Your body loses sodium when you sweat. Researchers sampled 38 random tennis players. The average sodium loss was 500 milligrams per pound and the standard deviation was 62 milligrams per pound. Construct and interpret a 99% confidence interval to estimate the mean loss in sodium in the population.

Answer:

Step-by-step explanation:

Check attachment for solution

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:

Step-by-step explanation:

The detailed steps and appropriate workings is as shown in the attached file.

The question asks about reducing a second-order ODE to first-order given a known solution, which is achieved using the method of reduction of order to find a new function v(y) leading to a first-order ODE.

To reduce a second-order ordinary differential equation (ODE) to a first-order ODE, given that y1 is a solution, you can apply the method of reduction of order. This involves introducing a new function v(y) such that y can be expressed as a product of the known solution y1 and this new function v(y). Essentially, you substitute y = y1*v into the original second-order ODE and differentiate as necessary to obtain an equation in terms of v and its derivatives only, effectively transforming the equation into first-order.

The steps include (i) expressing y in terms of y1 and v, (ii) differentiating this expression to find the derivatives of y, and (iii) substituting all of this into the equation that y1 obeys. The resultant first-order equation will only involve the function v and its first derivative, v'. By solving this simplified equation, you can find v and thus the second solution to the original second-order ODE.

Answer:

20 m

Step-by-step explanation:

We are given that

Initial horizontal speed ,[tex]u_x=15 m/s[/tex]

Time, t=2 s

Initial vertical velocity, [tex]u_y=0[/tex]

We know that

[tex]h=u_yt+\frac{1}{2}gt^2[/tex]

Where [tex]g=-9.8m/s^2[/tex]

Using the formula

[tex]h=0(2)+\frac{1}{2}(-9.8)(2)^2[/tex]

[tex]h=-19.6 m\approx -20 m[/tex]

The negative sign indicates that the displacement of stone  is in downward direction.

Hence, the height of the cliff is closest to 20 m

Answer:

[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]  

[tex]p_v =P(z<-1.252)=0.105[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas

Step-by-step explanation:

Data given and notation n  

n=180 represent the random sample taken

X=157 represent the students who came from Arkansas or a bordering state

[tex]\hat p=\frac{157}{180}=0.872[/tex] estimated proportion of students who came from Arkansas or a bordering state

[tex]p_o=0.9[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher or not than 0.9.:  

Null hypothesis:[tex]p\geq 0.9[/tex]  

Alternative hypothesis:[tex]p < 0.9[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-1.252)=0.105[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas

Answer:

The zeros of this function are 3 and -5.

Step-by-step explanation:

One note: The question says "Find the zeros of EACH function", but there seems to be only one function in the question. I hope I answered it well.

To complete the function provided, you need to get both terms inside both parenthesis to equal zero. So, if its (x-3)(x+5), the X's need to equal the opposite of the number that is being added or subtracted.

For the first parenthesis, (x-3), the number is -3, so the zero of that side is positive 3. (3-3 = 0)

For the second parenthesis, (x+5), the number is positive 5, so the zero of that side is -5. (-5+5 = 0)

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:

"s" is the river speed  16 miles is the one-way distance

Time = distance / velocity  

2.6  =  [16 / (13 + s) ]   +  [16 / (13 -s) ]

That solves to s = 3 miles per hour

Step-by-step explanation:

Answer: the speed of the current is 3 mph

Step-by-step explanation:

Let x represent the speed of the river current.

Bob can row 13mph in still water. Assuming that while rowing upstream, he rowed against the current, this means that his total speed upstream is (13 - x) mph

Assuming that while rowing downstream, he rowed with the current, this means that his total speed downstream is (13 + x) mph.x

If the total distance downstream and back is 32 miles. Assuming the distance upstream and downstream is the same, then the distance upstream is 32/2 = 16 miles. Distance downstream is also 16 miles.

Time = Distance/speed

Time taken to row upstream is

16/(13 - x)

Time taken to row downstream is

16/(13 + x)

If total time spent is 2.6 hours, it means that

16/(13 - x) + 16/(13 + x) = 2.6

Cross multiplying, it becomes

16(13 + x) + 16(13 - x) = 2.6(13 + x)(13 - x)

= 208 + 16x + 208 - 16x = 2.6(169 - 13x + 13x - x²)

416 = 2.6(169 - x²)

416/2.6 = 169 - x²

160 = 169 - x²

x² = 169 - 160 = 9

x =√9

x = 3

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:

k=1

P(12)=256

Step-by-step explanation:

The model for population involving birth and death is given as:

[tex]\frac{dP}{dt}=(b-d)P[/tex]

where b=birth rate and d=death rate.

If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]

[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.

Substituting b and d into our model

[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant

Simplifying using indices

[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]

Next, we Separate Variables and Integrate both sides

[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]

[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]

[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration

[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]

When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]

[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.

(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169

[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]

[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]

Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.

[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]

[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]

Therefore the population after a year is 256.

The population of fish after 1 year would be 285.

Let P be the population of fish at any time t months.

We are given:

We need to determine the population after 12 months.

The logistic growth model can be expressed as:

dP/dt = rP(1 - P/K)

where r is the intrinsic growth rate and K is the carrying capacity of the environment. Here, since the birth and death rates are inversely proportional to √P, we can infer that the growth rate might be proportional to √P, leading us to use a simplified logistic growth equation.

Given P(6) = 169, let’s find the intrinsic growth rate , r. From population changes between 0 and 6 months:

Let’s assume the logistic growth function in the form:

P(t) = [tex]P_0 e^{rt}[/tex]

We solve for r:

169 = [tex]100 e^{6r}[/tex]

1.69 = [tex]e^{6r}[/tex]

Taking natural logarithms:

ln(1.69) = 6r

0.524 = 6r

r = 0.0873/month

Now, to find P(12):

P(12) = [tex]100 e^{0.0873*12}[/tex]

= [tex]100 e^{1.048}[/tex]

= 100 * 2.853

= 285.3 fish

So, the population after 1 year is approximately 285 fish.

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:

Please find attached file for complete answer solution and explanation of same question.

Step-by-step explanation:

Answer:

The point estimate of population mean is 21.6.

Step-by-step explanation:

A point estimate is a numerical value that is used to estimate the value of the parameter under study.

The point estimate is calculated using the sample values.

For example, sample ([tex]\bar x[/tex]) is the point estimate of population mean (μ), sample standard deviation (s) is the point estimate of population standard deviation (σ), sample proportion ([tex]\hat p[/tex]) is the point estimate of population proportion (p).

It is provided that the sample mean of 100 items is,

[tex]\bar x=21.6[/tex]

Then the point estimate of population mean is:

[tex]\mu=\bar x=21.6[/tex]

Thus, the point estimate of population mean is 21.6.

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:

[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]

In order to find the variance we need to find first the second moment given by:

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

And replacing we got:

[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]

The variance is calculated with this formula:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]

And the standard deviation is just the square root of the variance and we got:

[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

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

And replacing we got:

[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]

In order to find the variance we need to find first the second moment given by:

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

And replacing we got:

[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]

The variance is calculated with this formula:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]

And the standard deviation is just the square root of the variance and we got:

[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]

The required  value of mean 0.330 and standard deviation 0.5545 for the defective transistors.

Given that,

A company has found that in carton of transistors: 92% contain no defective transistors,

3% contain one defective transistor, 3% contain two defective transistors,

and 2% contain three defective transistors.

We have to find,

Calculate the mean and variance for the defective transistors. Mean = Variance.

According to the question,

Let,  X the random variable who represent the number of defective transistors.

The following probability distribution for X,

X         0           1           2         3

P(X)   0.92     0.03    0.03     0.02

To calculate the expected value with the following formula:

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

On substitute all the values in the formula,

[tex]E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15[/tex]

To find the variance first the second moment given by:

[tex]E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)[/tex]

On substitute all the values in the formula;

[tex]E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330[/tex]

Therefore, The variance is calculated with this formula:

[tex]Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075[/tex]

And the standard deviation is just the square root of the variance,

[tex]Standard \ deviation = \sqrt{0.0375} = 0.5545[/tex]

Hence, The required  value of mean 0.330 and standard deviation 0.5545 for the defective transistors.

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Answer:

It would take 3 painters, a total number of 8 hours, to paint the house

Step-by-step explanation:

Let the number of hours be denoted by N, and the number of painters be denoted by P. The inverse relationship can be expressed as:

N α [tex]\frac{1}{P}[/tex]

Removing the proportionality symbol (α), introduces a constant of proportionality, which can be assumed to be C, in this example.

So, N = [tex]\frac{C}{P}[/tex]

When N = 6, P = 4

6 = [tex]\frac{C}{4}[/tex]

C = 24

So that, N = [tex]\frac{24}{P}[/tex]. This is the formula to use for determination of the other variable when one of them is known.

Now, given 3 painters, the number of hours needed can be obtained as follows:

N = [tex]\frac{24}{3}[/tex] = 8 hours

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:

(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.

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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:

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:

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:

30.85% probability of one can less than 12 oz

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 12.1, \sigma = 0.2[/tex]

a) What is the probability of one can less than 12 oz

This is the pvalue of Z when X = 12. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{12 - 12.1}{0.2}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085

30.85% probability of one can less than 12 oz

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.

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

Using the combinations formula C(n, r), the number of ways Maya can choose the food and drinks for the bridal shower is computed as: C(12,3) for drinks * C(12,3) for appetizers * C(10,3) for desserts.

The subject of this question is combinations in Mathematics. Maya has choices of 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts. She wants to pick 3 of each, and the order of selection does not matter. We can use the combination formula to calculate the number of ways she can do this:

So the total number of different ways Maya can pick the food and drinks to serve at the bridal shower is C(12,3) * C(12,3) * C(10,3).

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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

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:

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:

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:

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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