Answer:
[tex]z=\frac{0.427 -0.5}{\sqrt{\frac{0.5(1-0.5)}{110}}}=-1.531[/tex]
[tex]p_v =P(z<-1.531)=0.0629[/tex]
If we compare the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of tails is significantly less than 0.5.
Step-by-step explanation:
1) Data given and notation
n=110 represent the random sample taken
X=47 represent the number of tails obtained
[tex]\hat p=\frac{47}{110}=0.427[/tex] estimated proportion of tails
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion of tails is lower than 0.5:
Null hypothesis:[tex]p\geq 0.5[/tex]
Alternative hypothesis:[tex]p < 0.5[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.427 -0.5}{\sqrt{\frac{0.5(1-0.5)}{110}}}=-1.531[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-1.531)=0.0629[/tex]
If we compare the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of tails is significantly less than 0.5.
When women take birth control pills, some of the hormones found in the pills eventually make their way into lakes and waterways. In one study, a water sample was taken from various lakes. The data indicate that as the concentration of estrogen in the lake water goes up, the fertility level of fish in the lake goes down. The estrogen level is measured in parts per trillion (ppt) and the fertility level is recorded as the percent of eggs fertilized.a. What are the cases in this study?b. How many variables are mentioned in the description?c. What are the variables?d. Classify each variable as either qualitative or quantitative.
Answer:
There are two variables in the description, the estrogen level and the fertility level. Both are continuous variable.
Step-by-step explanation:
We are given the following in the question:
A water sample was taken from various lakes. The data indicate that as the concentration of estrogen in the lake water goes up, the fertility level of fish in the lake goes down.
The estrogen level is measured in parts per trillion (ppt) and the fertility level is recorded as the percent of eggs fertilized.
a) Case in study
The case in study is to find the effect on estrogen level on fertility level in fish.
As the estrogen level increases, the fertility level in fish decreases.
b) Variables in description.
There are two variables.
The estrogen levelFertility leveld) Types of variable
The estrogen level is measured in parts per trillion (ppt) and the fertility level is recorded as the percent of eggs fertilized. thus, both are expressed in numerical values. Thus, they are a quantitative variables.
Also, both the estrogen level and fertility level are measured and not counted. Both can take any value within an interval and can be expressed in decimals. Thus, they bot are continuous variable.
Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. nbsp 90 % confidence; the sample size is 3200 comma of which 15 % are successes 90% confidence; the sample size is 3200, of which 15% are successes The margin of error Eequals=nothing. (Round to four decimal places as needed.)
Answer: The margin of error E = 0.0104
Step-by-step explanation:
The formula to find the margin of error that corresponds to the given statistics and confidence level for population proportion is given by :-
[tex]E=z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where
n= Sample size
[tex]\hat{p}[/tex] = Sample proportion
z* = critical value.
As per given , we have
n= 3200
[tex]\hat{p}=0.15[/tex]
Confidence level : 90%
The critical z-value for 90% confidence is z* = =1.645[By z-table]
Substitute all values in the formula , we get
[tex]E=(1.645)\sqrt{\dfrac{0.15(1-0.15)}{3200}}[/tex]
[tex]E=(1.645)\sqrt{0.00003984375}[/tex]
[tex]E=(1.645)(0.00631219058648)=0.0103835535148\approx0.0104[/tex]
Hence, the margin of error E = 0.0104
Human blood pressure levels are normally distributed. If you measured an individual's blood pressure and found the blood pressure level to have a z-score of 2.1, what would you conclude about that person?
a. The individual's blood pressure is unusually high, compared to others.
b. The individual's blood pressure is 2.1 times higher than the average person.
c. The individual's blood pressure level is about average, compared to others.
d. Since the z-score is positive we know that the individual has normal blood pressure, compared to others.
Answer:
a. The individual's blood pressure is unusually high, compared to others
Step-by-step explanation:
The individual's blood pressure is unusually high, compared to others. Since the z- score is positive, it implies that the individual's blood pressure is higher than the average blood pressure of others. And also given that the z-score is as high as 2.1 (higher than 95% confidence interval which is 1.96) implies that the individual's blood pressure is extremely higher than the average blood pressure of others.
in a class of 26 students, 15 of them like maths, 13 of them like english and 9 of them like neither. find the probability that a student chosen at random likes english but not maths.
venn diagram doesn't need to be completed,, some working out would help because this is exam revision. thanks!
Answer:
0.0769
Step-by-step explanation:
Let x be the number of student that offer both subjects
15 - x + x + 13 - x + 9 = 26
-x + 37 = 26
-x = -11
x = 11
Number of student that offer english but not math = 13 - 11
= 2
The probability of english but not math = 2/26
The probability that a student is chosen at random likes English but not maths is [tex]\dfrac{1}{13}[/tex].
Given information:
In a class of 26 students, 15 of them like maths, 13 of them like English and 9 of them like neither.
Now, the number of students who like either maths or English will be,
[tex]26-9=17[/tex]
Now, out of 17, 15 students like maths and 13 students like English.
So, the number of students who like both the subjects will be,
[tex]E\cap M=13+15-17\\=11[/tex]
Now, 11 students like both the subjects and 13 students like English.
So, the number of students who like English but not maths will be,
[tex]13-11=2[/tex]
Thus, the probability that a student chosen at random likes English but not maths will be calculated as,
[tex]P=\dfrac{2}{26}\\=\dfrac{1}{13}[/tex]
Therefore, the probability that a student is chosen at random likes English but not maths is [tex]\dfrac{1}{13}[/tex].
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When a new charter school opened in 1992, there were 440 students enrolled. Write a formula for the function N(t), representing the number of students
Answer:
N(t) = m(X)+c
Step-by-step explanation:
Here, M is slope, also known as gradient, while X is the number which is variable and can keep on changing according to the number of increase, decrease or any other modification to the number of children in the school.
An finally, C is the constant which remains the same whatever the variable is. Therefore C would be equal to 440.
I hope this helps you.
What was the annual growth rate for Egypt in 1990? What is the estimated annual growth rate in 2010? Be sure to show your work for each answer.
Answer:
A. The annual growth rate in Egypt in 1990 was 31 - 9 = 22 per 1,000 people. The estimated annual growth rate in 2010 is 25 - 5 = 20 per 1,000 people.
B. I feel like Egypt is in the third stage (Late Developing) in the Demographic transition model because the birth and death rate was higher but then as time progressed, they both started to level out. In order for Egypt to advance in the model, they will have to have a developed country (urbanization) and lower their birth/death rates through the use of birth controls and healthcare.
Step-by-step explanation:
Consider the linear system 211 + 3x2 - 5.23 = b 7.01 + 2.02 + 813 = b2 -X1 + 12 - 5.23 = b3
(a) Find the echelon form of the augmented matrix of the above system.
(b) Find the conditions on b1,b2, b3 for which this system has a solution.
(c) Do you see the shape of the points (61, 62, 63) for which the above system has a solution?
(d) If you randomly picked a (61, 62, 63) in R3, do you expect the above system to have a solution?
Answer:
The answers are shown in the step by step explanation that is attached
Step-by-step explanation:
The step by step calculation is as shown in the attachment below
What is the forecast for this year using exponential smoothing with trend if alpha=0.5 and alpha=0.3?
Assume the forecast for last year was 21,000 and the forecast for two years ago was 19,000, and that the trend estimate for last years forecast was 1,500.
a)18,750
b)19,500
c)21,000
d)22,650
e)22,800
Answer:
i. the forecast for this year using alpha = 0.5 is 11250
ii. using alpha = 0.3 is 7350
Step-by-step explanation:
using exponential smoothing,the formula is given as: F(t+1) =αAt + (1-α)Ft, where F(t+1) is the new forecast or required forecast, α is the alpha, At is the each date or observation and Ft is the current trend.
i. using alpha= 0.5, the year forecast = (0.5 x 21000) + (1-0.5) x 1500 = 11250
ii. using alpha = 0.3,the year forecast = (0.3 x 21000) + (1-0.3) x 1500 = 7350.
this year forecast using alpha = 0.5 and 0.3 are 11250 and 7350 respectively
World grain production was 1241 million tons in 1975 and 2048 million tons in 2005, and has been increasing at an approximately constant rate.
Answer:
The equation of line is [tex]P(t)=26.9t+1241[/tex]
Step-by-step explanation:
Consider the provided information.
World grain production was 1241 million tons in 1975 and 2048 million tons in 2005,
We need to find the linear function.
The difference of the year is: 2005-1975=30
The function represents the world grain production at time t years after 1975, Thus, the first points is (0,1241) and the second points is (30,2048)
Find the slope of the line by using the formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Substitute the respective values in the above formula.
[tex]m=\dfrac{2048-1241}{30-0}\\\\m=\dfrac{807}{30}\\\\m=26.9[/tex]
The slope of the linear function is 26.9
The slope intercept form is: [tex]y=mx+b[/tex]
Where b is the y intercept. y intercept is a point where the value of x is 0.
Therefore y intercept of the linear function is 1241 because the first points is (0,1241).
Hence, the equation of line is [tex]P(t)=26.9t+1241[/tex]
Given the data, the approximated rate of increase in world grain production from 1975 to 2005 is about 32.28 million tons per year. Increases might be driven by population growth and rising food demands. Future food supply forecasts point to significant increases by 2050.
Explanation:The subject of your question appears to be related to linear growth over time, specifically in the context of world grain production. Given the data provided, we can calculate the approximate rate of increase in grain production by taking the difference in amounts (2048 million tons in 2005 minus 1241 million tons in 1975) and divide it by the difference in years (2005 minus 1975). This calculation results in approximately 32.28 million tons increase per year.
It's also noteworthy to consider that grain production increases might be driven by the increasing global population and rising food demands. Additionally, future projections for food supply such as milk and meat production, estimated to face significant increases by 2050, might also apply to grain production, further emphasizing the importance of understanding and managing these growth rates for global food security.
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In a trend line based on five observations if the average of Y is 100 and the slope of line is 22, then the intercept is?
a)32
b)34
c)36
d)38
The value of intercept will be 34.
What is Equation of line?The equation of line in point-slope form passing through the points
(x₁ , y₁) and (x₂, y₂) with slope m is defined as;
⇒ y - y₁ = m (x - x₁)
Where, m = (y₂ - y₁) / (x₂ - x₁)
Given that;
In a trend line based on five observations,
The average of Y = 100
And, The slope of line = 22
Now,
Since, The equation of line is,
⇒ Y = mx + c
Where, m is slope and c is y - intercept.
When x = 1;
⇒ Y = m + c
When x = 2;
⇒ Y = 2m + c
When x = 3;
⇒ Y = 3m + c
When x = 4;
⇒ Y = 4m + c
When x = 5;
⇒ Y = 5m + c
Here, The average of Y is 100.
So, We get;
⇒ (m + c) + (2m + c) + (3m + c) + (4m + c) + (5m + c) / 5 = 100
⇒ 15m + 5c / 5 = 100
⇒ 3m + c = 100
Substitute m = 22;
⇒ 3 × 22 + c = 100
⇒ 66 + c = 100
⇒ c = 100 - 66
⇒ c = 34
Thus, The value of intercept is,
⇒ c = 34
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A consumer advocate claims that 80 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue. (a) Suppose that the advocate's claim is true, and suppose that a random sample of 7 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 5 or more subscribers in the sample are not satisfied with their service. (Do not round intermediate calculations. Round final answer to p in 2 decimal place. Round other final answers to 4 decimal places.) Binomial, n
Answer: 0.8520
Step-by-step explanation:
Given : The probability that cable television subscribers are not satisfied with their cable service is 80%=0.80.
We assume that each subscriber is independent from each other, so we can apply Binomial distribution.
In binomial distribution, the probability of getting success in x trials is given by :-
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]
, where n is the total number of trials , p is the probability of getting success in each trial .
Let x be the number of subscribers in the sample are not satisfied with their service..
So, p=0.8
Sample size : n=7
The probability that 5 or more subscribers in the sample are not satisfied with their service will be :-
[tex]P(x\geq5)=P(5)+P(6)+P(7)\\\\=^7C_5(0.8)^5(0.2)^2+^7C_6(0.8)^6(0.2)^1+^7C_7(0.8)^7(0.2)^0\\\\=\dfrac{7!}{5!(7-5)!}(0.0131072)+(7)(0.0524288)+(1)(0.2097152)\ \[\because\ ^nc_r=\dfrac{n!}{r!(n-r)!}]\\\\=0.2752512+0.3670016+0.2097152\\\\=0.851968\approx0.8520[/tex]
Hence, the probability that 5 or more subscribers in the sample are not satisfied with their service is 0.8520 .
Final answer:
The detailed answer explains how to calculate the probability of 5 or more subscribers not satisfied out of a sample of 7 using the binomial formula.
Explanation:
Binomial Probability Calculation:
Given:
Probability of dissatisfaction (p) = 0.80
Sample size (n) = 7
Calculate the probability that 5 or more subscribers are not satisfied using the binomial formula: P(X >= 5) = 1 - P(X < 5)
Use appropriate formula:
P(X < 5) = (7C0 * (0.80)^0 * (0.20)^7) + (7C1 * (0.80)^1 * (0.20)^6) + (7C2 * (0.80)^2 * (0.20)^5) + (7C3 * (0.80)^3 * (0.20)^4) + (7C4 * (0.80)^4 * (0.20)^3)
On a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.
Answer:
In 80% of the tests the number of points scored was less than 49 and in 20% of the tests the number of points scored was higher than 49.
Step-by-step explanation:
When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.
On a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.
The interpretation is that in 80% of the tests the number of points scored was less than 49 and in 20% of the tests the number of points scored was higher than 49.
Final answer:
The 80th percentile indicates a student scored higher than 80 percent of peers, with a score of 49 out of 60 on the assignment.
Explanation:
The 80th percentile for a 60-point written assignment, where the score was 49, means that 80 percent of the students earned 49 points or less on the assignment. Conversely, it also means that 20 percent of the students earned more than 49 points. Therefore, a student who scored 49 points on this assignment performed better than 80 percent of their peers.
Which of the following measures of dispersion are based on deviations from the mean?
A. Variance
B. Standard deviation
C. Mean deviation
D. All of the choices are correct.
E. None of the choices are correct
Answer:
D.
Step-by-step explanation:
The mean deviation is the measure of dispersion used to evaluate the spread of the data calculated by taking deviation from mean. The mean deviation formula is
[tex]M.D=\frac{sum|x-xbar|}{n}[/tex]
|x-xbar| are known as absolution deviations.
So, the mean deviation is the measures of dispersion that is based on deviations from the mean.
Standard deviation is also computed by computing mean deviation first i.e.
[tex]s=\sqrt\frac{sum(x-xbar)^2}{n-1}[/tex]
Variance is also computed by mean deviation first
[tex]variance=s^2=\frac{sum(x-xbar)^2}{n-1}[/tex]
Note: All formula for sample are considered and formulas for population also results in the same conclusion.
Hence, variance, standard deviation and mean deviation all are based on deviation from mean.
The measures of dispersion that are based on deviations from the mean are variance, standard deviation, and mean deviation.
Explanation:The measures of dispersion that are based on deviations from the mean are variance, standard deviation, and mean deviation.
Therefore, the correct answer is D. All of the choices are correct.
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A pomegranate is thrown from ground level straight up into the air at time t=0 with velocity 32 feet per second. Its height at time t seconds is f(t)=−16t2+32t. Find the time tg it hits the ground and the time th it reaches its highest point. What is the maximum height h? Enter the exact answers.
Answer:
It hit the ground at t=2
It reaches its highest point at t=1
Its maximum height =16
Step-by-step explanation:
The graph of the function of the height will be a parapola opening downward where it's x intercepts = the time where the object hit the ground and its maximum point = the point with the maximum point
The x intercepts are 0 when the object was thrown and 2 when it landed
The x coordinate of the vertex is in the middle of the two x intercepts so it will be 1
Substituting the x coordinate of the vertex gives us the y coordinate of the vertex which is the maximum hight 16
The pomegranate hits the ground at time t=2 seconds (tg=2) and reaches its maximum height at time t=1 second (th=1). The maximum height (h) is 16 feet.
Explanation:The given function f(t) = -16t2 + 32t represents the height of the pomegranate at any time t. The pomegranate hits the ground when its height is zero, so set f(t) = 0 and solve for t.
0 = -16t2 + 32t
0 = t(-16t + 32)
Thus, t = 0 (the initial time) and t = 2 (when it hits the ground). Therefore, tg = 2.
The pomegranate reaches its maximum height when the derivative of f(t) is zero. This will give us the time th.
f'(t) = -32t + 32
Set f'(t) = 0 and solve for t. Then -32t + 32 = 0 implies t = 1. So, th = 1.
The maximum height h is simply f(th) = -16(1)2 + 32(1) = 16 feet.
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Determine the average rate of change over the interval -1 < x < 1 of the exponential function whose graph is shown below
A. -6
B. -3
C. 3
D. 6
Answer:
The correct answer is 3
Step-by-step explanation:
i just took the lesson
Answer:
the correct answer is B. -3
Step-by-step explanation:
i actually just did this in my class a week ago and i had trouble with it but then learned it and understood!! good luck!
Find the intersection of the line through (0, 1) and (4.3, 2) and the line through (2.1, 3) and (5.3, 0).
Answer: The intersection of the line through (0, 1) and (4.3, 2) and the line through (2.1, 3) and (5.3, 0) is (3.392, 1.789).
Step-by-step explanation:
We know that the equation of a line that passes through two points (a,b) and (c,d) is given by :-
[tex](y-b)=\dfrac{d-b}{c-a}(x-a)[/tex]
Similarly , the equation of line that passes through (0, 1) and (4.3, 2) would be:
[tex](y-1)=\dfrac{2-1}{4.3-0}(x-0)[/tex]
[tex](y-1)=\dfrac{1}{4.3-}(x)[/tex]
[tex]4.3(y-1)=x[/tex]
[tex]4.3y-4.3=x-----(1)[/tex]
Equation of line that passes through (2.1, 3) and (5.3, 0) would be:
[tex](y-0)=\dfrac{3-0}{2.1-5.3}(x-5.3)[/tex]
[tex]y=\dfrac{3(x-5.3)}{-3.2}-----(2)[/tex]
To find the intersection point (x,y) , we substitute the value of y from (2)in (1) , we get
[tex]4.3(\dfrac{3(x-5.3)}{-3.2})-4.3=x[/tex]
[tex]-4.03125(x-5.3)-4.3=x[/tex]
[tex]-4.03125x+21.365625-4.3=x[/tex]
[tex]-4.03125x+17.065625=x[/tex]
[tex]x+4.03125x=17.065625[/tex]
[tex]5.03125x=17.065625[/tex]
[tex]x=\dfrac{17.065625}{5.03125}\approx3.392[/tex]
Put value of x in (2) , we get
[tex]y=\dfrac{3(3.392-5.3)}{-3.2}[/tex]
[tex]y=\dfrac{3(-1.908)}{-3.2}\approx1.789[/tex]
Hence, the intersection of the line through (0, 1) and (4.3, 2) and the line through (2.1, 3) and (5.3, 0) is (3.392, 1.789).
This solution involves finding the equations of the two lines using the slope and y-intercept, setting the equations equal to each other to find the x-coordinate of the intersection, and substituting the x-value into one of the equations to find the corresponding y-coordinate of the intersection.
Explanation:To find the intersection of the two lines, first we need to find the equations of these lines. We can use the formula y = mx + c, where m is the slope and c is the y-intercept.
For the line passing through (0, 1) and (4.3, 2), we find the slope (m) first: m = (2-1) / (4.3-0) = 1/4.3. The line passes through the y-axis at (0,1), so c = 1. Thus, the equation is y = (1/4.3)x + 1.
For the line passing through (2.1, 3) and (5.3, 0), the slope m = (0-3) / (5.3-2.1) = -3/3.2. This line does not pass through the y-axis, so c is not 0. Substituting one of the points into the equation y = mx + c, we can find c: 3 = -3/3.2*2.1 + c, yields c = 3.984375. So, the equation is y = -3/3.2*x + 3.984375.
Now, we set these two equations equal to each other and solve for x: (1/4.3)x + 1 = -3/3.2*x + 3.984375. This will yield the x-coordinate of the intersection. Substitute the x-coordinate into any of the line equations to get the y-coordinate. These coordinates give us the intersection point of the two lines.
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Find the value of 15.0 NN in pounds. Use the conversions 1slug=14.59kg1slug=14.59kg and 1ft=0.3048m1ft=0.3048m.
Express your answer in pounds to three significant figures.
3.37 lb
Step-by-step explanation:
The question requires you to convert weight in Newtons to weight in pounds.
Given 15.0 N to convert to pounds, remember the conversion rate where;
1 Newton = 0.224809 pound-force
1 N= 0.224809 lb
15 N= ?
Perform cross-product
=15*0.224809
=3.37 lb
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The following are weights in pounds of a college sports team: 165, 171, 174, 180, 182, 188, 189, 192, 198, 202, 202, 225, 228, 235, 240 Find the weight that is 2 standard deviations below the mean. Round your answer to the nearest pound.
Answer:
There are no values in the data that is two standard deviations below the mean.
Step-by-step explanation:
We are given the following data set in question:
165, 171, 174, 180, 182, 188, 189, 192, 198, 202, 202, 225, 228, 235, 240
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{2971}{15} = 198.06[/tex]
The mean of sample is 198.06 pounds.
Sum of squares of differences = 7984.93
[tex]S.D = \sqrt{\dfrac{7984.93}{14}} = 23.88[/tex]
The sample standard deviation is 23.88 pounds.
We have to find the weight that is 2 standard deviations below the mean.
[tex]x < \bar{x}- 2s\\x < 198.06 -2(23.88)\\x < 150.3[/tex]
Thus, we have to find a value less than 150.3.
Sorted data: 165, 171, 174, 180, 182, 188, 189, 192, 198, 202, 202, 225, 228, 235, 240
There are no values in the data that is less than 150.3
please help and explain
Answer:
The value of x = 5.
The length of KJ = 29 units.
Step-by-step explanation:
Given L and M are the mid points of the lines.
So, LM becomes the mid segment.
Also, [tex]$ \textbf{LM} = \frac{\textbf{GH + KJ}}{\textbf{2}} $[/tex]
Here, the length of LM = 25 units.
Length of GH = 2x + 11 units.
Length of KJ = 6x - 1 units.
Therefore, we have: [tex]$ LM = \frac{2x + 11 + 6x - 1}{2} $[/tex]
= [tex]$ \frac{8x + 10}{2} $[/tex]
[tex]$ \implies 25 = 4x + 5 $[/tex]
[tex]$ \implies 4x = 20 $[/tex]
⇒ x = 5
Therefore, KJ = 6(5) - 1
= 29 units.
Hence, the answer.
Show whether the following signals are even, odd or neither. For the signals which are neither odd nor even, find and plot their even and odd components.
a. x[n] = u[n] - u[n-6] (plot the even and odd parts)
b. x[n] = n^2 cos(pi n/3)
c. x[n] = ne^-|n|
Answer:
a) it is neither even nor odd
b) it is an even signal
c) it is an odd signal
Step-by-step explanation:
A function f(x) or a signal is said to be even if its satisfies the condition of f(-x) = f(x). this implies that the graph of such a function or signal has a symmetrical relationship with respect to the y-axis.
A function f(x) or a signal is said to be odd if its satisfies the condition of f(-x) = - f(x). this implies that the graph of such a function or a signal has a skew-symmetrical relationship with respect to the y-axis.
from the first option ; a) x[n] = u[n] - u [n-6], from the conditions attached to even and odd functions, it can be inferred that the first option does not satisfy the conditions for even and odd functions hence, it is neither even nor odd.
The attachements below gives a detailed explanation of the second and third option.
Please help... I have no clue
Answer:
OPTION C: Sin C - Cos C = s - r
Step-by-step explanation:
ABC is a right angled triangle. ∠A = 90°, from the figure.
Therefore, BC = hypotenuse, say h
Now, we find the length of AB and AC.
We know that: [tex]$ \textbf{Sin A} = \frac{\textbf{opp}}{\textbf{hyp}} $[/tex]
and [tex]$ \textbf{Cos A} = \frac{\textbf{adj}}{\textbf{hyp}} $[/tex]
Given, Sin B = r and Cos B = s
⇒ [tex]$ Sin B = r = \frac{opp}{hyp} = \frac{AC}{BC} = \frac{AC}{h} $[/tex]
⇒ [tex]$ \textbf{AC} = \textbf{rh} $[/tex]
Hence, the length of the side AC = rh
Now, to compute the length of AB, we use Cos B.
[tex]$ Cos B = s = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{AB}{h} $[/tex]
⇒ [tex]$ \textbf{AB} = \textbf{sh} $[/tex]
Hence, the length of the side AB = sh
Now, we are asked to compute Sin C - Cos C.
[tex]$ Sin C = \frac{opp}{hyp} $[/tex]
⇒ [tex]$ Sin C = \frac{AB}{BC} $[/tex]
[tex]$ = \frac{sh}{h} $[/tex]
= s
Sin C = s
[tex]$ Cos C = \frac{adj}{hyp} $[/tex]
[tex]$ \implies Cos C = \frac{AC}{BC} $[/tex]
⇒ Cos C = [tex]$ \frac{rh}{h} $[/tex]
Therefore, Cos C = r
So, Sin C - Cos C = s - r, which is OPTION C and is the right answer.
The average length of the fish caught in Lake Amotan is 12.3 in. with a standard deviation of 4.1 in. Assuming normal distribution, find the probability that a fish caught there will be longer than 18in.
Answer:
0.082 is the probability that a fish caught will be longer than 18 inch.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12.3 inch
Standard Deviation, σ = 4.1 inch
We are given that the distribution of average length of the fish is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P( longer than 18 inch)
P(x > 18)
[tex]P( x > 18) = P( z > \displaystyle\frac{18 - 12.3}{4.1}) = P(z > 1.39)[/tex]
[tex]= 1 - P(z \leq 1.39)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 18) = 1-0.918 =0.082[/tex]
0.082 is the probability that a fish caught will be longer than 18 inch.
Answer: the probability that a fish caught there will be longer than 18in is 0.08226
Step-by-step explanation:
Assuming a normal distribution for length of the fishes caught in Lake Amotan, the formula for normal distribution is expressed as
z = (x - µ)/σ
Where
x = length of fishes
µ = mean length
σ = standard deviation
From the information given,
µ = 12.3 in
σ = 4.1 in
We want to find the probability that a fish caught there will be longer than 18in. It is expressed as
P(x > 18) = 1 - P(x ≤ 18)
For x = 18,
z = (18 - 12.3)/4.1 = 1.39
Looking at the normal distribution table, the probability corresponding to the z score is 0.91774
P(x > 18) = 1 - 0.91774 = 0.08226
Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly Normal distribution with mean 336 days and standard deviation 6 days. Let X X
Answer:
a) [tex] \mu -3\sigma = 336-3*6=318[/tex]
[tex] \mu+-3\sigma = 336+3*6=354[/tex]
b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%
Step-by-step explanation:
Assuming this question : "Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "
(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?
First we need to remember the concept of empirical rule.
From this case we assume that [tex] X\sim N(\mu = 336. \sigma =6)[/tex] where X represent the random variable "length of horse pregnancies from conception to birth"
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:
[tex] \mu -3\sigma = 336-3*6=318[/tex]
[tex] \mu+-3\sigma = 336+3*6=354[/tex]
(b) What percent of horse pregnancies are longer than 342 days?
For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%
The problem is applying the concept of normal distribution in statistics to describe the length of horse pregnancies, which are said to follow a normal distribution with a mean of 336 days and a standard deviation of 6 days.
Explanation:The student is being asked to deal with a problem that relates to the normal distribution concept in statistics applied to horse pregnancies. If X represents the length of horse pregnancies, it's stated that it follows a normal distribution with a mean (average) of 336 days and a standard deviation of 6 days.
The normal distribution, also known as the Gaussian or bell curve, is a function that describes the probability distribution of many kinds of data, in this case, the horse gestation period. The distribution is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
In practical terms, it means that most horses will have a gestation period near the 336 days (mean value), with few horses having gestation periods significantly shorter or longer. The standard deviation (in this case, 6 days) gives an indication of how much the gestation period is expected to vary from the mean.
Learn more about Normal Distribution here:https://brainly.com/question/34741155
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Given a set of data sorted from smallest to largest, define the first, second, and third quartiles.
a. The first quartile is the area within one standard deviation of the mean.
The second quartile is the area within two standard deviations of the mean.
The third quartile is the area within three standard deviations of the mean.
b. The first quartile is the mean of the lower half of the data below the median.
The second quartile is the median
The third quartile is the mean of the upper half of the data above the median.
c. The first quartile is the minimum value.
The second quartile is the median.
The third quartile is the maximum value.
d. The first quartile is the median of the lower half of the data below the overall median.
The second quartile is the overall median
The third quartile is the median of the upper half of the data above the overall median.
e. The first quartile is the area that contains the 25% of all values that are closest to the mean.
The second quartile is the area that contains the 50% of all values that are closest to the mean.
The third quartile is the area that contains the 75% of all values that are closest to the mean.
Answer:
Step-by-step explanation:
Given a set of data sorted from smallest to largest, i.e. arranged in ascending order we are to find out the median, I and III quartiles
We know that the median is the middle entry of data arranged in ascending order
Q1 is the entry below which 25% lie and Q3 is one above which 25% lie
Hence proper definition would be
d. The first quartile is the median of the lower half of the data below the overall median.
The second quartile is the overall median
The third quartile is the median of the upper half of the data above the overall median.
Option b is wrong becuase mean is not necessary here. Option a is wrong because this has nothing to do with std deviation
Option c is wrong since minimum value cannot be q1
Option e is wrong because we have exactly 25% lie below Q1
put these fractions in order 1/5 2/5 3/5 4/5
Step-by-step explanation:
All fractions have the same denominator = 5.
The larger the counter, the greater the fraction.
Therefore
[tex]\dfrac{1}{5}<\dfrac{2}{5}<\dfrac{3}{5}<\dfrac{4}{5}[/tex]
sampling units described in part c. Refer to the primary b and describe a sampling plan that results in a simple random sample b. When the Bureau of Labor Statistics conducts a survey, it begins by partitioning the United States adult popula- tion into 2,007 groups called primary sampling units Assume that these primary sampling units all contain the same number of adults. If you randomly select one adult from each primary sampling unit, is the result a simple random sample? Why or why not?
Since the population is first divided into various groups and the sampling was done later.
It is not the example of a simple random sampling as it does not allow for selecting more than 1 adult from each group that would have been possible in all simple random sampling.
The 2,007 adults should be selected without regard to the groups The groups are selected with no restrictions on the number chosen from every group.Learn more about the sampling units described in part.
brainly.com/question/13822477.
Suppose a certain species of fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean kilograms and standard deviation kilograms. Let x be the weight of a fawn in kilograms. Convert the following z interval to a x interval.
Answer:
[tex] z<3.95[/tex]
Step-by-step explanation:
Assuming this complete question:
"Suppose a certain species of fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean [tex]\mu =26[/tex] kilograms and standard deviation [tex]\sigma=4.2[/tex] kilograms. Let x be the weight of a fawn in kilograms. Convert the following z interval to a x interval.
[tex] X<42.6[/tex]"
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 weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(26,4.2)[/tex]
Where [tex]\mu=26[/tex] and [tex]\sigma=4.2[/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]
We know that the Z scale and the normal distribution are equivalent since the Z scales is a linear transformation of the normal distribution.
We can convert the corresponding z score for x=42.6 like this:
[tex] z=\frac{42.6-26}{4.2}=3.95[/tex]
So then the corresponding z scale would be:
[tex] z<3.95[/tex]
A die is rolled n times. what is the probability that at least 1 of the 6 values never apears?
Answer:
Zero
Step-by-step explanation:
The possible outcomes from rolling a die are 1,2,3,4,5 and 6. The probability of getting one of these numbers when the die is rolled
=1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6
=1
The probability of not getting one of the number plus the probability of getting one of them = 1
Therefore;
probability of not getting one of the number + 1 = 1
probability of not getting one of the number = 0
A company had 80 employees whose salaries are summarized in the frequency distribution below. Find the
mean salary.
Salary ($) Employees
5,001-10,000 16
10,001-15,000 14
15,001-20,000 15
20,001-25,000 17
25,001-30,000 18
Answer:
[tex] \bar X = \frac{1435040}{80}=17938[/tex]
Step-by-step explanation:
Since we have a groued data for this case we can construct the following table to find the expected value.
Interval Frequency(fi) Midpoint(xi) xi*fi
5001-10000 16 7500.5 120008
10001-15000 14 12500.5 175007
15001-20000 15 17500.5 262507.5
20001-25000 17 22500.5 382508.5
25001-30000 18 27500.5 495009
Total 80 1435040
And we can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n f_i x_i}{n}[/tex]
Where [tex] n=\sum_{i=1}^n f_i = 80[/tex]
And if we replace we got:
[tex] \bar X = \frac{1435040}{80}=17938[/tex]
Suppose a baseball player had 229 hits in a season. In the given probability distribution, the random variable x represents the number of hits the player obtained in a game. Round one decimal place. show work.
x...........0......1......2......3......4......5
P(x)..0.1712...0.4886....0.2389....0.0706.....0.0256.......0.0051
a.) Compute and interpret the mean of the random variable x
?x =
Which of the following interpretations is correct?
1.) As the number of trials n decreases, the mean of the observations will approach the mean of the random variable.
2.) As the number of trials n increases, the mean of the observations will approach the mean of the random variable.
3.) The observed value of the random variable will almost always be less than the mean of the random variable.
4.) The observed value of the random variable will almost always be equal to the mean of the random variable.
b.) Compute the standard deviation of the random variable x.
?x =
Answer:
a) summation of p(x)/n ie. (0.1712+...+0.0051)/6=0.16675
b)1
c).var=summation (x-mean) squared /n ie (0.1712-0.16675)squared +...+(0.0051-0.16675)squared/n=0.027351948
SD =square root of variance =0.16538
Step-by-step explanation: