Which expression is not a perfect square trinomial?

Answer:

The probability is 0.4841.

Step-by-step explanation:

The provided table is:

From above table, it is known that

Number of subjects are 1008.

The probability that the person chosen is a woman given that the person is a light smoker can be calculated as:

[tex]P(Woman| Light smoker)=\frac{P(Woman and light smoker)}{P(Light smoker)} \\P(Woman| Light smoker)= \frac{\frac{76}{1008} }{\frac{157}{1008} } = 0.4841[/tex]

Thus, required probability is 0.4841.

The probability that a randomly selected person is a woman given that the person is a light smoker is [tex]\[ 0.400} \][/tex]

To find the probability that a randomly selected person is a woman given that the person is a light smoker, we need to use conditional probability. The formula for conditional probability [tex]\( P(A|B) \)[/tex] is:

[tex]\[P(A|B) = \frac{P(A \cap B)}{P(B)}\][/tex]

Where:

[tex]\( P(A|B) \)[/tex] is the probability of event [tex]\( A \)[/tex] occurring given that [tex]\( B \)[/tex] has occurred.

[tex]\( P(A \cap B) \)[/tex] is the probability of both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] occurring.

[tex]\( P(B) \)[/tex] is the probability of an event [tex]\( B \)[/tex] occurring.

Let's define the events:

[tex]\( A \)[/tex] : The person chosen is a woman.

[tex]\( B \)[/tex] : The person chosen is a light smoker.

We need the number of light smokers and the number of women who are light smokers. Suppose we have the following data:

Total number of subjects: 1008

Number of light smokers: [tex]\( n_{\text{light smokers}} \)[/tex]

A number of women who are light smokers: [tex]\( n_{\text{women and light smokers}} \)[/tex]

Given that the number of women who are light smokers is [tex]\( n_{\text{women and light smokers}} \)[/tex] and the total number of light smokers is [tex]\( n_{\text{light smokers}} \)[/tex], the probability can be calculated as follows:

[tex]\[P(\text{woman | light smoker}) = \frac{n_{\text{women and light smokers}}}{n_{\text{light smokers}}}\][/tex]

If we don't have the exact numbers, we'll need those to calculate the probability. However, let's assume the following values (hypothetically for the purpose of illustration):

Total number of light smokers: 150

Number of women who are light smokers: 60

The probability that a randomly selected person is a woman given that the person is a light smoker is:

[tex]\[P(\text{woman | light smoker}) = \frac{60}{150} = 0.4\][/tex]

Rounding to the nearest thousandth:

[tex]\[0.4 = 0.400\][/tex]

Answer:

76cm²

Step-by-step explanation:

The triangle from all indication is an isosceles triangle:

Let x represent the side

∴ x = 4 - 6 = -2

x =  -2

x + 2 = 0

Using the fomular, as area of triangle, that is:

Area = √s(s - a)(s - b)(s - c)

s = a + b+ c/2, where a = x + 2; b = x+ 2; c = x

∴ s = x + 2 + x + 2 + x/2 = 3x + 4/2

Area = √3x + 4/2( 3x + 4/2 - x - 2)(3x + 4/2 - x - 2)(3x + 4/2 - x)

= √3x + 4/2( x/2)(x/2)(x + 4/2)

= √3x + 4/2(x²/4)(x + 4)/2            

Let's assume x = 12

∴ Area = √5760 = 76cm²

Answer:

Its C.100

Area of a triangle b*h/2 (6+4=10*2=20) 20*10=200/2 will give you 100.

Step-by-step explanation:

Answer:

[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<-2.308)=0.0105[/tex]

Step-by-step explanation:

Previous concepts

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

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(150,13)[/tex]  

Where [tex]\mu=150[/tex] and [tex]\sigma=13[/tex]

We are interested on this probability

[tex]P(X<120)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<-2.308)=0.0105[/tex]

Answer:

The type 1 error here is a. Deciding that the absorption rates are different, when in fact they are not.

Step-by-step explanation:

A type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion).

More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis. By one common convention, if the probability value is below 0.05, then the null hypothesis is rejected.

In inferential statistics, the null hypothesis is a general statement or default position that there is nothing significantly different happening, like there is no association among groups or variables, or that there is no relationship between two measured phenomena.

Answer:

[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]    

[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]    

So on this case the 98% confidence interval would be given by (1.55;3.53)

Step-by-step explanation:

Previous concepts

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

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

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

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

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

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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

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

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

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

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

The mean calculated for this case is [tex]\bar X=2.54[/tex]

The sample deviation calculated [tex]s=1.110[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,9)".And we see that [tex]t_{\alpha/2}=2.82[/tex]

Now we have everything in order to replace into formula (1):

[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]    

[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]    

So on this case the 98% confidence interval would be given by (1.55;3.53)    

The 98% confidence interval would be given by (1.55;3.53)

A range of values that, with a particular level of confidence, is likely to encompass a population value is called a confidence interval. A population mean is typically stated as a percentage that falls between an upper and lower interval.

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

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

show the sample mean for the given sample.  

population mean (the relevant variable)

The sample standard deviation is denoted by s.

n stands for the number of samples.  

Resolution of the issue

The following formula produces the mean's confidence interval:

[tex]\bar x[/tex] ± [tex]\frac{t_a}{2} \frac{s}{\sqrt{n} }[/tex] ____________(1)

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

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

[tex]s = \sqrt{\frac{\sum^n_{i=1}(x_i-\bar x)}{n-1} }[/tex] ____________(3)

The mean calculated for this case is X = 2.54

The sample deviation calculated s = 1.110

t In order to calculate the critical value [tex]\frac{t_a}{2}[/tex] we need freedom, given by: to find first the degrees of

df = n-1=10-19

Since the Confidence is 0.98 or 98%, the value of a = 0.02 and a/2 = 0.01, and we can use excel, a calculator or a tabel to find the critical value.

[tex]\frac{t_a}{2}[/tex]  = 2.82

Now we have everything in order to replace into formula (1):

2.54 - 2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex] = 1.55

2.54 +  2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex]  = 3.53

Answer:

0.267

Step-by-step explanation:

p = 0.3 q = 0.7

10C3 × p³ × q⁷

0.266827932

Answer:

26.68% probability that exactly three workers take public transportation daily

Step-by-step explanation:

For each worker, there are only two possible outcomes. Either they take public transportation daily, or they do not. The probability of a worker taking public transportation daily is independent from other workers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

30% of workers take public transportation daily.

This means that [tex]p = 0.3[/tex]

In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily?

This is [tex]P(X = 3)[/tex] when [tex]n = 10[/tex]. So

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

[tex]P(X = 3) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668[/tex]

26.68% probability that exactly three workers take public transportation daily

a. The experimental​ unit:

The experimental units would be the lawyers that participate in this experiment. The experimental units are the subjects upon which the experiment is performed.

b. The response​ variable:

The response variable would be income. This is the variable that measures the response or outcome of the study.

c. The ​factors:

The factors are the variables whose levels are manipulated by the researcher. In this case, these would be the Mach score, classification and gender.

d. The levels of each​ factor:

The levels would include the levels of the Mach score (high, moderate, low) and the levels of gender (male, female).

e. The treatments:

The treatments are all the possible combinations of one level of each factor. Therefore, these are: High and male, high and female, moderate and men, moderate and female, low and male, low and female.

Answer:

The total compound return over the 3 years is 15.26%

Step-by-step explanation:

Let the initial investment sum be assumed to be X

The total return after each year can be calculated as follows:

After First year: X + (13% of X) = 1.13X

After Second year: 1.13X + (20% of 1.13X) = 1.13X + 0.226X = 1.356X

After Third year: 1.356X - (15% of 1.356X) = 1.356X - 0.2034X = 1.1526X

It is apparent from here that after the third year, the investment has increased the initial X, by 0.1526X, which is 15.26%.

The total compound return over the 3 years is 15.26%

Answer:

(a)

The slope of the equation = 0.02

Therefore T-intercept equals 15

(b)

Therefore the average global surface temperature in 2050 is 17°C.

Step-by-step explanation:

If a equation is in the form

y= mx+c........(1)

Then the slope of the equation is m.

Slope: The tangent of the angle which is made with the positive x-axis.

If θ be the angle , Then slope (m)= tanθ.

If a equation in the form

[tex]\frac{x}{a} +\frac{y}{b} =1[/tex]............(2)

Then x-axis intercept equals a and y-axis intercept equals b.

Given equation is

T=0.02t+15.0

Comparing with equation (1)

The slope of the equation = 0.02

Again we can rewrite the equation as

[tex]T-0.02t=15[/tex]

[tex]\Rightarrow \frac{T}{15} -\frac{0.02t}{15}=1[/tex]

Comparing with (2)

Therefore T-intercept equals 15

(b)

Here t= 2050-1950 =100

Putting t=100 in the given equation

T=0.02(100)+15 = 2+15 =17

Therefore the average global surface temperature in 2050 is 17°C.

Answer:

5 -2√10 = C

Step-by-step explanation:

√25 -√40

√25 = √5 x5 = √5² = 5

√40 = √5 x 8 = √5x 2x4 = √10x4 = √10x 2² = 2√10 (when the 2 comes back into the square root it become 2²)

√25 -√40 = 5 -2√10

Answer:

a) 0.931491

b) 0.995307

c) 0.994030

Step-by-step explanation:

a) Since all components must be working, the reliability of the computer is the product of the reliability of the three components:

[tex]R_1 = 0.97*0.97*0.99\\R_1=0.931491[/tex]

b) The resulting reliability is now the reliability of the first computer, added to the possibility of failure of the first computer multiplied by the reliability of the second computer:

[tex]R= R_1 +(1-R_1)*R_2\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99\\R=0.995307[/tex]

c) If a switch with reliability of 0.98 must be activated to turn on the second computer, the switch's reliability must be taken into account as follows:

[tex]R= R_1 +(1-R_1)*R_2*R_S\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99*0.98\\R=0.994030[/tex]

The reliability of the system is simply its probability of not failing

(a) The reliability of the computer

This is the product of the reliabilities of the three modules.

So, we have:

[tex]R = 0.97 \times 0.97 \times 0.99[/tex]

[tex]R = 0.931491[/tex]

Approximate

[tex]R = 0.9315[/tex]

Hence, the reliability of the computer is 0.9315

(b) The reliability when a backup is used

In (a), the reliability of the computer is 0.9315

When the computer fails, the reliabilities of the other two are 1 - 0.9315 and 1 - 0.9315.

So, the reliability when a backup is used is calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9315)][/tex]

[tex]R = 0.99530775[/tex]

Approximate

[tex]R = 0.9953[/tex]

Hence, the reliability of the computer when a backup is 0.9953

(c) The overall reliability of the system

In (a), the reliability of the computer is 0.9315.

Also, the reliability of the switch is 0.98

So, the reliability of the backup is:

[tex]R = 0.9315 \times 0.98[/tex]

[tex]R = 0.9129[/tex]

So, the overall system has:

The reliability of the overall system is then calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9129)][/tex]

[tex]R = 0.9940[/tex]

Hence, the reliability of the overall system is 0.9940

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Answer:

ooh i just learned this,  not 100% sure but the amount he should have to deposit is $3835.12

if the weekly one means depositing money every week then  it would be $36.88 I think.

Step-by-step explanation:

p=?

r=.08

n=52

t=2

4500 = P(1+.08/52)^(52 x 2)

divide both sides by (1+.08/52)^(52 x 2)

and you are left with $3835.12

if i take into account that Marc is depositing the money every week the i would divide it by 104  (that is 52 x 2 because 52 weeks in a year and it says 2 years) you would be left with $36.88.

Hope I was any help.

Answer: Marc should deposit $39.87 weekly.

Step-by-step explanation:

We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as

S = R[{(1 + r)^n - 1)}/r][1 + r]

Where

S represents the future value of the investment.

R represents the regular payments made(could be weekly, monthly)

r = represents interest rate/number of interval payments.

n represents the total number of payments made.

From the information given,

S = $4500

Assuming there are 52 weeks in a year, then

r = 0.08/52 = 0.0015

n = 52 × 2 = 104

Therefore,

4500 = R[{(1 + 0.0015)^104 - 1)}/0.0015][1 + 0.0015]

4500 = R[{(1.0015)^104 - 1)}/0.0015][1.0015]

4500 = R[{(1.169 - 1)}/0.0015][1.0015]

4500 = R[{(0.169)}/0.0015][1.0015]

4500 = R[112.67][1.0015]

4500 = 112.839R

R = 4500/112.839

R = 39.87

Answer:

6.08333

Step-by-step explanation:

6ft 0.8inches

Answer:

6 ft 1 in

Step-by-step explanation:

Answer:

The probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.

Step-by-step explanation:

Let X = number of Americans who believe that statistics teachers know the true meaning of life.

The probability of the random variable X is,

P (X) = 0.574

The event of a person not believing that statistics teachers know the true meaning of life is the complement of the event X.

The probability of the complement of an event, E is the probability of that event not happening.

[tex]P(E^{c})=1-P(E)[/tex]

Compute the value of [tex]P(X^{c})[/tex] as follows:

[tex]P(X^{c})=1-P(X)\\=1-0.574\\=0.426[/tex]

Thus, the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.

Answer:

Probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life = 0.426 .

Step-by-step explanation:

We are given that a poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life.

Let the above probability that % of Americans who believe that statistics teachers know the true meaning of life = P(A) = 0.574

Now, probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is given by = 1 - Probability of randomly selecting someone who believe that statistics teachers know the true meaning of life = 1 - P(A)

So, required probability = 1 - 0.574 = 0.426 .

The probability that Team A wins and Team B loses is 0.112.

Given that,

Suppose Team A has a 0.75 probability to win their next game and Team B has a 0.85 probability to win their next game.

Assume these events are independent.

We have to determine,

What is the probability that Team A wins and Team B loses?

According to the question,

In an independent event and probability, the outcomes in an experiment are termed as events. Ideally, there are multiple events like mutually exclusive events, independent events, dependent events, and more.

Team A has a 0.75 probability to win their next game,

And Team B has a 0.85 probability to win their next game.

Therefore,

The probability that Team A wins and Team B loses is

[tex]\rm The \ probability \ of \ A \ wins \ and \ team \ B \ loses = Probability \ of \ team \ A winning \ game \times (1- Probability \ of \ team B \ lose \ the \ game)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses = 0.75 \times (1-0.85)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.75 \times 0.15\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.112[/tex]

Hence, The probability that Team A wins and Team B loses is 0.112.

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Final answer:

The probability that Team A wins and Team B loses is calculated by multiplying the probability of Team A winning (0.75) with the probability of Team B losing (1 - 0.85). The result is 0.1125 or 11.25%.

Explanation:

To calculate the probability that Team A wins and Team B loses, we use the rules of independent events. The event of Team A winning has a probability of 0.75, and the event of Team B losing is the complement of Team B winning, which has a probability of 0.85. Since these are independent events, we multiply the probabilities:

P(Team A wins and Team B loses) = P(Team A wins) × P(Team B loses)

P(Team B loses) = 1 - P(Team B wins) = 1 - 0.85 = 0.15

Therefore, P(Team A wins and Team B loses) = 0.75 × 0.15 = 0.1125.

The probability that Team A wins and Team B loses is 0.1125, or 11.25%.

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Estimated  second point is (t,p) =(5,9320)

b

The rate is 180 thousand employer per year

c

The percentage rate of change of the function is 1.939%

Step-by-step explanation:

Looking at the graph

  First is to obtain the scale of the graph what i mean is what the distance between each line segment

    Considering the y-axis  each line segment is

                              [tex]\frac{9300-9200}{5} = \frac{100}{5} =20[/tex]

So this means that after 9300 the next line segment is 9320

Considering the x-axis each line segment is

                            [tex]\frac{4-2}{4} = \frac{2}{4} = 0.5[/tex]

What this means is that 2 line segment after 4 is 4 +2 ×(0.5) =5

So looking this two points (new_t,new_p) = (5 , 9320) = we see that they form a coordinate

       B) The labeled point that we are to consider are

   [tex](t_1,p_2) = (4.8, 9284) \ (t_2,p_2) = (5, 9320)[/tex]

The rate change

                 [tex]= \frac{p_2-p_1}{t_2-t_1}=\frac{9320-9284}{5-4.8} = \frac{36}{0.2} = 180[/tex]

So the rate is 180 thousand employer per year

C)

 So to obtain the percentage rate of change of the function

   Now

       [tex]f(4.8) = 9284 \ Thousand[/tex]

      [tex]f'(4.8) = 180 \ Thousand[/tex]    

Note: This is so because differentiation is the same as  slope of the graph

    Hence the percentage  rate of change

                          [tex]\frac{f'(4.8)}{f(4.8)} *\frac{100}{1} = \frac{180 \ 000}{9284 \ 000} * \frac{100}{1}[/tex]

                           = 1.939%

The question seems to be related to calculus, discussing concepts of tangents and rates of change for a function. To find another point on a tangent line or calculate the rate of change, we need the function and additional details. The respective formulas for these calculations are mentioned, but we can't provide a factual answer without more information.

Given the question, it seems this is related to the field of calculus, specifically addressing concepts of tangents and rates of change of a function. However, to find another point on the tangent line or calculate the rate of change at a specific point, we need more context or the actual function.

Usually, for the first question, you can use the formula slope = (y2-y1) / (x2-x1), assuming (t, p) is a point on the tangent and you know the slope of the tangent. For the second question, the rate of change at a point is the derivative of the function at that point. For the third question, the percentage rate of change at a point is the derivative at that point divided by the function value at that point, multiplied by 100 to get percentage.

Without the function and some additional information, we cannot arrive at a factual answer. You might want to check the question again and include the necessary details.

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Answer:

Step-by-step explanation:

given that after scoring a touchdown, a football team may elect to attempt a two-point conversion, by running or passing the ball into the end zone. If successful, the team scores two points.

X=1 if successful and

X=0 if not

pdf of X is

X      1       0

p    0.8   0.2

E(x) = [tex]1(0.8)+0(0.2)\\=0.8[/tex]

[tex]E(x^2) = 1^2(0.8) = 0.8[/tex]

Var(x) = 0.8-0.8*0.8

= 0.16

Now let us consider Y.

Y is the no of points scored.

Y      2       0

p     0.8    0.2

E(Y) = [tex]2(0.8)+0(0.2)\\=1,.6[/tex]

[tex]2^2(0.8)+0(0.2)\\= 3.2[/tex]

The mean and variance for X and Y are calculated using their definitions in probability theory. For X, the mean is 0.80 and variance is 0.16. For Y, the mean is 1.60 and variance is 0.64.

The random variable X is a Bernoulli random variable because it has only two outcomes: success (X=1) and failure (X=0). The mean and variance for a Bernoulli random variable can be found using the formulas: Mean (E[X]) = p and Variance (Var[X]) = p(1-p).

 For X:

The mean E[X] = p = 0.80. So, X = 1 with probability 0.80.

The variance Var[X] = p(1-p) = 0.80(1-0.80) = 0.16

The random variable Y represents the number of points scored which can either be 0 or 2. We let p be the probability of scoring 2 points i.e., p = 0.80. Hence, if the two-point conversion is successful, we will score 2 points, otherwise, we score 0.

For Y:

The mean E[Y] = 0*(1-p) + 2*p = 0 + 2*0.80 = 1.60

The variance Var[Y] = (0-E[Y])^2*(1-p) + (2-E[Y])^2*p = (0-1.60)^2*(1-0.80) + (2-1.60)^2*0.80 = 0.64

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Answer: B

Step-by-step explanation:

The sampling distribution of the proportion of damaged trees is approximately Normal

Answer:B. The sampling distribution of the proportion of damaged trees is approximately Normal.

Step-by-step explanation: Sampling distribution is a probability distribution of a statistic which is gotten through the collection of a large number of samples from the population of interest.

A normal distribution is a distribution of the samples symmetrically around the mean, when a Statistical metric shows that the outcome is distributed around the central region it shows that the distribution is normal.

FOR A FORTY PERCENT INCIDENCE, THE SAMPLING DISTRIBUTION OF THE PROPORTION OF DAMAGED TREES IS APPROXIMATELY NORMAL AS IT IS CLOSE TO 50% WHICH IS THE MEAN OF 100%.

The data set is S = {10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40}

A stem-and-leaf plot is a method to represent the data in tabular form.

The stem consist of the first digits of the data values arranged in ascending order.

The leaf consist of the remaining digits.

The data provided is:

Stem | Leaf

    1  | 0 1  5

    2 | 1  4 4 7 9

    3 | 3 5 5 7 9

    4 | 0 1  

The original data is:

10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40

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The original set of data consists of number sequences as follows: 10 12 4 7 11 4 3 10 0, 10 4 14 11 13 2 4 6, 12 6 9 10, 5 13 4, 10 14 12 11, and 6 10 11 0 11 13 2.

The original set of data is:

Answer:

d = 80 ft

Step-by-step explanation:

Start with angle BCA, tan(BCA) = 30/15 so BCA = 63.43 degrees.

BCA = ECD since they are opposite angles.

tan ECD = d/40

40 tan(63.4degrees) = d

d = 80 ft

Alternatively, you can use similar triangles since the angles are the same, BCA = ECD, CED = CAB, and CBA = EDC. In that case use proportions to get 30/15 = d/40 so 2 = d/40 and d = 80.

Answer:

I code an example question with answer.

Step-by-step explanation:

A study on educational aspirations of High School Students ( S. Crysdale, International Journal of Comparative Sociology, Vol 16, 1975, pp 19-36) measured aspirations using the scale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when family income was middle, the counts were (11, 52, 23, 22); when family income was high, the counts were (9, 41, 12, 27)

A. Construct a suitable contingency table for the above data.

B. Find the conditional distribution on aspirations for those whose family income was high.

C. Conduct a Chi-square test of Independence between educational aspirations and income levels.

D. Explain what further analyses you could do that would be more informative than a chi-squared test.

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Chi-Squared Test for Homogeneity of Several Categorical Populations

Null Hypothesis: populations of people are homogeneous with respect to the four levels of education (low, med, high income groups educated same)

Alternative Hypothesis: populations not homogeneous.

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Some Grad

High High Some Grad

School School College College Total

Low Income

9 44 13 10 76 [observed counts]

8.07 38.14 13.36 16.42 [expected counts]

0.106 0.901 0.010 2.513 [Chi-Square contr]

Medium Income

11 52 23 22 108

11.47 54.20 18.99 23.34

0.019 0.089 0.847 0.077

High Income

9 41 12 27 89

9.45 44.66 15.65 19.23

0.022 0.300 0.851 3.135

Total 29 137 48 59 273

Chi-Sq = 8.871, DF = 6, P-Value = 0.181

"P-Value" = 0.181 > 0.10 [90% confidence interval]; thus Null Hypothesis of homogeneity should be rejected (low, med, high income groups educated same). Alternative Hypothesis should be accepted (education dependent upon income level)

The task is a statistical analysis using SAS/R to test the independence of two variables, economic status and educational aspirations. This involves chi-square and likelihood ratio tests, investigating standardized residuals, and performing more powerful tests for comprehensive results.

The question involves a statistical analysis task using SAS/R to determine the independence between two variables: economic status and educational aspirations. The Chi-square (X2) and likelihood ratio (G2) tests can be utilized to analyze the independence. Standardized residuals can help diagnose potential patterns and significance of divisions, while more powerful tests such as Fisher's exact test or Monte Carlo simulation could provide further insights.

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Let X be the random variable representing monthly trainee income. X is distributed with mean $1100 and standard deviation $150. You want to find the proportion of trainees that earn less than $900 per month, or Pr(X < 900).

Using normalcdf (on a TI calculator, for instance), you would compute

normalcdf(-1E99, 900, 1100, 150)

to get a proportion of approximately 0.09121, or 9.121%.

That is, the syntax for normalcdf is

normalcdf(lower limit, upper limit, mean, standard deviation)

In this case, you pick a very large negative number for "lower limit" in order to simulate negative infinity.

Answer:

Step-by-step explanation:

Hello!

The researcher's objective is to compare the effectiveness of a water softener when used with a filtering process against its effectiveness when used without filtering.

To do so 90 locations were randomly divided into two equal groups.

Group A locations used the water softener with filtering.

Group B locations used the water softener without filtering.

At the end of three months, a water sample was taken of each location and its level of softness was registered (Scale 1 to 5, 5 represents the softest water)

X₁: Softness of water of a location from group A

n₁= 45 locations

X[bar]₁= 2.1

S₁= 0.7

X₂: Softness of water of a location from group B

n₂= 45 locations

X[bar]₂= 1.7

S₂= 0.4

To compare the effectiveness of the softener with and without a filtering process the parameter of interest is the difference between both population means:

Parameter: μ₁ - μ₂

The point estimation of the difference between the population means is the difference of the sample means: X[bar]₁ - X[bar]₂= 2.1-1.7= 0.4

Since the objective is to test if there is any difference with or without the filtering process, the hypothesis test to make is two-tailed:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Since there is no information about the distribution of both variables, you have to apply the central limit theorem and approximate the distribution of X[bar]₁ and X[bar]₂ to normal. Once both samples mean distribution is approximate to normal you can use the statistic:

[tex]Z= \frac{(X[bar]₁ - X[bar]₂)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

[tex]Z_{H_0}= \frac{(2.1-1.7)-0}{\sqrt{\frac{0.49}{45} +\frac{0.16}{45} } } = 3.3282[/tex]

As said before, this test is two-tailed, so you will have two critical values:

Critical value 1: [tex]Z_{\alpha /2}= Z_{0.025}= -1.95[/tex]

Critical value 2: [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

The p-value of this test is also two tailed, you can calculate it as:

P(Z≥3.33) + P(Z≤3.33)= (1 - P(Z≤3.33))+P(Z≤-3.33)= (1-0.999566)+0.000434= 0.000868

p-value: 0.000868

This value means that 0.0868% of the sample size 45 taken from this population will provide natural evidence that there is no difference between the population means of the effectiveness of the water softener used with and without a filtering process.

A little reminder, the p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Both using the critical value method and the p-value method the decision is to reject the null hypothesis. This means that with a 5% level of significance there is a difference between the true average of the effectiveness of the water softener used with a filtering process and the true average effectiveness of the water softener used without a filtering process.

I hope it helps!

Answer:

1. 27x+18  =  x+x+x+x+x......+x  + 18

You sum "x" 27 times.

2. [tex](36,\infty)[/tex]

3. [tex]285/2 = 142.4[/tex]

4. 2*(12  1/3 )*(8  1/2)  + 2*(12  1/3 )*(6  3/4)+2*(6  3/4 )*(8  1/2)

Step-by-step explanation:

1. Remember that multiplication is a simplification of the sum, so, when you say for example, 4*3, that actually means 3+3+3+3, similarly, when you say, 27x, that means x+x+x...+x  27 times.

2. From the image you can see that  

The 36 is NOT taken, and then you go all the way to infinity, therefore we say [tex](36,\infty)[/tex].  Suppose that 36 was taken, then we would say [tex][36,\infty)[/tex].

3. From the attached photo

you can see that we can compute first the area of the rectangle with length = 15 and height = 7, and also note that at the top a triangle with base 15 and height 5 is formed, so the area of the whole figure would be the area of the rectangle at the bottom plus the area of the triangle on top. That would be 7*15+(15*5)/2 = 285/2

4. Remember that in general the formula for surface area would be

                                              [tex]2lw +2lh+2wh[/tex]

Where   l = length   ,  w = wide,   h = height.  In this case  l = 12  1/3   , w = 8  1/2   and h =  6  3/4

Answer: $4,184.14

Step-by-step explanation:

If a car depreciates, it mean the the car is losing it's marketable worth and sometimes at a constant rate. The worth of the car after some years does not remain the same.

The formula for this depreciation when at a constant rate is denoted as:

D = P × [1 - r/100]^n

Where:

D=the Depreciated value of the car after n period which is what is being determined.

P = initial value of the car before depreciation is considered and in this case, P = $12,000

r = constant Rate Of depreciation and in this case = 10%

n = period being considered, which in this case, n = 10years.

Hence,

D = 12,000 × [1 - 10/100]^10

D = $4,184.14

Answer:

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Because by definition:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] E(\bar X) = \mu[/tex]

[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]

And for this case we have this:

[tex] \mu_{\bar X}= \mu = 100[/tex]

[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]

Step-by-step explanation:

Previous concepts

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

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Because by definition:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] E(\bar X) = \mu[/tex]

[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]

And for this case we have this:

[tex] \mu_{\bar X}= \mu = 100[/tex]

[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]

Answer: The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596

Step-by-step explanation:

Since the weights of catfish are assumed to be normally distributed,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = weights of catfish.

µ = mean weight

σ = standard deviation

From the information given,

µ = 3.2 pounds

σ = 0.8 pound

The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as

P(x ≤ 3 ≤ 5.4)

For x = 3

z = (3 - 3.2)/0.8 = - 0.25

Looking at the normal distribution table, the probability corresponding to the z score is 0.401

For x = 5.4

z = (5.4 - 3.2)/0.8 = 2.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.997

Therefore,.

P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596

To determine the probability that a catfish will weigh between 3 and 5.4 pounds, we calculate the z-scores for these weights and find the corresponding probabilities. The probability is approximately 0.5957 or 59.57%.

The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds can be calculated using the standard normal distribution.

First, we convert the weights to z-scores using the formula:

Z = (X - μ) / σ

For 3 pounds:

Z = (3 - 3.2) / 0.8 = -0.25

For 5.4 pounds:

Z = (5.4 - 3.2) / 0.8 = 2.75

Next, we look up these z-scores on the z-table or use a calculator with normal distribution functions to find the probabilities.

P(Z < 2.75) = 0.9970 (rounded to four decimal places)

P(Z < -0.25) = 0.4013 (rounded to four decimal places)

Then we find the difference to determine the probability of a catfish weighing between these two values:

Probability = P(Z < 2.75) - P(Z < -0.25)

Probability = 0.9970 - 0.4013 = 0.5957

The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is approximately 0.5957 or 59.57%.

Answer:

a) 15.87% probability that a single car of this model fails to meet the NOX requirement.

b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

Step-by-step explanation:

We use the normal probability distribution and the central limit theorem to solve this question.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 0.25, \sigma = 0.05[/tex]

a. What is the probability that a single car of this model fails to meet the NOX requirement?

Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So

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

[tex]Z = \frac{0.3 - 0.25}{0.05}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8417.

1 - 0.8413 = 0.1587.

15.87% probability that a single car of this model fails to meet the NOX requirement.

b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?

Now we have [tex]n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025[/tex]

The probability is 1 subtracted by the pvalue of Z when X = 0.3. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{0.3 - 0.25}{0.025}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

The probability that a single car of this model fails to meet the NOX requirement is 15.87%.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 0.25 g and a standard deviation of 0.05 g/mi

a) For > 0.3:

z = (0.3 - 0.25)/0.05 = 1

P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587

b) For > 0.3, sample size = 4

z = (0.3 - 0.25)/(0.05 ÷√4) = 2

P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228

The probability that a single car of this model fails to meet the NOX requirement is 15.87%.

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Answer: the store sold 31 1/3 pounds

Step-by-step explanation:

The store sold 15 2/3 pounds of carrots. Converting to improper fraction, it becomes 47/3 pounds.

The store sold 12 1/3 pounds of asparagus. Converting to improper fraction, it becomes 37/3 pounds.

The store sold 3 1/3 pounds of cabbage. Converting to improper fraction, it becomes 10/3 pounds.

Therefore, the total number of pounds that the store sold is

47/3 + 37/3 + 10/3 = 94/3 = 31 1/3 pounds

Answer:

34%

Step-by-step explanation:

Amount of anesthetic in the 25% solution + amount of anesthetic in the 40% solution = amount of anesthetic in the mixed solution

0.25 (40) + 0.40 (60) = x (40 + 60)

10 + 24 = 100x

x = 0.34

The concentration of the mixed solution is 34%.