Suppose that a fungal disease originates in the middle of an orchard, initially affecting only one tree. The disease spreads out radially at a constant speed of 45 feet per day.
(a) What area will be affected after 2 days?
(b) What area will be affected after 4 days?
(c) What area will be affected after 8 days?
(d) Write a formula for the affected area as a function of time, measured in days. Use t as your variable for time, in days.

The probability that a student selected at random majors in engineering is 30% which is 0.3.

The probability  that the student both majors in engineering and play club sports is 10% which is 0.1.

For a student who is selected at random to be one who majors in engineering, there are two possible ways.

The student majors in engineering OR the Student both majors in engineering and plays club

The Probabilty that the student majors in Enginnering =

The probability that the student majors in engineering plus the  probability that Student both majors in engineering and plays club sport

= 0.3+0.1= 0.4

Answer:

Explanation:

The event landing on a three or four on a rolling is independent of the event  of landing on a three or four on any other rolling.

Thus, for independent events you calculate the joint probability as the product of the probabilities of each event:

Here, P(A) = P(B) = P(C) = P(D) = P(E) = probability of landing on a three or four.

Now to calculate the probability of landing on a three or four, you follow this reasoning:

Finally, probability of landing on a three or four on all five rolls: (1/3)⁵ = 1/243 ≈ 0.004 ← answer

Answer:

A

Step-by-step explanation:

The relative frequency is calculated by dividing the frequency of that class to the sum of frequencies. It can be represented as

[tex]Relative frequency=\frac{f}{sum(f)}[/tex]

Hence, the relative frequency of class is the percentage or proportion of data lies in that class.

The cumulative frequency of a class is computed by adding the frequency of the respective class to the frequencies of all previous classes. The cumulative frequency of first class will always be equal to the frequency of first class.

Hence, the cumulative frequency  for a class is the sum of frequency for that class and the frequencies of all previous classes.

Relative frequency is the proportion of total data that falls into a class, while cumulative frequency is the sum of frequencies of that class and all previous classes. Therefore, relative frequency illustrates the percentage of data in a certain class, while cumulative frequency shows the accumulation of data up to that point.

The terms relative frequency and cumulative frequency both pertain to statistics; however, they represent different concepts. Relative frequency of a class is the percentage or proportion of the whole set of data that falls in that class. It's calculated by dividing the frequency of that class by the total number of data points.

On the other hand, cumulative frequency of a class is the sum of the frequencies of that class and all previous classes in a dataset. Essentially, it accumulates counts as you proceed through the dataset.

For instance, if you have five classes with frequencies of 1, 2, 3, 4 and 5, the relative frequencies would be 0.0667, 0.1333, 0.2, 0.2667, and 0.3333 respectively (assuming we divide each frequency by the total data points, which is 15 in this case). In contrast, the cumulative frequencies would be 1, 3, 6, 10 and 15, indicating the total count up to each class.

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Answer:

Nominal level of measurement is the one where data cannot be arranged in an ordering scheme.

Explanation:

This Nominal measurement level is generally characterized by various data that consist of categories of various types,labels of various kinds and also names.

The data that are involved in this type of measurement usually cannot be arranged in an orderly manner or in an ordering scheme.

One of the best example of these type of measurement includes survey responses such as choosing between Yes or No or sometimes undecided.

Thus we can consider nominal level measurement as the answer.

Answer:

nominal

Step-by-step explanation:

Final answer:

The equation that describes the situation is g + 300 - 1600 = 1500. Simplifying this equation gives g = 2800.

Explanation:

To write the equation that describes the situation, we start with the initial weight of the mound as g pounds. Throughout the day, 300 pounds are added to the mound, so the new weight is g + 300 pounds. Two orders of 800 pounds each are sold, so 1600 pounds are removed from the mound. At the end of the day, the weight of the mound is 1500 pounds. The equation that describes the situation is:

g + 300 - 1600 = 1500

Simplifying this equation, we have:

g - 1300 = 1500

Adding 1300 to both sides, the equation becomes:

g = 2800

Answer:

A) 8%

Step-by-step explanation:

A confidence interval of proportions has an upper end and a lower end.

The margin of error is the difference between these points divided by two. Also, it is the upper end subtracted by the estimated proportion, or the estimated proportion subtracted by the lower end.

In this problem, we have that:

Upper end: 0.68

Lower end: 0.52

Margin of error

[tex]M = \frac{0.68 - 0.52}{2} = 0.08[/tex]

The margin of error is 8%.

So the correct answer is:

A) 8%

Answer:

Length of shadow = 35.56 ft

Step-by-step explanation:

∠ BAC = 68°

Height of building = BC = 88 ft

Length of shadow = AC = ?

[tex]tan(68^{o})=\frac{BC}{AC}\\\\tan(68^{o})=\frac{88}{b}\\\\b=\frac{88}{tan(68^{o})}\\\\b=\frac{88}{2.475}\\\\b=35.56 \,ft[/tex]

The Length of shadow  is 35.56 ft.

Given that,

Based on the above information, the calculation is as follows:

[tex]tan 68 ^{\circ} = BC \div AC\\\\tan 68 ^{\circ} = 88 \div b\\\\b = 88 \div tan 68 ^{\circ} \\\\= 88 \div 2.475[/tex]

= 35.56 ft.

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Answer:

A) The data generated is a cross-sectional data because time is not a factor .Secondly. 2,300(variables) people are selected at a particular time.

B) It is a ratio level of measurement because zero has meaning and it is clearly defined that is if the customer did not wait and was attended to immediately then the value is zero. Secondly there is a greater than and less than relationship that exits between the data for example  if a  customer wait for 20 minutes and another customer waits for 25 minutes, 25 is greater than 20.  Hence the greater than and less than relation between data.

 C)The level of measurement achieve from the data collected from this question is ordinal and the data is a qualitative. It’s ordinal because the relationship between the data set is focused on the order at which the data value are arranged. Secondly it’s a qualitative data set in the data value represents the difference in quality and not necessarily how much the difference is.

Step-by-step explanation:

Step One: Definition Of Cross-sectional data,

This can be defined as a type of data that is been collected by noting or observing many variables at a given period of time

Step Two:Definition Of  Time series data

       This can be defined as a type of that that is been collected by noting or observing a particular variable at different points of time. They are usually collected at fixed intervals

Step Three: Difference between cross-sectional data and time series data

The main difference between time series data and cross-sectional data is that for time series data the focus of observation for collecting the data s on the same variable over a period of time while for cross sectional data the focus is on the different variable at the same point in time

Step Four: Definition Of  Level of measurement

 This defines the relationship that exits among data values. The of level of measurement includes nominal, ordinal, interval, ratio

Step Five:Definition Of  Ratio level of measurement

The ratio level of measurement or the ratio scale gives a description of the order and the exact difference that exit between data values, it also has a clear definition for zero value.

Step Six:Definition Of Ordinal level of measurement

For this level of measurement the main focus is the order of the data values, but the exact difference between each data values is not really specified looking at our question part c we know that a #7 is better than a #6 but we cannot say how much better it is.

The data from the wireless service provider's survey is cross-sectional, as it captures data at a single point in time. The question about hold time collects quantitative continuous data, while the service rating question provides ordinal data, which is qualitative.

The data generated from the study conducted by the wireless service provider's support center would be considered cross-sectional because it captures data at a single point in time rather than over multiple periods. Cross-sectional data is collected by observing many subjects at the same point in time or without regard to differences in time.

The level of data measurement obtained from the question asking customers about how long they had been on hold is quantitatively continuous. This is because the time spent on hold can be measured in units (such as minutes) that can take on any value within a given range, including fractions of a minute.

Regarding the customer service rating on a scale of 1-7, the level of data measurement is ordinal. This type of data is qualitative, as it categorizes data into different levels with a meaningful order, but the differences between the levels are not mathematically uniform or meaningful. It captures the order of the ratings but does not quantify the difference between each level.

Answer: 332640 anagrams

Step-by-step explanation:

Given the word:

'masslessness'

Total number of letters= 12

Number of S = 6

E =2

The rest are one each.

The number of possible arrangements can be written as;

N = 12!/(6!2!)

N = 332640

The number of anagrams that can be created from the word 'masslessness' is 239500800, calculated using the formula for permutations with repetition.

This question is an application of the concept of permutations from combinatorics in mathematics. In this case, the word 'masslessness' consists of 12 characters including 3 's', 2 's', and 2 'l'. The formula for calculating permutations when there are repeating members within a set is n!/(r1! * r2! * ... * rn!), where n is the total number of members and r is the number of repeating members. Applying this formula, the number of anagrams of 'masslessness' would be 12!/(3! * 2! * 2!) = 239500800.

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Answer:

d. The median and the IQR

Step-by-step explanation:

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median. This means that the median is the appropriate numerical measure to describe the center.

The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.

The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.

The quartiles are used to measure the spread of a data-set.

The interquartile range(IQR), is Q3-Q1.

So the correct answer is:

d. The median and the IQR

Appropriate numerical measures of centre and spread of the distribution are : D) Median and IQR respectively.

Median is the positional average showing the mid point of a data. Median divides the data into two equal halves, both containing 50% of the data.

Quartiles are also a positional measure, dividing the data into 4 equal quarters, Q1 & Q3 covering 25% & 75% of the data respectively.

Interquartile Range shows the difference between Quartile 1 & Quartile 3. It is a measure of dispersion of data, denoting the spread & distribution of data.

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Answer:

f'(π) = (-5/3)

Step-by-step explanation:

for the equation

∫₀ˣ (3f(t)+5t)dt=sin(x)

3*∫₀ˣ f(t) dt+5*∫₀ˣ t*dt=sin(x)

then

∫₀ˣ t*dt = (t²/2) |₀ˣ = x²/2 - 0²/2 = x²/2

thus

3*∫₀ˣ f(t) dt + 5* x²/2 =  sin(x)

∫₀ˣ f(t) dt =  1/3*sin(x) - 5/6*x²

then applying differentiation

d/dx ( ∫₀ˣ f(t) dt) = f(x) - f(0)  ( from the fundamental theorem of calculus)

d/dx (1/3*sin(x) - 5/6*x²) = 1/3*cos(x) - 5/3*x

therefore

f(x) - f(0) =  1/3*cos(x) - 5/3*x

f(x) =  1/3*cos(x) - 5/3*x + f(0)

applying differentiation again (f'(x) =df(x)/dx)

f'(x) =  -1/3*sin(x) - 5/3

then

f'(π) =  -1/3*sin(π) - 5/3 = 0 - 5/3 = -5/3

f'(π) = (-5/3)

Given expression is:

→ [tex]\int x_0 (3f(t)+5t) dt = sin(x)[/tex]

or,

→ [tex]3\times \int x_0 f(t) dt+5\times \intx_0 t\times dt = sin(x)[/tex]

then,

→ [tex]\int x_0 t\times dt = (\frac{t^2}{2} ) x_0[/tex]

      [tex]\frac{x^2}{2} -\frac{0^2}{2} = \frac{x^2}{2}[/tex]

thus,

→ [tex]3\times \int x_0 f(t) +5\times \frac{x^2}{2} = sin(x)[/tex]

  [tex]\int x_0 f(t) dt = \frac{1}{3}\times sin(x) - \frac{5}{6}\times x^2[/tex]

By applying differentiation, we get

→ [tex]\frac{d}{dx} (\int x_0 f(t) dt) = f(x) - f(0)[/tex]

  [tex]\frac{d}{dx}(\frac{1}{3} sin(x)) - \frac{5}{6}x^2= \frac{1}{3} cos(x) - \frac{5}{3} x[/tex]

Therefore,

→ [tex]f(x) -f(0) = \frac{1}{3}cos (x) - \frac{5}{3} x[/tex]

  [tex]f(x) = \frac{1}{3}cos (x) -\frac{5}{3} x+f(0)[/tex]

By differentiating again, we get

→ [tex]f'(x) = -\frac{1}{3}sin(\pi) -\frac{5}{3}[/tex]

           [tex]= 0-\frac{5}{3}[/tex]

           [tex]= -\frac{5}{3}[/tex]

Thus the response above is correct.

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Explanation on determining which type of cameras sold more in 2001 and when digital camera sales surpassed film camera sales.

(a) To show that more film cameras than digital cameras were sold in 2001, we need to substitute t = 0 into both functions:

For digital cameras: f(0) = 3.06(0) + 6.84 = 6.84 million units
For film cameras: g(0) = -1.85(0) + 16.48 = 16.48 million units
Therefore, more film cameras (16.48 million units) were sold in 2001 than digital cameras (6.84 million units).

(b) To find when digital camera sales first exceeded film camera sales, we need to set the two functions equal to each other and solve for t:
3.06t + 6.84 = -1.85t + 16.48
4.91t = 9.64
t ≈ 1.96 years, which is approximately at the end of 2003.

Answer:

ljgiuipuiopu

Step-by-step explanation:

Answer:

$2 640

Step-by-step explanation:

Principal = $16 500; Time = 2 years Rate = 8%

Simple Interest = P × R × T/100

= $16 500 × 8 × 2/100

=$264 000/100

Simple Interest = $2 640

Answer:

a) Range = 25.6

b) Variance = 48.1446

c) s = 6.9386

d) See below

Step-by-step explanation:

(a)  the sample range

To obtain the sample range, we must sort the data increasingly :

23.7, 26.3, 28.3, 28.7, 29.6, 29.8, 30.1, 31, 33.5, 49.3

Then, find the distance between the largest and the lowest

Range = 49.3 - 23.7 = 25.6

(b) the sample variance [tex]s^2[/tex] from the definition

In order to find the variance from the definition, we need first the mean. The mean is defined as the average

[tex]\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}[/tex]

where the [tex]x_i[/tex] are the values of the data collected and n=10 the size of the sample.

So, the mean is

[tex]\bar x=31.03[/tex]

Now, the variance of the sample is defined as  

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

and we have that the variance is

[tex]s^2=48.1446[/tex]

(c) the sample standard deviation

The sample standard deviation is nothing but the square root of the variance

[tex]s=\sqrt{48.1446}=6.9386[/tex]

d) [tex]s^2[/tex]  using the shortcut method

The shortcut method figures out the variance without having to compute the mean. The formula is

[tex]s^2=\frac{n\sum x_i^2-(\sum x_i)^2}{n(n-1)}[/tex]

where n=10 is the sample size .

So, using the shortcut method,

[tex]s^2=\frac{10*10,061.91-96,286.09}{10*9}=48.1446[/tex]

Answer: c. Let d represent Enrique's distance from home (in miles).

e. Let w represent Enrique's weight in pounds

Step-by-step explanation:

A variable is a particular type of value that denotes an unknown quantity.

To define a variable , we need define a amount or quantity with units and in case of length an exact position is required.

a. Let w represent Enrique's weight

It is not a variable because no units is defined.

b. Let d represent Enrique's distance (in miles).

It is not a variable because the points to establish distance traveled by Enrique is not given .

c. Let d represent Enrique's distance from home (in miles).

It is a variable as both units and position as mentioned.

d. Let d represent Enrique's distance from home.

It is not a variable because no units is defined.

e. Let w represent Enrique's weight in pounds.

It is a variable because unit of weight is mentioned.

Hence, the correct statements that properly define a variable are  :

c. Let d represent Enrique's distance from home (in miles).

e. Let w represent Enrique's weight in pounds

Answer:

a. Standard error of mean

Step-by-step explanation:

The standard error of mean is computed by dividing the standard deviation to the square root of sample size. It can be represented as

[tex]Standard error=\frac{standard deviation}{\sqrt{n}}[/tex]

The standard deviation of the sampling distribution is known as the standard error of mean because it measures the accuracy of sample mean as compared to population mean

The standard deviation of the sampling distribution of the mean is referred to as the Standard Error of the Mean, which measures the dispersion of sample means around the true population mean.

The standard deviation of the sampling distribution of the mean is known as the Standard Error of the Mean. When we say sampling distribution of the mean, we're referring to the distribution of means for all possible samples of a given size from the same population. The Standard Error of the Mean measures the dispersion of these sample means around the true population mean.

For example, if you have a population of test scores with a mean of 70 and a standard deviation of 10, and you were to take many samples of 30 scores from this population, the standard error of the mean would be the standard deviation of all those sample means.

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Answer: True

Step-by-step explanation:

A subset is a term used in the sets topic In mathematics to describe a Part of A larger values being considered.

Sample In experimental procedures is known as Part of what is to be used to determine the reaction of a larger mass of specimen.

Taking these two definitions by side, it is quite evident that both of them represent taking part of a larger picture to form a smaller picture which does not totally deviate from the larger picture being considered.

Answer:

Algebraic Theorem Proofs

Step-by-step explanation:

We have to prove the following in the question:

Theorems used:

[tex]XX' = 0\\X + 1 = X\\X + 0 = X\\X+X'=1\\XX = X\\X+X=X[/tex]

(a) X(X′ + Y) = XY

[tex]X(X'+ Y) \\=XX'+XY\\=0+XY\\=XY[/tex]

(b) X + XY = X

[tex]X + XY \\=X(1+Y)\\=XY[/tex]

(c) XY + XY′ = X

[tex]XY + XY'\\=X(Y+Y')\\=X(1)\\=X[/tex]

(d) (A + B)(A + B′) = A

[tex](A + B)(A + B')\\=AA + AB' +BA + BB'\\=A + A(B+B')+0\\=A+A(1)\\=A[/tex]

I would be happy to provide step-by-step solutions to each of these theorems.

(a) Suppose we have X(X′ + Y) = XY. If we were to simplify using the law of distribution, we would have X(X′) + XY. However, since X′ is the complement of X or the negation of X, X(X′) becomes zero and the equation simplifies to 0 + XY which is XY. If we equate the two expressions X(X′ + Y) and XY, they are not equivalent.

(b) Now we have X + XY = X. On the left side, the expression can be rewritten as X(1 + Y). But observing it closely, the factor of X is multiplied by the bracket. However, bringing X out in front results in an equation that isn't equivalent to X.

(c) Then we move on to XY + XY′ = X. First, we attempt to simplify the left side of the equation using the law of distribution. It becomes X(Y + Y′). It is important to remember that Y + Y′ equates to 1 based on Boolean algebra. Hence, our equation simplifies to X(1), or simply "X". But considering the fact that Y' means the complement or the negation of Y, XY' would be zero (because anything multiplied by zero is zero). Therefore, the equation is not equivalent to X.

(d) Finally, (A + B)(A + B′) = A. The left-hand side of the equation can be expanded using distributive law into AA′ + AB + AB' + BB′. Looking at it closely, AA′ and BB′ become zero because of the property that states an element and its complement's product is always zero. Coming to AB and AB', since A' is the complement of A, AB + AB' simplifies to 0 + 0 which is equal to 0. But 0 doesn't equate to A making the equation non-equivalent.

In summary, all four theorems failed to prove equivalent under Boolean algebra rules.

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Answer:

[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]

Step-by-step explanation:

For this case we can construct the following table in order to find the mean for the grouped data:

  Interval           Frequency (fi)    Midpoint (Xi)      Xi *fi

5001-10000              27                   7500.5          202513.5

10001-15000             23                  12500.5         287511.5

15001-20000             12                  17500.5         210006

20001-25000            18                  22500.5        405009  

25001-30000            30                 27500.5         825015

Total                          110                                         1930055

And the mean is calculated with the following formula:

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

Where [tex] n = \sum_{i=1}^n f_i = 110[/tex]

So then if we replace into the formula we got:

[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]

The dimensions of the parcel are 80 ft (width) and 88 ft (length).

To find the dimensions of the parcel, we can set up a system of equations using the given information. Let's denote the width of the parcel as x ft. Since the parcel is 8 ft longer than it is wide, the length can be represented as x + 8 ft. We can use the Pythagorean theorem to find the length of the diagonal: x^2 + (x + 8)^2 = 232^2. Solving this equation will give us the width and length of the parcel.

Expanding the equation, we get: x^2 + (x^2 + 16x + 64) = 53824. Combining like terms, we have: 2x^2 + 16x + 64 = 53824. Rearranging the equation, we have a quadratic equation: 2x^2 + 16x + (-53760) = 0. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.

By factoring, we find that the dimensions of the parcel are 80 ft (width) and 88 ft (length).

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Answer:

a) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

The individual pprobabilities are:

[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]

[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]

[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]

And if we replace we got:

[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]

b) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

And we can calculate this with the following excel formula:

[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]

And we got:

[tex] P(X\geq 3) = 0.023115[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=300000000, p=\frac{1}{500000000})[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

(a) Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

The individual probabilities are:

[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]

[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]

[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]

And if we replace we got:

[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]

(b) Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]

And we can calculate this with the following excel formula:

[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]

And we got:

[tex] P(X\geq 3) = 0.023115[/tex]

The best point estimate for the number of excess pounds that the overweight men weighed is the mean value given, which is 35 pounds. Furthermore, the standard deviation of 3.5 pounds indicates that majority of these men's excess weight is within 3.5 pounds of this mean value.

In this problem, you have been given a random sample of 90 overweight men with a mean weight being overweight by 35 pounds. This mean (35 pounds) represents the best point estimate of the amount of excessive weight that the men of your study hold. What this means is that our best guess, considering this sample data, of an overweight man's excess weight is 35 pounds. We compute this mean value by adding up all the data values and dividing by the total number, in this case, 90. The standard deviation tells us somewhat about how much variation exists in the population's weights from this mean value. In this scenario, the standard deviation is 3.5 pounds, which means that the majority of the men in this study were within 3.5 pounds of the mean excess weight of 35 pounds.

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The best point estimate of the number of excess pounds that the sample of 90 overweight men weighed is the mean of their weights, which is given as 35 pounds.

In this problem, the best point estimate of the number of excess pounds that the men weighed would be the mean of their weights. Since we are given that the mean number of excess pounds that they were overweight was 35, the best point estimate would also be 35. This is because in statistics, the mean of a sample is used as the best point estimate of the population mean. The fact that they mentioned the standard deviation of 3.5 pounds does not affect the best point estimate, as that is related more to the dispersion or spread of the data around the mean, and not the central point (which is the mean).

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Answer: Experimental study

Step-by-step explanation:

This is an experimental study because it involves determination of the effect of each treatment on separate individuals. It involves selecting the 81 individuals randomly to be treated with a particular treatment (either of the two treatment). In experimental study the researcher apply separate treatments on a different groups and measure its effect on each of the groups.

This is an experimental study where participants were randomly assigned to either naturopathic care or standardized psychotherapy to assess their effects on anxiety scores.

This is an example of an experimental study. In an experimental study, researchers assign participants to different groups and manipulate variables to determine cause-and-effect relationships. In this case, Cooley et al. randomly assigned individuals to either naturopathic care or standardized psychotherapy, which are the two treatments being compared. They then measured the effects of these treatments on anxiety scores.

The fact that participants were randomly assigned to the two treatments suggests that this study was experimental rather than observational.

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Answer:

The answer to your question is 7345 l

Step-by-step explanation:

Data

Convert 7kl to 345 L

Process

- To solve this question we must know the equivalence of kl to L.

                    1 kl -----------------  1000 l

- Now, use proportions and cross multiplication to solve it

                    1 kl ----------------  1000 l

                    7 kl----------------   x l

                    x = (7 x 1000) / 1

                    x = 7000 / 1

                    x = 7000 l

- Calculate the result

                   7000 + 345 = 7345 l

Answer:

The correct question is

Let [tex] t^2y''+10ty'+8y=0 [/tex]. Find all values of r such that [tex] y = t^r [/tex] satisfies the differential equation for t > 0. If there is more than one correct answer, enter your answers as a comma separated list.

The answer is r = 8, r = 1

Step-by-step explanation:

By substituting     [tex] y = t^r [/tex]  into the given equation [tex] t^2y''+10ty'+8y=0 [/tex].

That is substituting [tex] y = t^r [/tex] , [tex] y' = rt^{r-1} [/tex], [tex] y'' = r(r-1)t^{r-2} [/tex] into the given equation, we have

[tex] t^2 r(r-1)t^{r-2} +10t rt^{r-1}+8 t^r =0 [/tex]

Implies that [tex] r(r-1)t^r +10rt^r + 8 t^r =0 [/tex]

Implies that  [tex] [r(r-1) +10r + 8]t^r = 0[/tex]

Implies that  [tex] [r^2 - r +10r + 8]t^r = 0[/tex]

Implies that  [tex] [r^2 - 9r + 8]t^r = 0[/tex]

Implies that  [tex] r^2 - 9r + 8 = 0[/tex], assuming [tex] t^r \ne 0[/tex]  being the assumed non- trivial solution.

The by solving this quadratic equation [tex] r^2 - 9r + 8 = 0[/tex] using factorization method, we have

[tex] r^2[/tex]  – r – 8r  + 8 = 0

Implies that r(r – 1) – 8(r – 1) = 0

Implies that (r – 8)(r – 1) = 0

Implies that r – 8 = 0 or r – 1 = 0

Implies that r = 8 or r = 1

Therefore, the value of r that satisfies the given equation is r = 8, r = 1.

Final answer:

The values of r such that y = t^r satisfies the given differential equation are r = -8 and r = -1.

Explanation:

The student is tasked with finding values of r such that y = t^r satisfies the given differential equation t^2y'' + 10ty' + 8y = 0 for t > 0. To solve this, we will substitute y = tr into the equation and find the characteristic equation that the values of r must satisfy.

Using the proposed solution y = t^r, we calculate the first and second derivatives of y:

y' = rt^{r-1}

y'' = r(r-1)t^{r-2}

Substituting these derivatives back into the original differential equation, we get:

t^2(r(r-1)t^{r-2}) + 10t(rt^{r-1}) + 8t^r = 0

Simplifying, we find the characteristic equation:

r(r-1) + 10r + 8 = 0

Which factors to:

(r + 8)(r + 1) = 0

Thus, the possible values of r are:

r = -8

r = -1

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

[tex]P(A)+P(B)+P(A\cap B)=P(A\cup B)[/tex]

We know that

Maximum value of [tex]P(A\cup B)[/tex]=1 and minimum value of [tex]P(A\cup B)[/tex]=0

[tex]0\leq P(A\cup B )\leq 1[/tex]

[tex]0\leq P(A)+P(B)-P(A\cap B)\leq 1[/tex]

[tex]0\leq 0.4+0.7-P(A\cap B)\leq 1[/tex]

[tex]0\leq 1.1-P(A\cap B)\leq 1[/tex]

[tex]0\leq 1.1-P(A\cap B)[/tex]

[tex]P(A\cap B)\leq 1.1[/tex]

It is not possible that [tex]P(A\cap B)[/tex] is equal to 1.1

[tex]1.1-P(A\cap B)\leq 1[/tex]

[tex]-P(A\cap B)\leq 1-1.1=-0.1[/tex]

Multiply by (-1) on both sides

[tex]P(A\cap B)\geq 0.1[/tex]

Again, [tex]P(A\cup B)\geq P(B)[/tex]

[tex]0.4+0.7-P(A\cap B)\geq 0.7[/tex]

[tex]1.1-P(A\cap B)\geq 0.7[/tex]

[tex]-P(A\cap B)\geq -1.1+0.7=-0.4[/tex]

Multiply by (-1) on both sides

[tex]P(A\cap B)\leq 0.4[/tex]

Hence, [tex]0.1\leq P(A\cap B)\leq 0.4[/tex]

Answer:

Step-by-step explanation:

1. sinФ = cos 25

25 ° is in the between 0 and 90°

therefore it can simply represent

cosФ= sin (90-Ф) = sin (90 - 25) = sin 65

2. sin(Ф/3 + 10) = cos Ф

cos Ф = sin (90 -Ф)

sin(Ф/3 + 10) = cos Ф

sin(Ф/3 + 10) = cos Ф = sin(90-Ф)

Ф/3+10=90-Ф

10Ф+3/30 = 90-Ф

10Ф+3 = 30(90-Ф)

10Ф+3 = 2700-30Ф

10Ф+30Ф=2700-3

40Ф = 2697

Ф = 2697 / 40 = 67.425 ≅ 67.4°

Answer:

False

Step-by-step explanation:

The mentioned statement is not true because the F sampling distribution is used for variance estimator in the Analysis of the variance. The Analysis of variance technique used to compare two different variance estimate to check the equality of population mean. The chi-square statistic involve the comparison of variance with specified variance whereas F-statistic is calculated by taking the ratio of two variances.

The sampling distribution for the variance estimator in Analysis of Variance (ANOVA) is Chi-square if key assumptions such as random selection, equal standard deviations, and normally distributed populations are met. If not, further statistical analysis is required.

In Analysis of Variance (ANOVA), the sampling distribution for the variance estimator is indeed Chi-square, but only if the data meets certain assumptions. These assumptions are:

If these assumptions are met, we can proceed to calculate the F ratio, which is our test statistic for Analysis of Variance (ANOVA). However, if any of these assumptions aren't met, further statistical testing is required. Even in such cases, it's important to know the distribution of sample means, the sums, and the F Distribution to make accurate statistical conclusions.

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