Answer:
Step-by-step explanation:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 6 15 20 15 6 1
we use these for the expansion of (5x² + 2y³)⁶
1(5x²)⁶(2y³)⁰ + 6(5x²)⁵(2y³)¹ + 15(5x²)⁴(2y³)² + 20(5x²)³(2y³)³+ 15(5x²)²(2y³)⁴+ 6(5x²)¹(2y³)⁵ + 1(5x²)⁰(2y³)⁶
78125ₓ¹²+187500ₓ¹⁰ y³ +37500ₓ⁸y⁶+20000ₓ⁶y⁹+6000x⁴y¹²+960x²y¹⁵+2y¹⁸
a.)a = 6, b = 9. the coefficient of xᵃyᵇ ( 20000ₓ⁶y⁹) = 20000
b) a = 2, b = 15. the coefficient of xᵃyᵇ ( 960x²y¹⁵) = 960
c) a = 3, b = 12. the coefficient of xᵃyᵇ is not present
d) a = 12, b = 0 the coefficient of xᵃyᵇ ( 78125ₓ¹²) = 78125
e) a = 8, b = 9. the coefficient of xᵃyᵇ is not present
The coefficients of xᵃyᵇ for respective given values of a and b have been provided below.
We are given the expression;
(5x² + 2y³)⁶
Using online binomial expansion calculator gives us;
15625x¹² + 37500x¹⁰y³ + 37500x⁸y⁶ + 20000x⁶y⁹ + 6000x⁴y¹² + 960x²y¹⁵ + 64y¹⁸
We want to find the coefficient of xᵃyᵇ in the binomial expansion;
1) When a = 6 and b = 9, the coefficient is 20000
2) When a = 2, b = 15; the coefficient is 960
3) When a = 3 and b = 12; there is no coefficient
4) When a = 12 and b = 0; the coefficient is 15625
5) When a = 8 and b = 9; there is no coefficient.
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Please help!! It’s due @ midnight
How much should Marc deposit weekly into an account at 8% compounded weekly in order to have
$4500 available for a round trip plane ticket, hotel, and spending money for a trip to Sweden in 2 years?
Please give step by step!
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
1. Write an equivalent expression for 27x+18
2. Write the inequality this number line represents
3.erin is going to paint a wall in her house she needs to find the area of the wall so she knows how much paint to purchase what is the area of her wall
4.walt received a package that is 12 1/3 inches long 6 3/4 inches high and 8 1/2 inches wide what is the surface area of the package
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
The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and standard deviation 0.05 g/mi. government regulations call for NOX emissions no higher than 0.3 g/mi.
a. What is the 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?
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%.
What is z score?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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It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains a sample of 16 students’ IQ scores from a statistics class at UT. What is the shape of this sampling distribution?
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]
The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leaves on time in 0.24. What is the probability that the train arrives on time, given that it leaves on time?
Answer:
0.9524
Step-by-step explanation:
Note enough information is given in this problem. I will do a similar problem like this. The problem is:
The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?
Solution:
This is conditional probability.
Given:
Probability train arrive on time and leave on time = 0.8 Probability train arrive on time = 0.84 Probability train leave on time = 0.86Now, according to conditional probability formula, we can write:
[tex]P(Leave \ on \ time | arrive \ on \ time)[/tex] = P(arrive ∩ leave) / P(arrive)
Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"
and
P(arrive) means probability arriving on time given as 0.84, so:
0.8/0.84 = 0.9524
This is the answer.
Two hikers come to a ravine and want to know how wide it is. They set up two similar triangles as shown in the diagram. How far is it across the ravine?
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.
An investment earns 13% the first year, earns 20% the second year, and loses 15% the third year. The total compound return over the 3 years was ______.
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%
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. What is the probability that Team A wins and Team B loses
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.
For more details refer to the link given below.
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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%.
The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. 2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8 Find a 98% confidence interval for the true mean.
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
A poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life. What is the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life. Report your answer as a decimal accurate to 3 decimal places.
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 .
In human resource management, performance of employees is measured as a numerical score which is assumed to be normally distributed. The mean score is 150 and the standard deviation 13. What is the probability that a randomly selected employee will have a score less than 120?
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]
A study on educational aspirations of high school students (Crysdale, Int. J. Comp. Sociol., 16, 19-36, 1975) measured aspirations using the sale (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 the family income was middle, the counts were (11, 52, 23, 22); when the family income was high, the counts were (9, 41, 12, 27). a. Use SAS/R to test whether the aspirations and family income are independent, reporting both the X2 and G2 statistics. b. No matter your answer in part a, do the standardized residuals suggest any interesting patterns? c. Using SAS/R, conduct a more powerful test than those in part a
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.
Explanation: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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A store sells 15 2/3 pounds of carrots, 12 1/3 pounds of asparagus, and 3 1/3 of cabbage. How many pounds did the store sell altogether?
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
Convert the measurement as indicated.
73 inches = ? ft ?in
Answer:
6.08333
Step-by-step explanation:
6ft 0.8inches
Answer:
6 ft 1 in
Step-by-step explanation:
Absorption rates are important considerations in the creation of a generic version of a brand-name drug. A pharmaceutical company wants to test if the absorption rate of a new generic drug (G) is the same as its brand-name counterpart (B).They run a small experiment to test H subscript 0 : space mu subscript G minus mu subscript B equals 0 against the alternative H subscript A : space mu subscript G minus mu subscript B not equal to 0 . Which of the following is a Type I error?a. Deciding that the absorption rates are different, when in fact they are not.b. The researcher cannot make a Type I error, since he has run an experiment.c. Deciding that the absorption rates are different, when in fact they are.d. Deciding that the absorption rates are the same, when in fact they are.e. Deciding that the absorption rates are the same, when in fact they are not.
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.
Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is γ = 3.5 and that the excess X − 3.5 over the minimum has a Weibull distribution with parameters α = 2 and β = 1.5.
(a) What is the cdf of X?
F(x) = 0 x < 3.5
1−e^−((x−3.5)2.5)2 x ≥ 3.5
(b) What are the expected return time and variance of return time? [Hint: First obtain
E(X − 3.5)
and
V(X − 3.5).]
(Round your answers to three decimal places.)
E(X) = 10^−1 weeks
V(X) = (10^−1 weeks)2
(c) Compute
P(X > 6).
(Round your answer to four decimal places.)
Answer:
Step-by-step explanation: see attachment for solution
If X is the time in 10 to 1 week of the shipment of a defective product until the customer returns the product.
Now suppose the mini returns y = 3.5 and the excess X is 3.5 over the mini has a Weibull distribution with parameters Alfa 2 and Beta 1.5.The shipment of the defective product will be
A. 1-e x(25)2. B. 5/4 .Expected value is E (X) = 5.7125.
C. P(X>5) = 0.3679.Learn more about the shipment of a defective product
brainly.com/question/17125592.
Some scientists believe that the average surface temperature of the world has been rising steadily. The average surface temperature can be modeled by T = 0.02t + 15.0. where T is temperature in ∘C and t is years since 1950.
(a) What do the slope and T-intercept represent?
(b) Use the equation to predict the average global surface temperature in 2050.
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.
Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.
Function Point
w = y3 − 9x2y
x = es, y = et
s = −5, t = 10
Evaluate each partial derivative at the given values of s and t.
Answer:
The value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex]The value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Step-by-step explanation:
Given that the Function point are [tex]w=y^3-9x^2y[/tex]
[tex]x=e^s[/tex], [tex]y=e^t[/tex] and s = -5, t = 10
To find [tex]\frac{\partial w}{\partial s}[/tex] and [tex]\frac{\partial w}{\partial t}[/tex]using the appropriate Chain Rule :[tex]w=y^3-9x^2y[/tex]
Substitute the values of x and y in the above equation we get
[tex]w=(e^t)^3-9(e^s)^2(e^t)[/tex]
[tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to s by using chain rule we have[tex]\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)[/tex]
[tex]=e^{3t}-18e^{2s}.(e^t)[/tex]
[tex]=e^{3t}-18e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex][tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to t by using chain rule we have[tex]\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)[/tex]
[tex]=3e^{3t}-9e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :
[tex]\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}[/tex]
[tex]=e^{3(10}-18e^{2(-5)+10}[/tex]
[tex]=e^{30}-18e^{-10+10}[/tex]
[tex]=e^{30}-18e^0[/tex]
[tex]=e^{30}-18[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex][tex]\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}[/tex]
[tex]=3e^{3(10)}-9e^{2(-5)+10}[/tex]
[tex]=3e^{30}-9e{-10+10}[/tex]
[tex]=3e^{30}-9e{0}[/tex]
[tex]=3e^{30}-9[/tex]
[tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Using the Chain Rule, it is found that:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18[/tex]
[tex]\frac{\partial W}{\partial s}(-5,10) = 3e^{30} - 9[/tex]
w is a function of x and y, which are functions of s and t, thus, by the Chain Rule:
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
Then, the derivatives are:
[tex]\frac{\partial W}{\partial x} = -18xy[/tex]
[tex]\frac{\partial x}{\partial s} = e^s[/tex]
[tex]\frac{\partial W}{\partial y} = 3y^2 - 9x^2[/tex]
[tex]\frac{\partial y}{\partial s} = 0[/tex]
Then
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial s} = -18xy(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^se^t(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s}e^t[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s + t}[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18e^{2(-5) + 10} = -18[/tex]
Then, relative to t:
[tex]\frac{\partial x}{\partial t} = 0[/tex]
[tex]\frac{\partial y}{\partial t} = e^t[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
[tex]\frac{\partial W}{\partial t} = (3y^2 - 9x^2)e^t[/tex]
[tex]\frac{\partial W}{\partial t} = (3e^{2t} - 9e^{2s})e^t[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = (3e^{20} - 9e^{-10})e^{10} = 3e^{30} - 9[/tex]
A similar problem is given at https://brainly.com/question/10309252
16. A car depreciates in value at a rate of 10%. The car currently has a value of $12,000.
What will be its value in 10 years?
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
Determine the original set of data. 1 0 1 5 2 1 4 4 7 9 3 3 5 5 5 7 9 4 0 1 Legend: 1|0 represents 10The originat set of the data is?
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
Learn ore about stem Leaf graph here:
https://brainly.com/question/31998860
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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:
10 12 4 7 11 4 3 10 010 4 14 11 13 2 4 612 6 9 105 13 410 14 12 116 10 11 0 11 13 2In a study of the effects of acid rain, a random sample of 100 trees from a particular forest is examined. Forty percent of these show some signs of damage. Which of the following statements is correct?
a.None of the above
b. The sampling distribution of the proportion of damaged trees is approximately Normal
c.This is a comparative experiment.
d.If a sample of 1000 trees was examined, the variability of the sample proportion would be larger than in a sample of 100 trees
e.If we took another random sample of trees, we would find that 40% of these would show some signs of damage
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%.
If one of the 1008 subjects is randomly selected, find the probability that the person chosen is a woman given that the person is a light smoker. Round to the nearest thousandth.
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]
Researchers wanted to compare the effectiveness of a water softener used with a filtering process with a water softener used without filtering, Ninety locations were randomly divided into two groups of equal size. Group A locations used a water softener and the filtering process, while group B used only the water softener. At the end of three months, a water sample was tested at each location for its level of softness. (Water softness was measured on a scale of 1 to 5, with 5 being the softest water.) The results were as follows. x1-2.1 s1-0.7 x2-1.7 82 0.4 State the null and alternate hypothesis. Graph and shade the critical region. Find the critical value, the point estimate for the difference in population means given by these samples, and it's test statistic. Label these values and areas on your graph above. Find and explain the meaning of the P-value. Shade a graph showing the area equal to the p-value. Clearly state your initial and final conclusion
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!
Simplify.
25
−
40
A) 3
10
B) 10
10
C) 5 − 2
10
D) 2
10
− 5
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
To prepare for surgery, Anne mixes an anesthetic solution using two different concentrations: 40 mL of 25% solution and 60 mL of 40% solution.
What is the concentration of the mixed solution?
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%.
The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is ________.
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%.
Explanation: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%.
Compute the area of triangle, if x equals 4 more than 6. A) 10 B) 50 C) 100 D) 200 E) 400
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:
Solve using normalcdf
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.
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. For a certain football team, the probability that this play is successful is 0.80.1. Let X = 1 if successful, X = 0 if not. Find the mean and the variance of X. Round the answers to two decimal places.The mean of X is = .The variance of X is =2. Let Y be the number of points scored. Find the mean and variance of Y. Round the answers to two decimal places.The mean of Y is =The variance of Y is =
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.
Explanation: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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A study of Machiavellian traits in lawyers was performed. Machiavellian describes negative character traits such as manipulation, cunning, duplicity, deception, and bad faith. A Mach rating score was determined for each in a sample of lawyers. The lawyers were then classified as having high comma moderate comma or low Mach rating scores. The researcher investigated the impact of both Mach score classification and gender on the average income of a lawyer. For this experiment, identify Bold a. the experimental unit, Bold b. the response variable, Bold c. the factors, Bold d. the levels of each factor, and Bold e. the treatments.
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.