A sorting algorithm takes 1 second to sort n =1000 items.

1) How many operations will be performed if the sorting algorithm is O(n2) (approximately)?

2) How long will it take to sort 10,000 items if the sorting algorithm is O(n2)?

3) How much time will one operation take if the sorting algorithm is O(n2)?

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

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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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".

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.

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.

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

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]

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.

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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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.

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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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

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

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.

Learn more about the sampling units described in part.

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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.

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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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.

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.

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]

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.

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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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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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:

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.

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!

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

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.

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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The value of intercept will be 34.

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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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%

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)

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.

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

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.

d) 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.