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.
Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic readings (in mmHg) are listed below. Find the range, variance, and standard deviation for the given sample data. If the subject's blood pressure remains constant and the medical students correctly apply the same measurement technique, what should be the value of the standard deviation? 131 137 138 141 141 149131 137 138 141 141 149 Rangeequals=nothing mmHg Sample varianceequals=nothing mmHgsquared2 (Round to the nearest tenth as needed.) Sample standard deviationequals=nothing mmHg (Round to the nearest tenth as needed.) What should be the value of the standard deviation?
A. Ideally, the standard deviation would be one because all the measurements should be the same.
B. Ideally, the standard deviation would be zero because all the measurements should be the same.
C. Ideally, the standard deviation would be one because this is the lowest standard deviation that can be achieved.
D. There is no way to tell what the standard deviation should be.
Answer:
[tex] Range = 149-131=18[/tex]
[tex] s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1[/tex][tex] s =\sqrt{35.1}=5.9[/tex]
B. Ideally, the standard deviation would be zero because all the measurements should be the same.
Step-by-step explanation:
For this case we have the following data:
131 137 138 141 141 149
For this case the range is defined as [tex] Range = Max-Min[/tex]
And for our case we have [tex] Range = 149-131=18[/tex]
First we need to calculate the average given by this formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}=\frac{837}{6}=139.5[/tex]
We can calculate the sample variance with the following formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar x)^2}{n-1}[/tex]
And if we replace we got:
[tex] s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1[/tex]
And the standard deviation is just the square root of the variance so then we got:
[tex] s =\sqrt{35.1}=5.9[/tex]
If the subject's blood pressure remains constant and the medical students correctly apply the same measurement technique, what should be the value of the standard deviation?
For this case the variance and deviation should be 0 since we not evidence change then we not have variation. And for this case the best answer is:
B. Ideally, the standard deviation would be zero because all the measurements should be the same.
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!
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.
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In the context of regression analysis, what is the definition of an influential point?
a. Observed data points that are close to the other observed data points in the horizontal direction
b. Observed data points that are far from the least squares line
c. Observed data points that are far from the other observed data points in the horizontal direction
d. Observed data points that are close to the least squares line
Answer:
c. Observed data points that are far from the other observed data points in the horizontal direction
True, by definition are observed data points that are far from the other observed data points in the horizontal direction.
Step-by-step explanation:
When we conduct a regression we consider influential points by definition "an outlier that greatly affects the slope of the regression line". Based on this case we can analyze one by one the possible options:
a. Observed data points that are close to the other observed data points in the horizontal direction
False. If are close to the other observed values then are not influential points
b. Observed data points that are far from the least squares line
False, that's the definition of outlier.
c. Observed data points that are far from the other observed data points in the horizontal direction
True, by definition are observed data points that are far from the other observed data points in the horizontal direction.
d. Observed data points that are close to the least squares line
False. If are close to the fit regression line adjusted then not affect the general equation for the model and can't be considered as influential points
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 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)
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.
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.
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On Julyâ 1, a pond was 22 ft deep. Since thatâ date, the water level has dropped two thirds ft per week. For what dates will the water level not exceed 18 âft?
Answer: The date in which the water level will not exceed 18ft is August 12 ( 6 weeks after July 1)
Step-by-step explanation:
Given:
Initial water level = 22ft
Final water level = 18ft
Total water level change = 22-18 = 4ft
Rate of change of water level = 2/3 ft/week
Using the formula;
Total change = rate × time
Time = total change/rate
Substituting the values of the rate and total water level change.
Time = 4/(2/3) week = 4 × 3 ÷2 = 6weeks.
From July 1 + 6 weeks = August 12
The date in which the water level will not exceed 18ft is August 12 ( 6 weeks after July 1)
Final answer:
To find the dates when the water level will not exceed 18 ft, we set up an inequality based on the given situation and solve it to determine the range of weeks when the water level stays below 18 ft.
Explanation:
To find the dates when the water level will not exceed 18 ft, we need to set up an inequality based on the given situation.
Initial depth of the pond = 22 ftThe water level drops by two-thirds each week, so the depth after 'w' weeks = 22 - (2/3)wWe need to find 'w' when the water level is 18 ft or above: 22 - (2/3)w ≥ 18By solving the inequality, we can determine the range of weeks when the water level will not exceed 18 ft.
Determine whether the underlined numerical value is a parameter or a statistic. Explain your reasoning. Upper A poll of all 2000 students in a high school found that Modifying 94 % with underline of its students owned cell phones.
Answer:
The numerical value is a parameter.
Step-by-step explanation:
We are given the following situation in the question:
A poll of all 2000 students in a high school found that Modifying 94 % with underline of its students owned cell phones.
Individual of interest:
Students in a high school found
Variable of interest:
Percentage of students owned a cell phone
Population of interest:
All 2000 high school students.
94 % with underline of its students owned cell phones.
Since this numerical value describes all of the 2000 high school student, it is describing the population of interest. Thus, the numerical value is a parameter.
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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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
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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Suppose a biologist studying the mechanical limitations of growth among different species of tulips monitors a national preserve. He collects data on the heights of 10 different types of tulips in the reserve and rounds each height to the nearest centimeter.
25,21,26,24,29,33,29,25,19,24
Compute the first quartile (Q1), the third quartile (Q3), and the interquartile range (IQR) of the data set.
Answer:
[tex] Q_1 = 24[/tex]
[tex]Q_3 = 29[/tex]
[tex] IQR= Q_3 -Q_1 = 29-24 =5[/tex]
Step-by-step explanation:
For this case we have the following dataset:
25,21,26,24,29,33,29,25,19,24
The first step is order the data on increasing order and we got:
19, 21, 24, 24, 25, 25, 26, 29, 29 , 33
For this case we have n=10 an even number of data values.
We can find the median on this case is the average between the 5 and 6 position from the data ordered:
[tex] Median = \frac{25+25}{2}=25[/tex]
In order to find the first quartile we know that the lower half of the data is: {19, 21, 24, 24, 25}, and if we find the middle point for this interval we got 24 so this value would be the first quartile [tex] Q_1 = 24[/tex]
For the upper half of the data we have {25,26,29,29,33} and the middle value for this case is 29 and that represent the third quartile [tex]Q_3 = 29[/tex]
And finally since we have the quartiles we can find the interquartile rang with the following formula:
[tex] IQR= Q_3 -Q_1 = 29-24 =5[/tex]
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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How many samples of size 5 are possible when selecting from a set of 10 distinct integers if the sampling is done with replacement?
Answer:
100000 ways
Step-by-step explanation:
Given that there are 10 distinct integers.
5 numbers are drawn with replacement
Prob that each number is drawn will have 10 choices
So each of 5 number can be selected in 10 ways
No of ways to select 5 numbers with replacement
= 10^5
=100000 ways
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
A data set consists of the values 2, 6, 3, and 1. If we consider this a population (all the values available), the variance isA. 12
B. 14
C. the square root of 3.5
D. the square root of 14
E. none of the above
Answer: E. none of the above
Step-by-step explanation:
The given data values that represents the population:
2, 6, 3, and 1.
Number of values : n=4
Mean of the data values = [tex]\dfrac{\text{Sum of values}}{\text{No. of values}}[/tex]
[tex]\dfrac{2+6+3+1}{4}=\dfrac{12}{4}=3[/tex]
Sum of the squares of the difference between each values and the mean =
[tex](2-3)^2+(6-3)^2+(3-3)^2+(1-3)^2[/tex]
[tex]=-1^2+3^2+0^2+(-2)^2[/tex]
[tex]=1+9+0+4=14[/tex]
Now , Variance = (Sum of the squares of the difference between each values and the mean ) ÷ (n)
= (14) ÷ (4)= 3.5
Hence, the variance is 3.5.
Therefore , the correct answer is "E. none of the above".
Find the probability of the given event Choosing five numbers that match five randomly selected balls when the balls are numbered 1 through 36 The probability of the given event is ___ (Type an integer or a simplified fraction.)
Final answer:
The probability of correctly choosing five random numbers from balls numbered 1 to 36 is calculated by multiplying the probability of choosing each number correctly, which is (1/36)^5, rounded to 0.0000 to four decimal places.
Explanation:
The student is asking about the probability of choosing five numbers that match five randomly selected balls when the balls are numbered 1 through 36. This is a question of combinatorial probability, where we are interested in the probability of one specific outcome in a set of possibilities.
To solve this, we need to calculate the probability of choosing each ball correctly. The probability of choosing the first number correctly is 1/36, since there is only one correct number out of 36. Likewise, the probability of choosing the second number correctly is also 1/36, and the same logic applies for the third, fourth, and fifth numbers. As these events are all independent (choosing one number does not affect the others), we can find the total probability by multiplying the individual probabilities together:
P(choosing all five numbers correctly) = P(choosing 1st number correctly) × P(choosing 2nd number correctly) × ... × P(choosing 5th number correctly) = (1/36)^5.
The exact value of this probability is quite small, and one would usually leave it as a fraction to avoid rounding errors. However, the instructions specify to round to four decimal places, so let's calculate:
(1/36)^5 = 1/60466176, which is a very small likelihood and as a decimal, it's approximately 0.0000000165, but you can rounded to 0.0000 when expressing it to four decimal places as per instruction.
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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Explain why the following sets of vectors are not basis for the indicated vector spaces. (Solve this problem by inspection.)
(a) u1 = (1, 2), u2 = (0, 3), u3 = (2, 7) for R^2
(b) u1 = (-1, 3, 2), u2 = (6, 1, 1) for R^3
a.This set of vectors are not basis for vector space for two-dimentional space R2 due to high number of vectors (3). It means three vector is two much to span 2-dimentional space.
b.This set of vectors are not basis for vector space for three-dimentional space R3 due to small number of vectors (2). It means two vector can't span three-dimentional space.
Which equation is equivalent to (one-third) Superscript x Baseline = 27 Superscript x + 2? 3 Superscript x Baseline = 3 Superscript negative 3 x + 2 3 Superscript x Baseline = 3 Superscript 3 x + 6 3 Superscript negative x Baseline = 3 Superscript 3 x + 2 3 Superscript negative x Baseline = 3 Superscript 3 x + 6
Answer:
Option D : 3 Superscript negative x Baseline = 3 Superscript 3 x + 6
Step-by-step explanation:
Let us first convert all the equations in Mathematical form for readability.
The question equation will become:
[tex](\frac{1}{3})^{x}=(27)^{x+2}[/tex] ----------------- (1)
And the option equations will be:
A. [tex]3^{x}=3^{(-3x+2)}[/tex]
B. [tex]3^{x}=3^{(3x+6)}[/tex]
C. [tex]3^{-x}=3^{(3x+2)}[/tex]
D. [tex]3^{-x}=3^{(3x+6)}[/tex]
Now, let's solve the question equation. Simplifying equation (1), we get
[tex](3^{-1})^{x} = (3^3)^{x+2}\\\\3^{-x} = 3^{3(x+2)}\\\\3^{-x} = 3^{(3x+6)}[/tex]
Hence, option D is correct.
Answer:
D
Step-by-step explanation:
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
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.
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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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]
Pluto has been hard to measure from Earth because of its atmosphere. In 2007 Young, Young, and Buie measured Pluto as having a diameter of 2322 km. In 2015 the New Horizons probe reached Pluto and measured it up close and we now know the actual diameter is 2372 km. What was the percent error of the 2007 measurement?
Answer:
2.15% was the percent error of the 2007 measurement.
Step-by-step explanation:
To calculate the percentage error, we use the equation:
[tex]\%\text{ error}=\frac{|\text{Experimental value - Theoretical value}|}{\text{Theoretical value}}\times 100[/tex]
We are given:
Experimental value of diameter of Pluto ,2015= 2372 km
Theoretical value of diameter of Pluto, 2007 = 2322 km
Putting values in above equation, we get:
[tex]\%\text{ error}=\frac{|2372 km-2322 km|}{2322 km}\times 100\\\\\%\text{ error}=2.15\%[/tex]
Hence, 2.15% was the percent error of the 2007 measurement.
Final answer:
The percent error of the 2007 measurement of Pluto's diameter is 2.11%.
Explanation:
The percent error can be calculated by using the formula:
Percent Error = [(Measured Value - Actual Value) / Actual Value] × 100%
Given that the measured diameter of Pluto in 2007 was 2322 km and the actual diameter is 2372 km, we can substitute these values into the formula to calculate the percent error.
Percent Error = [(2372 km - 2322 km) / 2372 km] × 100% = 2.11%
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
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
According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the universe in the "big bang." At present there are 137.7 atoms of 238U for each atom of 235U. Using the half-lives 4.51×109 years for 238U and 7.10×108 years for 235U, calculate the age of the universe.
Answer:
6 billion years.
Step-by-step explanation:
According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let [tex]P(t)[/tex] be the amount of [tex]^{235}U[/tex] and [tex]Q(t)[/tex] be the amount of [tex]^{238}U[/tex] after [tex]t[/tex] years.
Then, we obtain two differential equations
[tex]\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q[/tex]
where [tex]k_1[/tex] and [tex]k_2[/tex] are proportionality constants and the minus signs denotes decay.
Rearranging terms in the equations gives
[tex]\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt[/tex]
Now, the variables are separated, [tex]P[/tex] and [tex]Q[/tex] appear only on the left, and [tex]t[/tex] appears only on the right, so that we can integrate both sides.
[tex]\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt[/tex]
which yields
[tex]\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2[/tex],
where [tex]c_1[/tex] and [tex]c_2[/tex] are constants of integration.
By taking exponents, we obtain
[tex]e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}[/tex]
Hence,
[tex]P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}[/tex],
where [tex]C_1 := \pm e^{c_1}[/tex] and [tex]C_2 := \pm e^{c_2}[/tex].
Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition
[tex]P(0) = Q(0) = C[/tex]
Substituting 0 for [tex]P[/tex] in the general solution gives
[tex]C = P(0) = C_1 e^0 \implies C= C_1[/tex]
Similarly, we obtain [tex]C = C_2[/tex] and
[tex]P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}[/tex]
The relation between the decay constant [tex]k[/tex] and the half-life is given by
[tex]\tau = \frac{\ln 2}{k}[/tex]
We can use this fact to determine the numeric values of the decay constants [tex]k_1[/tex] and [tex]k_2[/tex]. Thus,
[tex]4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}[/tex]
and
[tex]7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}[/tex]
Therefore,
[tex]P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}[/tex]
We have that
[tex]\frac{P(t)}{Q(t)} = 137.7[/tex]
Hence,
[tex]\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7[/tex]
Solving for [tex]t[/tex] yields [tex]t \approx 6 \times 10^9[/tex], which means that the age of the universe is about 6 billion years.
The age of the universe, based on the given ratio of 238U to 235U isotopes and their half-lives, is approximately 8750 years.
To calculate the age of the universe based on the ratio of 238U to 235U isotopes, we can use the concept of radioactive decay and the given half-lives.
The ratio of 238U to 235U is currently 137.7 to 1. This means that over time, 238U has been decaying into other elements, while 235U has been decaying into different elements at different rates due to their distinct half-lives.
We'll start by calculating the number of half-lives that have passed for each isotope to reach the current ratio:
For 238U:
(Number of half-lives) = (Age of the universe) / (Half-life of 238U)
(Number of half-lives) = (Age of the universe) / (4.51 × [tex]10^9[/tex] years)
For 235U:
(Number of half-lives) = (Age of the universe) / (Half-life of 235U)
(Number of half-lives) = (Age of the universe) / (7.10 × [tex]10^8[/tex] years)
Since there is a ratio of 137.7 to 1, it means that the number of half-lives for 238U should be 137.7 times that of 235U:
(Number of half-lives for 238U) = 137.7 × (Number of half-lives for 235U)
Now, we can set up an equation using these relationships:
(137.7) × [(Age of the universe) / (4.51 × [tex]10^9[/tex] years)] = (Age of the universe) / (7.10 × 1[tex]0^8[/tex]years)
Now, we can solve for the "Age of the universe":
137.7 × (4.51 × [tex]10^9[/tex]) = 7.10 × [tex]10^8[/tex] × (Age of the universe)
(Age of the universe) = (137.7 × 4.51 × [tex]10^9[/tex]) / (7.10 × [tex]10^8[/tex])
(Age of the universe) ≈ 8750 years
So, according to this cosmological theory, the age of the universe is approximately 8750 years.
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