Answer:
[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]
e. P(5<X<8)=0.251
Step-by-step explanation:
Given that boiler follows a Poisson distribution,$\sim$
[tex]X $\sim$Poi(\mu=5)[/tex]
Poisson Distribution formula is given by the expression:-
[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]
Our probabilities will be calculated as below:
a.
[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]
b.
[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]
c.
[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]
d.
[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]
e.P(5<X<8)
[tex]P(X=6)+P(X=7)\\=0.251[/tex]
The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The students must qualify by meeting certain standards in terms of GPA and standardized test scores. The goal of this new policy is to increase the geographic diversity of students from states beyond Arkansas and its border states. Historically, 90% of all new students came from Arkansas or a bordering state. Ginger, a student at the U of A, sampled 180 new students the following year and found that 157 of the new students came from Arkansas or a bordering state. Does Ginger’s study provide enough evidence to indicate that this new policy is effective with a level of significance 10%? What would be the correct decision?
a. Reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansasb. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
c. Reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
d. Fail to reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansas
Answer:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
Step-by-step explanation:
Data given and notation n
n=180 represent the random sample taken
X=157 represent the students who came from Arkansas or a bordering state
[tex]\hat p=\frac{157}{180}=0.872[/tex] estimated proportion of students who came from Arkansas or a bordering state
[tex]p_o=0.9[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher or not than 0.9.:
Null hypothesis:[tex]p\geq 0.9[/tex]
Alternative hypothesis:[tex]p < 0.9[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
A small social network contains seven people who are network friends with six other people in the network, one person who is network friends with five other people in the network, and five people who are network friends with four other people in the network. The rest are network friends with three other people in the network. The network contains 50 pairs of network friends.
(a) How many people are network friends with three other people in the network?
(b) How many people are in the network?
Answer:
(a) 11 people
(b) 24 people
Step-by-step explanation:
Part (a)
Let X be the people who are network friends with three other people in the network.
The total pairings = 50
Since [tex]v[/tex]∈[tex]V[/tex],
[tex](7)(6) + (1)(5) + (5)(4) + (X)(3) = (2)(50)[/tex]
Solve for X to find the people who are network friends with three other people,
[tex]67 + 3X = 100[/tex]
[tex]X = 11[/tex]
Answer: There are 11 people who are network friends with three other people in the network.
Part (b)
Total people in the network,
[tex]= 7 + 1 + 5 + X\\ = 13 + 11\\ = 24[/tex]
Answer: There are a total of 24 people in the network.
There are 6 people who are network friends with three other people in the network. The total number of people in the network is 19.
Explanation:This problem involves understanding the concept of pairs in network friends. Each friendship forms a pair. So, if one person has 6 friends, this would contribute 6 pairs. Here, there are 7 people with 6 friends, 1 person with 5 friends, and 5 people with 4 friends. Therefore, these people contribute (7x6) + (1x5) + (5x4) = 42 + 5 + 20 = 67 pairs. But, we're told there are only 50 pairs in total. The reason for this discrepancy is that each pair is being counted twice. When we divide the total pairs (67) by 2, we get 33.5 but since the number of pairs must be an integer, we subtract this half pair to get 33 pairs for the listed people. That means, there must be 50 - 33 = 17 pairs remaining for the people who are network friends with three other people. Given that each of these people contributes 3 pairs, there must be 17 / 3 = 5.67 people, but since we're talking about whole people, we need to round this up to 6. Therefore, there are 6 people who are network friends with three other people in the network. Now, adding up all the people in the network, we have 7 (people with 6 friends) + 1 (person with 5 friends) + 5 (people with 4 friends) + 6 (people with 3 friends) = 19 people in total.
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a rectangular box with two sqaure opposite ends is to hold 8000 cubic inches. find the dimensions of the cheapest box if tge recangular sides cost 15 times more as much per square inche as the top, bottom, and square ends
Answer:
the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in
Step-by-step explanation:
since the box has a volume V
V= x*y*z = b=8000 in³
since y=z (square face)
V= x*y² = b=8000 in³
and the cost function is
cost = cost of the square faces * area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides
C = a* 2*y² + a* 2*x*y + 15*a* 2*x*y = 2*a* y² + 32*a*x*y
to find the optimum we can use Lagrange multipliers , then we have 3 simultaneous equations:
x*y*z = b
Cx - λ*Vx = 0 → 32*a*y - λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ
Cy - λ*Vy = 0 → (4*a*y + 32*a*x) - λ*2*x*y = 0
4*a*32/λ + 32*a*x - λ*2*x*32*a/λ = 0
128*a² /λ + 32*a*x - 64*a*x = 0
32*a*x = 128*a² /λ
x = 4*a/λ
x*y² = b
4*a/λ * (32*a/λ)² = b
(a/λ)³ *4096 = 8000 m³
(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in
then
x = 4*a/λ = 4*5/4 in = 5 in
y=32*a/λ = 32*5/4 in = 40 in
then the box has dimensions 5 in x 40 in x 40 in
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack _____________.
Answer:
Reliability
Step-by-step explanation:
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack reliability.
Reliability can be defined as the extent to which an experiment or test provides the same result on repeated trials.
The huge difference in Dylan's IQ test scores in a short period of time suggests that the Whaddayanno IQ test is not reliable.
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25∘ north of east. The road then turns due north for another 3.0 mi to the finish line. In miles, what is the straight-line distance from the starting point to the end of the race? Express your answer in miles.
Answer:
6.79 miles
Step-by-step explanation:
Consider the triangle ABC attached where A is the starting point and C is the finish point.
|AB|=5.2 miles
|BC|=3.0 Miles
≺ABC = 90+75 =165°
We are given the length of two sides and an angle not opposite any of the given sides.
The rule for solving this scenario is referred to as the Cosine Rule.
Note that the side opposite each angle is labelled using the corresponding small letter.
The Cosine Rule States that:
b²=a²+c²-2acCosB
|AC|²=3²+5.2²-(2X5.2XCos165°)
|AC|²=9+27.04-(-10.05)
|AC|²=9+27.04+10.05=46.09
|AC|=√46.09=6.79 miles
Therefore the distance on a straight line from A to C is 6.79 miles
Find the zeros of each function
f(x)=(x-3)(x+5)
Answer:
The zeros of this function are 3 and -5.
Step-by-step explanation:
One note: The question says "Find the zeros of EACH function", but there seems to be only one function in the question. I hope I answered it well.
To complete the function provided, you need to get both terms inside both parenthesis to equal zero. So, if its (x-3)(x+5), the X's need to equal the opposite of the number that is being added or subtracted.
For the first parenthesis, (x-3), the number is -3, so the zero of that side is positive 3. (3-3 = 0)
For the second parenthesis, (x+5), the number is positive 5, so the zero of that side is -5. (-5+5 = 0)
Oslo Company's target quality characteristic, T, for one of its key components is set at 82. Using the Taguchi Quality Loss Function (QLF) the company has determined the cost coefficient, k, to be $6,000. What is the estimated loss, L(x), if the value of the quality characteristic, x, is 85
Answer:
The estimated loss is $5,400 if the value of the quality characteristic,x is 85.
Step-by-step explanation:
We are given the following in the question:
Target value,T = 82
Cost coefficient, k = $6,000
x = 85
The Taguchi Quality Loss Function (QLF) is given by:
[tex]L(x) = k(x -T)^2[/tex]
where L(x) is the loss, k is the cost coefficient, T is the target value or the mean value.
We have to estimate the loss if the value of the quality characteristic is 85.
We put x = 85
[tex]L(85) = 6000(85 -82)^2\\L(85) =54000[/tex]
Thus, the estimated loss is $5,400 if the value of the quality characteristic,x is 85.
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)
Answer:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
Given that,
A company has found that in carton of transistors: 92% contain no defective transistors,
3% contain one defective transistor, 3% contain two defective transistors,
and 2% contain three defective transistors.
We have to find,
Calculate the mean and variance for the defective transistors. Mean = Variance.
According to the question,
Let, X the random variable who represent the number of defective transistors.
The following probability distribution for X,
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
To calculate the expected value with the following formula:
[tex]E(X) = \sum^{n}_{i=1} X_i. P.(X_i)[/tex]
On substitute all the values in the formula,
[tex]E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15[/tex]
To find the variance first the second moment given by:
[tex]E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)[/tex]
On substitute all the values in the formula;
[tex]E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330[/tex]
Therefore, The variance is calculated with this formula:
[tex]Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance,
[tex]Standard \ deviation = \sqrt{0.0375} = 0.5545[/tex]
Hence, The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
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(1 point) On a piece of paper, sketch each of the following surfaces: (i) z = x2 +y2 +6 (ii) z = 3x2 Use your graphs to fill in the following descriptions of cross-sections of the surfaces. (a) For (i) (z = x2 2 + 6): Cross sections with x fixed give a downward opening parabola in a plane parallel to the yz-plane a downward opening parabola in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane Cross sections with y fixed give Cross sections with z fixed give (b) For (ii) (z = 3x2): Cross sections with x fixed give Cross sections with y fixed give Cross sections with z fixed give an empty set, or one or two vertical lines in a plane parallel to the yz-plane> an empty set, or one or two vertical lines in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane
Answer:
ht j5yh bgmn gmjdt ky
Step-by-step explanation:
w245358697809908
The cross sections of the given mathematical functions provide a range of shapes, from parabolas to circles or lines, depending on whether x, y or z are fixed.
Explanation:Interpreting and sketching the three-dimensional functions in question gives us an understanding of the cross-sections. For the first function i) z = x2 + y2 + 6, when x is fixed, the cross section gives us a parabola facing upwards in yz-plane. Moving ahead, when y is fixed, it gives us a similar parabola in the xz-plane. When z is fixed, we end up with a circle in xy-plane.
Going to the second function ii) z = 3x2, cross sections with x fixed results in a vertical line in yz-plane as z is not a function of y here. For y being fixed, it will again give us a vertical line but in the xz-plane because z and y are again unrelated. If z is fixed, we obtain a parabola opening upwards (or downwards depending on the sign) parallel to xy-plane.
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3. It is known that 80% of all college professors have doctoral degrees. If 10 professors are randomly selected, find the probability that a. five have a doctoral degree b. fewer than 4 have doctoral degrees. c. At least 6 have doctoral degrees. d. Between 5 and 7 (inclusive) have doctoral degrees. e. What is the expected number of college professors with doctoral degrees
Answer:
a) [tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
b) [tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
c) [tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
d) [tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
e) [tex] E(X) = n*p = 10*0.8 =8[/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=10, p=0.8)[/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]
Part a
We want to find this probability:
[tex] P(X=5)[/tex]
And using the pmf we got:
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
Part b
[tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
Part c
For this case we can use the complement rule like this:
[tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
Part d
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
Part e
The expected number is given by:
[tex] E(X) = n*p = 10*0.8 =8[/tex]
This answer provides step-by-step explanations for finding probabilities of professors with doctoral degrees and the expected number of professors with doctoral degrees.
Expected number of college professors with a doctoral degree:
Find the probability that five have a doctoral degree: Using the binomial probability formula, [tex]P(X = 5) = 10C5 * 0.8^5 * 0.2^5[/tex]
Find the probability that fewer than 4 have a doctoral degree: Calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Find the probability that at least 6 have a doctoral degree: Calculate P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)).
Find the probability that between 5 and 7 inclusive have a doctoral degree: Calculate P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7).
Expected number of professors with a doctoral degree: Multiply the total number of professors by the probability (0.8).
A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to
Answer:
20 m
Step-by-step explanation:
We are given that
Initial horizontal speed ,[tex]u_x=15 m/s[/tex]
Time, t=2 s
Initial vertical velocity, [tex]u_y=0[/tex]
We know that
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
Where [tex]g=-9.8m/s^2[/tex]
Using the formula
[tex]h=0(2)+\frac{1}{2}(-9.8)(2)^2[/tex]
[tex]h=-19.6 m\approx -20 m[/tex]
The negative sign indicates that the displacement of stone is in downward direction.
Hence, the height of the cliff is closest to 20 m
please someone help me!!!! SHOW ALL WORK! ive been stuck on this question for 2 days!! :((((((
Step-by-step explanation:
x² + y² − 8x + 10y − 8 = 0
Rearrange:
x² − 8x + y² + 10y = 8
To complete the square, take half of the x and y coefficients, square it, then add the result to both sides.
(-8/2)² = 16
(10/2)² = 25
x² − 8x + 16 + y² + 10y + 25 = 8 + 16 + 25
x² − 8x + 16 + y² + 10y + 25 = 49
Factor the squares:
(x − 4)² + (y + 5)² = 49
The center is (4, -5) and the radius is 7.
A sample of 100 items has a population standard deviation LaTeX: \:\sigma\:\:σof 5.1 and a sample mean of 21.6. What is the value of the point estimate for the population mean?
Answer:
The point estimate of population mean is 21.6.
Step-by-step explanation:
A point estimate is a numerical value that is used to estimate the value of the parameter under study.
The point estimate is calculated using the sample values.
For example, sample ([tex]\bar x[/tex]) is the point estimate of population mean (μ), sample standard deviation (s) is the point estimate of population standard deviation (σ), sample proportion ([tex]\hat p[/tex]) is the point estimate of population proportion (p).
It is provided that the sample mean of 100 items is,
[tex]\bar x=21.6[/tex]
Then the point estimate of population mean is:
[tex]\mu=\bar x=21.6[/tex]
Thus, the point estimate of population mean is 21.6.
A researcher claims that the proportion of people who are right-handed is greater than 70%. To test this claim, a random sample of 600 people is taken and its determined that 424 people are right-handed.
The following is the setup for this hypothesis test:
{H0:p=0.70
Ha:p>0.70
Find the test statistic for this hypothesis test for a proportion.
The test statistic for the given hypothesis test for a proportion is calculated to be 0.356.
Given that:
The claim is that the proportion of people who are right-handed is greater than 70%.
So, p > 0.7 is the alternative hypothesis.
From the setup of the hypothesis test, the test is a one-tailed test since the alternative hypothesis has a sign ">".
Sample size, n = 600
Number of people right-handed = 424
So, the observed sample proportion is:
[tex]\hat{\text{p}}=\frac{424}{600}[/tex]
[tex]=0.7067[/tex]
The standard deviation is:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n} }[/tex]
[tex]=\sqrt{ \frac{0.7(1-0.7)}{600} }[/tex]
[tex]=0.01871[/tex]
So, the test statistic is:
[tex]\text{z}=\frac{\hat{\text{p}}-\text{p}}{\sigma}[/tex]
[tex]=\frac{0.7067-0.7}{0.01871}[/tex]
[tex]=0.356[/tex]
Hence, the test statistic is 0.356.
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Sean, a high school wrestler, has agreed to participate in a study of cardiovascular conditioning. He is left somewhat confused when, at the first research session, he is asked to complete a questionnaire about commonly purchased grocery items. Sean's confusion indicates a lack of ________ regarding the task. Question 2
Answer:
Face validity.
Step-by-step explanation:
Face validity is considered the weakest form of validity as it does a superficial, subjective assessment which does not involve objective approach, revealing the deeper intent of the test being used. It involved a process similar to skimming the surface of an item or a book to make opinions. Though it is a weak form of validity, it is the easiest to apply to research. Face validity is an informal approach to identifying how suitable the content of a test is to the research.
It generally asks the question:
Is the content of the test suitable to its aims?
A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550. The value after that date was modeled as follows. V(t)equals 14 comma 550 e Superscript negative 0.158 t a) What is the rate of change in the value of the equipment on January 1, 2019
Answer:
[tex]\frac{dV(t)}{dt} =[/tex] - 1675.38
Step-by-step explanation:
In 2017, the vakue of the kitchen equipment was $14550
V(0)=$14550
Its value after then was modelled by [tex]V(t)=14550e^{-0.158t[/tex]
We are required to find the rate of change in value on January 1, 2019
[tex]V(t)=14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t[/tex]
[tex]\\Let u= -0.158t,\frac{du}{dt}=-0.158[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}[/tex]
[tex]\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}[/tex]
In 2019, i.e. 2 years after, t=2
The rate of change of the value
[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}[/tex]
=[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.316}[/tex]= - 1675.38
The rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).
Explanation:The rate of change in the value of the equipment on January 1, 2019, can be found by taking the derivative of the given model equation V(t) = 14,550e^(-0.158t).
The derivative of V(t) with respect to t is dV/dt = -0.158(14,550)e^(-0.158t), which simplifies to dV/dt = -2,302.9e^(-0.158t).
Therefore, the rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).
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Consider a with 3 × 3 grid where each cell contains a number of coins; for example, the following represents a possible configuration of coins on the grid (the integer in each cell is the number of coins in that cell):12 3 11 8 42 13 0This configuration is transformed in stages as follows: in each step, every cell sends a coin to all of its neighbors (horizontally or vertically, not diagonally), but if there aren’t enough coins in a cell to send one to each of its neighbors, it sends no coins at all. For example, the above would result in the following after one step:11 2 34 7 21 12 2a) Show that every staring configuration results in stable configuration (one that no longer changes in this process), or repeatedly cycles through ???? configurations for some positive integer ???? (i.e., those same ???? configurations appear repeatedly in the sequence over and over as the transformation is applied).b) In the case that the initial configuration eventually cycles through ???? configurations, what are the possible values of ?????c) Either prove that for some positive integer ????, every configuration will reach a stable configuration or a repetition of a ????-cycle in ???? or fewer steps, or prove there is no such B.
Answer:
Step-by-step explanation:
Check attachment for solution
A movie theater offers 6 showings of a movie each day. A total of 1000 people come to see the movie on a particular day. The theater is interested in the number of people who attended each of the six showings. How many possibilities are there for the tallies for each showing for that day?
Answer:
2.03 × 10¹⁴ different possibilities.
Step-by-step explanation:
We want to know the different ways 1000 people can come out to watch a movie at 6 different times of the day.
This is a permutation and combination problem.
Dividing 1000 people amongst 6 movie showings = 1005C5 = 1005!/(1005-5)!5!
= (1005×1004×1003×1002×1001×1000!)/(1000!5!) = (1005×1004×1003×1002×1001)/5! = 2.03 × 10¹⁴ different possibilities.
i really need help:(
Answer: the height of the building is 40ft
Step-by-step explanation:
Looking at the right angle triangle formed,
With angle P as the reference angle, the length shadow of the building on ground represents the adjacent side of the right angle triangle.
The height of the building represents the opposite side of the right angle triangle.
a) to determine the height of the building, x, we would apply the tangent trigonometric ratio which is expressed as
Tan θ, = opposite side/adjacent side. Therefore, the equation becomes
Tan P = x/50
b) 0.8 = x/50
x = 50 × 0.8
x = 40 ft
Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. Based on your knowledge of the empirical rule, within what limits would you expect to find the number of infested fields, with probability approximately 95%? (Round your answers to three decimal places.) 4.29 X fields to 25.71 x fields What might you conclude if you found that x = 45 fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.
A. Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent. O
B. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes.
C. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that there are an indefinite number of trials.
D. Based on the limits above, it is likely that we would see x = 45, so all of the characteristics of a binomial experiment are satisfied.
Answer:
a.[tex]\mu=15[/tex]
b.[tex]\mu=7.8586 \ and \ \mu=22.1414[/tex]
c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.
Step-by-step explanation:
a.Binomial distribution is defined by the expression
[tex]P(X=k)=C_k^n.p^k.(1-p)^{n-k}[/tex]
Let n be the number of trials,[tex]n=100[/tex]
and p be the probability of success,[tex]p=15\%[/tex]
The mean of a binomial distribution is the probability x sample size.
[tex]\mu=np=100\times0.15=15[/tex]
b.Limits within which p is approximately 95%
sd of a binomial distribution is given as:[tex]\sigma=\sqrt npq\\q=1-p[/tex]
Therefore, [tex]\sigma=\sqrt(100\times0.015\times0.85)=3.5707[/tex]
Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.
[tex]sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414[/tex]
Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).
c. [tex]x=45[/tex] is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.
Final answer:
Using the empirical rule, we can determine the limits within which we would expect to find the number of infested fields with a 95% probability. If we found a number of infested fields that is higher than expected, it suggests that one of the characteristics of a binomial experiment may not be satisfied. Correct answer is B.
Explanation:
In this problem, we are given that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. We are asked to determine the limits within which we would expect to find the number of infested fields with a probability of approximately 95%. This situation can be modeled using the binomial distribution, as each field can be considered as a separate trial with two possible outcomes, infested or not infested.
According to the empirical rule, for a binomial distribution, we can expect about 95% of the outcomes to fall within two standard deviations of the mean. The mean number of infested fields can be calculated as 15% of the total number of fields, which is 15. The standard deviation of the number of infested fields can be determined using the formula σ = [tex]\sqrt{(npq)}[/tex], where n is the number of trials, p is the probability of success, and q is the probability of failure. In this case, n = 100, p = 0.15, and q = 0.85.
Using these values, we can calculate the standard deviation as σ = [tex]\sqrt{(100 * 0.15 * 0.85)}[/tex] ≈ 3.150. Therefore, we would expect to find about 95% of the number of infested fields within two standard deviations of the mean, which is between 15 - (2 * 3.150) = 8.700 and 15 + (2 * 3.150) = 21.300.
If we found that x = 45 fields were infested, we can conclude that this number is higher than what we would expect based on the binomial distribution. It is unlikely to observe such a high number of infested fields if the trials were independent and had only two possible outcomes. Therefore, we might suspect that one of the characteristics of a binomial experiment is not satisfied in this experiment.
Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes. Therefore, the correct answer is B.
Bob can row 13mph in still water. The total time to travel downstream and return upstream to the starting point is 2.6 hours. If the total distance downstream and back is 32 miles. Determine the speed of the river (current speed)
Answer:
"s" is the river speed 16 miles is the one-way distance
Time = distance / velocity
2.6 = [16 / (13 + s) ] + [16 / (13 -s) ]
That solves to s = 3 miles per hour
Step-by-step explanation:
Answer: the speed of the current is 3 mph
Step-by-step explanation:
Let x represent the speed of the river current.
Bob can row 13mph in still water. Assuming that while rowing upstream, he rowed against the current, this means that his total speed upstream is (13 - x) mph
Assuming that while rowing downstream, he rowed with the current, this means that his total speed downstream is (13 + x) mph.x
If the total distance downstream and back is 32 miles. Assuming the distance upstream and downstream is the same, then the distance upstream is 32/2 = 16 miles. Distance downstream is also 16 miles.
Time = Distance/speed
Time taken to row upstream is
16/(13 - x)
Time taken to row downstream is
16/(13 + x)
If total time spent is 2.6 hours, it means that
16/(13 - x) + 16/(13 + x) = 2.6
Cross multiplying, it becomes
16(13 + x) + 16(13 - x) = 2.6(13 + x)(13 - x)
= 208 + 16x + 208 - 16x = 2.6(169 - 13x + 13x - x²)
416 = 2.6(169 - x²)
416/2.6 = 169 - x²
160 = 169 - x²
x² = 169 - 160 = 9
x =√9
x = 3
Which of the following represents a null hypothesis? Group of answer choices Class A high school basketball teams who employ a sports psychologist will have a higher proportion ofwins over the course of the season than comparable teams who do not employ a sports psychologist. There will be no difference in rate of skill improvement between college gymnasts who practice meditation and those who do not. Does incorporating relaxation exercises into the daily practice routine of college vocal majors enhance their performance confidence? None of the above
In factual tests, an invalid speculation recommends that there is no importance between factors in a bunch of noticed information. In the given decisions, the assertion shows no distinction in the pace of ability improvement between school gymnasts who practice contemplation and the people who don't address the invalid speculation.
Explanation:In factual theory testing, an invalid speculation is an assertion recommending that no measurable relationship and importance exists in a bunch of noticed information between factors. It expects that any noticed contrasts are because of possibility. So in the given choices, the invalid speculation is: 'There will be no distinction in the pace of expertise improvement between school gymnasts who practice contemplation and the people who don't.' This assertion proposes no effect of the variable (reflection) on the result (ability improvement), which is what an invalid theory addresses in a trial of importance.
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The cost of flying a passenger plane from point A to point B is $70 comma 00070,000. The airline flies this route four times per day at 7 AM, 10 AM, 1 PM, and 4 PM. The first and last flights are filled to capacity with 240240 people. The second and third flights are only half full. Find the average cost per passenger for each flight.
Answer:
I. $291.67 per passenger
II. $583.33 per passenger
Step-by-step explanation:
The cost of flying the passenger plane from A to B is $70,000.
If the Capacity of the plane is 240 people.
(I)At 7A.M and 4P.M., the average cost per passenger is given as:
Average Cost=Total Cost/Number of Passengers
=70000/240=$291.67 per passenger
(II)At 10AM and 1PM, the flight is only half full, it has 120 passengers.
Average Cost = 70000/120=$583.33 per passenger
Answer: The cost will be $291.70 and $583.40 for the first and the last and for the second and third flights, respectively.
Step-by-step explanation: The cost is $70,000, the full capacity of seats is 240 and half capacity is 120.
In the first and last flights, which have the full capacity, the average cost per passenger is:
C = [tex]\frac{70,000}{240}[/tex] = 291.70
In the second and third ones, the average cost is:
C = [tex]\frac{70,000}{120}[/tex] = 583.4
The average cost per passenger for the first and the last flight are $291.70, while for the second and third are $583.4
The number of hours needed to paint a house, h, varies inversely with the number of painters, n. Four painters need 6 hours to paint a house. How many hours would it take 3 painters to paint the house?
Answer:
It would take 3 painters, a total number of 8 hours, to paint the house
Step-by-step explanation:
Let the number of hours be denoted by N, and the number of painters be denoted by P. The inverse relationship can be expressed as:
N α [tex]\frac{1}{P}[/tex]
Removing the proportionality symbol (α), introduces a constant of proportionality, which can be assumed to be C, in this example.
So, N = [tex]\frac{C}{P}[/tex]
When N = 6, P = 4
6 = [tex]\frac{C}{4}[/tex]
C = 24
So that, N = [tex]\frac{24}{P}[/tex]. This is the formula to use for determination of the other variable when one of them is known.
Now, given 3 painters, the number of hours needed can be obtained as follows:
N = [tex]\frac{24}{3}[/tex] = 8 hours
(A) In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have:
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.
(B) Now that you have one of the two variables in Part (A) isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers..
Answer:
C = 17/7 or 2 3/7
D = -4/7
Step-by-step explanation:
3C+4D=5
2C+5D=2
2C = 2 - 5D
C = 1 - 2.5D
3(1 - 2.5D) + 4D = 5
3 - 7.5D + 4D = 5
-3.5D = 2
D = -4/7
C = 1 - 2.5(-4/7)
C = 1 + 10/7 = 17/7
In the last homework, you analyzed the equation ˙x = cx − x 3 graphically, for c positive, zero and negative. Now perform a formal linear stability analysis, and compare your results with a graphical analysis
Answer:
Please find attached file for complete answer solution and explanation of same question.
Step-by-step explanation:
Use mathematical induction to prove that if L is a linear transformation from V to W, then L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) g
Answer:
The proof is shown in the explanation below.
Step-by-step explanation:
Analysis:
The proof by induction focuses on n. In this case, let n = 1, and [tex]L^{1}[/tex] will be a linear operator since [tex]L^{1} = L[/tex]
The exercise will show that [tex]L^{n}[/tex] is a linear operator on V and that [tex]L^{n+1}[/tex] is also a linear operator on V. This, follows that:
[tex]L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})[/tex]
Answer/Step-by-step explanation:
For the mathematical induction,
We show that the equation
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) is true for
L = 1,
Assume it is true for L = n and show that it is true for L = n + 1.
If L = 1, the equation become
(α1v1 + α2v2 +· · ·+αnvn)= α1(v1) + α2 (v2)+· · ·+αn (vn). Therefore, the Right Hand side(RHS) = Left Hand side(LHS)
When L = n, we assume the following is true
(α1nv1 + α2nv2 +· · ·+αnvn)= α1n(v1) + α2n (v2)+· · ·+αn (vn)
Then, when L = n + 1,
n +1 (α1v1 + α2v2 +· · ·+αvn)= α1(n +1) (v1) + α2(n +1) (v2)+· · ·+αn(n + 1)(vn).
Open the bracket,
n(α1v1 + α2v2 +· · ·+αvn) + α1v1 + α2v2 +· · ·+αnvn = α1n (v1) + α2v2 +· · ·+αvn ) + α1(v1) + α2v2+· · ·+αn(vn)
Since we assume the the equation is true for L = n, and eliminating some terms, then
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn)
According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health sciences is $51,541. The average starting salary for new college graduates in business is $53,901 (National Association of Colleges and Employers website, January 2015). Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $15,000.
a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000?
b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000?
c. What is the probability that a new college graduate in health sciences will earn a starting salary of less than $40,000?
d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences?
Answer:
Part (a) : 0.2297
Part (b) : 0.1112
Part (c) : 0.1469
Part (d) : 77,171
Step-by-step explanation:
Given info on Health Sciences:
Mean = $51,541
Standard Deviation = $11,000
Given info on Business:
Mean = $53,901
Standard Deviation = $15,000
Part (a)
Let X represents the new college graduate in business,
P (X ≥ 65,000) = 1 - P (X < 65,000)
= 1 - P ( z < [tex]\frac{65,000 - 53,901}{15,000}[/tex] )
= 1 - P ( z < 0.74)
= 1 - 0.77035
= 0.2297
Part (b)
Let Y represents the new college graduate in Health Sciences,
P (Y ≥ 65,000) = 1 - P (Y < 65,000)
= 1 - P ( z < [tex]\frac{65,000 - 51,541}{11,000}[/tex] )
= 1 - P ( z < 1.22)
= 1 - 0.88877
= 0.1112
Part (c)
Let Y represents the new college graduate in Health Sciences,
P (Y < 40,000) = P (Y < [tex]\frac{40,000-51,541}{11,000}[/tex])
= P ( z < -1.05 )
= 0.1469
Part (d)
To have a starting salary higher than 99%, the z-score = 2.33. Let A represents the salary of a new college graduate in health sciences higher than 99% of all starting salaries.
[tex]2.33 = \frac{A - 51,541}{11,000}[/tex]
[tex]A = 77,171[/tex]
77,171 new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in health sciences.
A soda filling machine is supposed to fill cans of soda with 12 fluid ounces. Suppose that the fills are actually normally distributed with a mean of 12.1 oz and a standard deviation of 0.2 oz. a) What is the probability of one can less than 12 oz
Answer:
30.85% probability of one can less than 12 oz
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 12.1, \sigma = 0.2[/tex]
a) What is the probability of one can less than 12 oz
This is the pvalue of Z when X = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 12.1}{0.2}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085
30.85% probability of one can less than 12 oz
g Suppose that when a certain lake is stocked with fish, the birth and death rates ˇ and ı are both inversely propor- (a) Show that ????1 p????2 P.t/D 2ktC P0 ; where k is a constant. (b) If P0 D 100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?
Answer:
k=1
P(12)=256
Step-by-step explanation:
The model for population involving birth and death is given as:
[tex]\frac{dP}{dt}=(b-d)P[/tex]
where b=birth rate and d=death rate.
If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]
[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.
Substituting b and d into our model
[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant
Simplifying using indices
[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]
Next, we Separate Variables and Integrate both sides
[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]
[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]
[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration
[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]
When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.
(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]
Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]
Therefore the population after a year is 256.
The population of fish after 1 year would be 285.
Let P be the population of fish at any time t months.
We are given:
Initial population: [tex]P_0[/tex] = 100 fishPopulation after 6 months: P(6) = 169 fishWe need to determine the population after 12 months.
The logistic growth model can be expressed as:
dP/dt = rP(1 - P/K)
where r is the intrinsic growth rate and K is the carrying capacity of the environment. Here, since the birth and death rates are inversely proportional to √P, we can infer that the growth rate might be proportional to √P, leading us to use a simplified logistic growth equation.
Given P(6) = 169, let’s find the intrinsic growth rate , r. From population changes between 0 and 6 months:
Let’s assume the logistic growth function in the form:
P(t) = [tex]P_0 e^{rt}[/tex]
We solve for r:
169 = [tex]100 e^{6r}[/tex]
1.69 = [tex]e^{6r}[/tex]
Taking natural logarithms:
ln(1.69) = 6r
0.524 = 6r
r = 0.0873/month
Now, to find P(12):
P(12) = [tex]100 e^{0.0873*12}[/tex]
= [tex]100 e^{1.048}[/tex]
= 100 * 2.853
= 285.3 fish
So, the population after 1 year is approximately 285 fish.