}}=\frac{424}{600}[/tex]

  [tex]=0.7067[/tex]

The standard deviation is:

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

  [tex]=\sqrt{ \frac{0.7(1-0.7)}{600} }[/tex]

  [tex]=0.01871[/tex]

So, the test statistic is:

[tex]\text{z}=\frac{\hat{\text{p}}-\text{p}}{\sigma}[/tex]

 [tex]=\frac{0.7067-0.7}{0.01871}[/tex]

 [tex]=0.356[/tex]

Hence, the test statistic is 0.356.

Learn more about Test Statistics here :

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Answer:

I. $291.67 per passenger

II. $583.33 per passenger

Step-by-step explanation:

The cost of flying the passenger plane from A to B is $70,000.

If the Capacity of the plane is 240 people.

(I)At 7A.M and 4P.M., the average cost per passenger is given as:

Average Cost=Total Cost/Number of Passengers

=70000/240=$291.67 per passenger

(II)At 10AM and 1PM, the flight is only half full, it has 120 passengers.

Average Cost = 70000/120=$583.33 per passenger

Answer: The cost will be $291.70 and $583.40 for the first and the last and for the second and third flights, respectively.

Step-by-step explanation: The cost is $70,000, the full capacity of seats is 240 and half capacity is 120.

In the first and last flights, which have the full capacity, the average cost per passenger is:

C = [tex]\frac{70,000}{240}[/tex] = 291.70

In the second and third ones, the average cost is:

C = [tex]\frac{70,000}{120}[/tex] = 583.4

The average cost per passenger for the first and the last flight are $291.70, while for the second and third are $583.4

Answer:

mL

Step-by-step explanation:

Ml

Step-by-step explanation:

x² + y² − 8x + 10y − 8 = 0

Rearrange:

x² − 8x + y² + 10y = 8

To complete the square, take half of the x and y coefficients, square it, then add the result to both sides.

(-8/2)² = 16

(10/2)² = 25

x² − 8x + 16 + y² + 10y + 25 = 8 + 16 + 25

x² − 8x + 16 + y² + 10y + 25 = 49

Factor the squares:

(x − 4)² + (y + 5)² = 49

The center is (4, -5) and the radius is 7.

Answer:

the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in

Step-by-step explanation:

since the box has a volume V

V= x*y*z = b=8000 in³

since y=z (square face)

V= x*y² = b=8000 in³

and the cost function is

cost = cost of the square faces *  area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides

C = a* 2*y² +  a* 2*x*y + 15*a* 2*x*y =  2*a* y² +  32*a*x*y

to find the optimum we can use Lagrange multipliers , then we have 3  simultaneous equations:

x*y*z = b

Cx - λ*Vx = 0 → 32*a*y -  λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ

Cy - λ*Vy = 0  → (4*a*y + 32*a*x) - λ*2*x*y = 0

4*a*32/λ  + 32*a*x - λ*2*x*32*a/λ = 0

128*a² /λ +  32*a*x - 64*a*x = 0

32*a*x = 128*a² /λ

x  = 4*a/λ

x*y² = b

4*a/λ * (32*a/λ)² = b

(a/λ)³ *4096 =  8000 m³

(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in

then

x  = 4*a/λ = 4*5/4 in = 5 in

y=32*a/λ = 32*5/4 in = 40 in

then the box has dimensions 5 in x 40 in x 40 in

Answer:

The argument is valid.

Step-by-step explanation:

We are told "Loretta's hobby is stamp collecting."

"If her husband likes to fish, then Loretta's hobby is not stamp collecting."

Her hobby is stamp collecting, so her husband does not like to fish.

"If her husband does not like to fish, then Nathan likes to read."

Her husband does not like to fish; therefore, Nathan likes to read.

The argument is valid.

The argument is invalid as it fails the informal test of validity; the premises about Loretta and her husband's hobbies do not logically lead to the conclusion about Nathan's reading habits.

The task involves assessing the validity of an argument by applying the informal test of validity. Let's deconstruct the argument provided. It states that Loretta's hobby is stamp collecting, and presents two conditional statements involving her husband's hobbies and Nathan's interest. Finally, it concludes that Nathan likes to read.

To apply the informal test of validity, we must determine if it's possible for all premises to be true while the conclusion could still be false. In this argument, even if all the stated conditions about Loretta and her husband's hobbies are true, they don't logically necessitate the conclusion about Nathan's reading habits. Thus, the argument is invalid because it's possible to imagine a scenario where all premises are true, but Nathan does not like to read. The premises about Loretta and her husband have no logical connection to Nathan's interests, making it invalid.

A valid argument ensures that if the premises are true, the conclusion must also be true. However, this argument fails to meet that criterion, indicating its invalidity based on the informal test.

Answer:

6.79 miles

Step-by-step explanation:

Consider the triangle ABC attached where A is the starting point and C is the finish point.

|AB|=5.2 miles

|BC|=3.0 Miles

≺ABC = 90+75 =165°

We are given the length of two sides and an angle not opposite any of the given sides.

The rule for solving this scenario is referred to as the Cosine Rule.

Note that the side opposite each angle is labelled using the corresponding small letter.

The Cosine Rule States that:

b²=a²+c²-2acCosB

|AC|²=3²+5.2²-(2X5.2XCos165°)

|AC|²=9+27.04-(-10.05)

|AC|²=9+27.04+10.05=46.09

|AC|=√46.09=6.79 miles

Therefore the distance on a straight line from A to C is 6.79 miles

Answer:

Therefore the required probability is 0.6472.

Step-by-step explanation:

Poisson distribution: A Poisson distribution is discrete distribution.

Let X be a discrete variable the number of event. Let λ be the mean of X.

Then the probability of k event is

[tex]P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}[/tex]

Here mean of each page is =0.03

Mean of 100 pages = (0.03×100)= 3

λ = 3

The required probability

P(X≤3)  

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

[tex]=\frac{3^0e^{-3}}{0!}[/tex] [tex]+\frac{3^1e^{-3}}{1!}[/tex][tex]+\frac{3^2e^{-3}}{2!}[/tex][tex]+\frac{3^3e^{-3}}{3!}[/tex]

[tex]=e^{-3}(1+3+\frac{9}{2}+\frac{27}{6})[/tex]

[tex]=e^{-3}(13)[/tex]

≈ 0.6472

Therefore the required probability is 0.6472

Answer:

C = 17/7 or 2 3/7

D = -4/7

Step-by-step explanation:

3C+4D=5

2C+5D=2

2C = 2 - 5D

C = 1 - 2.5D

3(1 - 2.5D) + 4D = 5

3 - 7.5D + 4D = 5

-3.5D = 2

D = -4/7

C = 1 - 2.5(-4/7)

C = 1 + 10/7 = 17/7

Answer: the height of the building is 40ft

Step-by-step explanation:

Looking at the right angle triangle formed,

With angle P as the reference angle, the length shadow of the building on ground represents the adjacent side of the right angle triangle.

The height of the building represents the opposite side of the right angle triangle.

a) to determine the height of the building, x, we would apply the tangent trigonometric ratio which is expressed as

Tan θ, = opposite side/adjacent side. Therefore, the equation becomes

Tan P = x/50

b) 0.8 = x/50

x = 50 × 0.8

x = 40 ft

Answer:

[tex]\frac{dV(t)}{dt} =[/tex] - 1675.38

Step-by-step explanation:

In 2017, the vakue of the kitchen equipment was $14550

V(0)=$14550

Its value after then was modelled by [tex]V(t)=14550e^{-0.158t[/tex]

We are required to find the rate of change in value on January 1, 2019

[tex]V(t)=14550e^{-0.158t[/tex]

[tex]\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t[/tex]

[tex]\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t[/tex]

[tex]\\Let u= -0.158t,\frac{du}{dt}=-0.158[/tex]

[tex]\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}[/tex]

[tex]\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}[/tex]

In 2019, i.e. 2 years after, t=2

The rate of change of the value

[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}[/tex]

=[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.316}[/tex]= - 1675.38

The rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).

The rate of change in the value of the equipment on January 1, 2019, can be found by taking the derivative of the given model equation V(t) = 14,550e^(-0.158t).

The derivative of V(t) with respect to t is dV/dt = -0.158(14,550)e^(-0.158t), which simplifies to dV/dt = -2,302.9e^(-0.158t).

Therefore, the rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).

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Answer:

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex]  °C/m.

Step-by-step explanation:

Given that, the temperature at a point (x,y,z) is

[tex]T(x,y,z)=200e^{-x^2-3y^2-9z^2}[/tex].

Rate change of temperature at the point P(2,-1,2) in the direction toward the point Q (3,-3,3) is [tex]D_uT(2,-1,2)[/tex]

[tex]T_x= 200(-2x)e^{-x^2-3y^2-9z^2}[/tex][tex]=-400x200.2xe^{-x^2-3y^2-9z^2}[/tex]

[tex]T_y = 200.(-6y)e^{-x^2-3y^2-9z^2}[/tex][tex]=-1200ye^{-x^2-3y^2-9z^2}[/tex]

[tex]T_z=200(-18z)e^{-x^2-3y^2-9z^2}[/tex][tex]=-3600ze^{-x^2-3y^2-9z^2}[/tex]

The gradient of the temperature

[tex]\bigtriangledown T(x,y,z)= (-400x200.2xe^{-x^2-3y^2-9z^2},-1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2})[/tex]                   [tex]=-400e^{-x^2-3y^2-9z^2}(x,3y,9z)[/tex]

[tex]\bigtriangledown T(2,-1,2) = -400e^{-2^2-3(-1)^2-9.2^2}(2,3.(-1),9.2)[/tex]

                   [tex]=-400 e^{-43}(2,-3,18)[/tex]

V=[tex]\overrightarrow {PQ}= \vec{Q}-\vec{P}[/tex]=(3,-3,3)-(2,-1,2)=(1,-2,1)

The unit vector of V is [tex]\frac{(1,-2,1)}{\sqrt{1^2+(-2)^2+1^2}}[/tex]

                                   [tex]=\frac{1}{\sqrt6}(1,-2,1)[/tex]

Therefore,

[tex]D_uT(2,-1,2) = \bigtriangledown T(2,-1,2).\frac{1}{\sqrt6}(1,-2,1)[/tex]

                       [tex]=-400 e^{-43}(2,-3,18).\frac{1}{\sqrt6}(1,-2,1)[/tex]

                       [tex]=-\frac{400}{\sqrt6}e^{-43}(2.1+(-3).(-2)+18.1)[/tex]

                       [tex]=-\frac{10400}{\sqrt6}e^{-43}[/tex]  

                        [tex]=-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex]  °C/m.

Answer:

The estimated loss is $5,400 if the value of  the quality characteristic,x is 85.

Step-by-step explanation:

We are given the following in the question:

Target value,T = 82

Cost coefficient, k = $6,000

x = 85

The Taguchi Quality Loss Function (QLF) is given by:

[tex]L(x) = k(x -T)^2[/tex]

where L(x) is the loss, k is the cost coefficient, T is the target value or the mean value.

We have to estimate the loss  if the value of the quality characteristic is 85.

We put x = 85

[tex]L(85) = 6000(85 -82)^2\\L(85) =54000[/tex]

Thus, the estimated loss is $5,400 if the value of  the quality characteristic,x is 85.

Answer:

The probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court is [tex]\frac{14}{29}[/tex].

Step-by-step explanation:

The conditional probability of an events A given that another events B has already occurred is given by:

[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}[/tex]

The estimation of the past presidential election is provided.

Denote the events as follows:

R = a Republican candidate would be elected

D = a Democratic candidate would be elected

C = a conservative judge would be appointed to the Supreme Court

M = a moderate judge would be appointed to the Supreme Court

L = a liberal judge would be appointed to the Supreme Court

The information provided is:

[tex]P(R)=\frac{5}{7},\ P(D)=\frac{2}{7}[/tex]

[tex]P(C|R)=\frac{1}{7},\ P(M|R)=\frac{1}{7},\ P(L|R)=\frac{5}{7}[/tex]

[tex]P(C|B)=\frac{1}{3},\ P(M|D)=\frac{1}{6},\ P(L|D)=\frac{1}{2}[/tex]

It is provided that a conservative judge was appointed to the Supreme Court during the presidential term.

Compute the probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court as follows:

[tex]P(D|C)=\frac{P(C|D)P(D)}{P(C|D)P(D)+P(C|R)P(R)}[/tex]

             [tex]=\frac{(\frac{1}{3}\times \frac{2}{7})}{(\frac{1}{3}\times \frac{2}{7})+(\frac{1}{7}\times \frac{5}{7})}[/tex]

             [tex]=\frac{\frac{2}{21}}{\frac{2}{21}+\frac{5}{49}}[/tex]

             [tex]=\frac{2}{21}\div \frac{29}{147}[/tex]

             [tex]=\frac{14}{29}[/tex]

Thus, the probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court is [tex]\frac{14}{29}[/tex].

Answer:

a.[tex]\mu=15[/tex]

b.[tex]\mu=7.8586 \ and \ \mu=22.1414[/tex]

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

[tex]P(X=k)=C_k^n.p^k.(1-p)^{n-k}[/tex]

Let n be the number of trials,[tex]n=100[/tex]

and p be the probability of success,[tex]p=15\%[/tex]

The mean of a binomial distribution is the probability x sample size.

[tex]\mu=np=100\times0.15=15[/tex]

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as:[tex]\sigma=\sqrt npq\\q=1-p[/tex]

Therefore, [tex]\sigma=\sqrt(100\times0.015\times0.85)=3.5707[/tex]

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

[tex]sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414[/tex]

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c.  [tex]x=45[/tex] is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Final answer:

Using the empirical rule, we can determine the limits within which we would expect to find the number of infested fields with a 95% probability. If we found a number of infested fields that is higher than expected, it suggests that one of the characteristics of a binomial experiment may not be satisfied.  Correct answer is B.

Explanation:

In this problem, we are given that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. We are asked to determine the limits within which we would expect to find the number of infested fields with a probability of approximately 95%. This situation can be modeled using the binomial distribution, as each field can be considered as a separate trial with two possible outcomes, infested or not infested.

According to the empirical rule, for a binomial distribution, we can expect about 95% of the outcomes to fall within two standard deviations of the mean. The mean number of infested fields can be calculated as 15% of the total number of fields, which is 15. The standard deviation of the number of infested fields can be determined using the formula σ = [tex]\sqrt{(npq)}[/tex], where n is the number of trials, p is the probability of success, and q is the probability of failure. In this case, n = 100, p = 0.15, and q = 0.85.

Using these values, we can calculate the standard deviation as σ = [tex]\sqrt{(100 * 0.15 * 0.85)}[/tex] ≈ 3.150. Therefore, we would expect to find about 95% of the number of infested fields within two standard deviations of the mean, which is between 15 - (2 * 3.150) = 8.700 and 15 + (2 * 3.150) = 21.300.

If we found that x = 45 fields were infested, we can conclude that this number is higher than what we would expect based on the binomial distribution. It is unlikely to observe such a high number of infested fields if the trials were independent and had only two possible outcomes. Therefore, we might suspect that one of the characteristics of a binomial experiment is not satisfied in this experiment.

Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes. Therefore, the correct answer is B.

Answer:

16.38%

Step-by-step explanation:

Given

Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000

Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09

p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes

0.09 + p = 1

p = 1 - 0.09

p = 0.91

The frequency of the carrier individual is calculated by npq

Where n = mating partners = 2

Frequency = 2 * 0.09 * 0.91

Frequency = 0.1638 or 16.38%

Answer:

The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is

0.42 or 42 %

Step-by-step explanation:

To solve the question, we note that

p²+2pq+q²=1  and p + q =1

where

p = frequency of the dominant allele

q= frequency of the recessive allele

p² = frequency of homozygous dominant allele

q² = frequency of homozygous recessive allele

2·p·q = frequency of hetrozygous individuals

The frequency of the homozygous recessive individuals q² is 0.09

Therefore the frequency of q = √(0.09) = 0.3, therefore p = 1 - 0.3 =0.7

The  percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2×p×q = Ae = 2*0.3*0.7 = 0.42

→ 0.42× 100 = 42 %

The slope of the line is [tex]m=-4[/tex]

Explanation:

Given that the pair of points are [tex](1,-19)[/tex] and [tex](-2,-7)[/tex]

We need to determine the slope of the line.

The slope of the line can be determined using the formula,

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Substituting the points [tex](1,-19)[/tex] and [tex](-2,-7)[/tex] for the coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] in the above formula, we have,

[tex]m=\frac{-7-(-19)}{-2-1}[/tex]

Simplifying the terms, we get,

[tex]m=\frac{-7+19}{-2-1}[/tex]

Adding the terms, we have,

[tex]m=\frac{12}{-3}[/tex]

Dividing, we get,

[tex]m=-4[/tex]

Thus, the slope of the line between the pair of points is [tex]m=-4[/tex]

Answer:

(a) 11 people

(b) 24 people

Step-by-step explanation:

Part (a)

Let X be the people who are network friends with three other people in the network.

The total pairings = 50

Since [tex]v[/tex]∈[tex]V[/tex],

[tex](7)(6) + (1)(5) + (5)(4) + (X)(3) = (2)(50)[/tex]

Solve for X to find the people who are network friends with three other people,

[tex]67 + 3X = 100[/tex]

[tex]X = 11[/tex]

Answer: There are 11 people who are network friends with three other people in the network.

Part (b)

Total people in the network,

[tex]= 7 + 1 + 5 + X\\ = 13 + 11\\ = 24[/tex]

Answer: There are a total of 24 people in the network.

There are 6 people who are network friends with three other people in the network. The total number of people in the network is 19.

This problem involves understanding the concept of pairs in network friends. Each friendship forms a pair. So, if one person has 6 friends, this would contribute 6 pairs. Here, there are 7 people with 6 friends, 1 person with 5 friends, and 5 people with 4 friends. Therefore, these people contribute (7x6) + (1x5) + (5x4) = 42 + 5 + 20 = 67 pairs. But, we're told there are only 50 pairs in total. The reason for this discrepancy is that each pair is being counted twice. When we divide the total pairs (67) by 2, we get 33.5 but since the number of pairs must be an integer, we subtract this half pair to get 33 pairs for the listed people. That means, there must be 50 - 33 = 17 pairs remaining for the people who are network friends with three other people. Given that each of these people contributes 3 pairs, there must be 17 / 3 = 5.67 people, but since we're talking about whole people, we need to round this up to 6. Therefore, there are 6 people who are network friends with three other people in the network. Now, adding up all the people in the network, we have 7 (people with 6 friends) + 1 (person with 5 friends) + 5 (people with 4 friends) + 6 (people with 3 friends) = 19 people in total.

https://brainly.com/question/15332165

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Answer:

Step-by-step explanation:

Hello!

The objective is to study the semicircular canals (SC) of the inner ear of the three-toed sloth (Bradypus variegatus) to see if it is weak compared to faster animals.

The study variable is X: length to width of the anterior semicircular canal of a three-toed sloth.

A sample of 7 sloths was taken and the semicircular canal was measured:

1.52, 1.06, 0.93, 1.38, 1.47, 1.20, 1.16

∑X= 8.72

∑X²= 11.15

a.

[tex]S^2= \frac{1}{n-1} * [sumX^2-\frac{(sumX)^2}{n} ]= \frac{1}{6} *[11.15-\frac{(8.72)^2}{7} ][/tex]

S²= 0.047≅0.05

S=0.218≅ 0.22

Comparing the estimation of the variance of the length to width of the anterior semicircular canal of three-toed sloths with the known number of length to width of the anterior semicircular canal of faster animals (S=0.09), it appears that the variability os length to width of the anterior semicircular canal of sloths is greater than the length to width of the anterior semicircular canal of faster animals.

b. and c.

To estimate the most plausible range of values of the population standard deviation of the anterior semicircular canal of the sloths, you have to do an estimation per confidence interval.

To be able to make this estimation we have to assume that the variable of interest has a normal distribution. With this assumption, it is valid to use a Chi-Square statistic to estimate the population standard deviation.

[tex]X^2= \frac{(n-1)S^2}{Sigma^2} ~X^2_{n-1}[/tex]

I'll choose a confidence level of 95%

The formula for the interval is:

[tex][\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ][/tex]

[tex]X^2_{n-1;1-\alpha /2}= X^2_{6;0.975}= 14.449[/tex]

[tex]X^2_{n-1;\alpha /2}}= X^2_{6;0.025}= 1.2373[/tex]

[tex][\frac{6*0.05}{14.449} +\frac{6*0.05}{1.2373} ][/tex]

[0.2076;2.4246] ⇒This confidence interval is for the population variance, calculating the square root of each bond gives us the CI for the population standard deviation:

√[0.2076;2.4246]= [0.1441;1.5571]

The 95% CI [0.1441;1.5571] is expected to contain the true value of the population standard deviation of the length to width of the anterior semicircular canal of the three-toed sloths.

As you can see this interval does not contain the known value of the population standard deviation for faster animals, which leads to thinking there is a difference between the standard deviation of the anterior semicircular canal in both species.

I hope it helps!

Answer:

b. 10.4

c.  26.9

Step-by-step explanation:

Let the universal set U = 100% which is the total no of people in the American community

Let A = 14.6% which is the total no of people living below poverty line

Let B = 20.7% which is the total no of people speaking foreign Language

C = 4.2% no of people who both speak foreign language and live below poverty line

X = no of people who neither live below poverty line nor speak foreign language

P (A) = 14.6%

P (B) = 20.7%

P (C) = P (A ∩ B) = 4.2%

P (A – C) = P (A ∩ U) = 14.6 – 4.2 = 10.4%

P (B – C) = P (B ∩ U) = 20.7 – 4.2 = 16.5%

P (X) = P (A ᴜ B) c =100 – (10.4 + 4.2 + 16.5) = 68.9%

a. The venn diagram is as shown above

b. Percent of Americans who live below poverty line and Speak English at home(minus foreign lang speakers living below poverty line) that is A only

= A – C

= 14.6 – 4.2

= 10.4%

c. Percentage of Americans Living below poverty line or Speaking foreign language

= A only + B only

A only = A – C ( People living below poverty line only)

= 14.6 -4.2

= 10.4%

B only = B – C ( people speaking foreign languages only)

= 20.7 – 4.2

= 16.5%

Hence

A only + B only = 10.4 + 16.5 = 26.9%

Answer:

Step-by-step explanation:

Check attachment for solution