Answer:
Bryant: 2059.5 (RIP)
Durant:2353.5
Step-by-step explanation:
To find how many points Kevin Durant and Kobe Bryant scored, set up equations based on the given information. Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.
Explanation:The question asks us to determine how many points Kevin Durant and Kobe Bryant scored individually during the 2012-2013 NBA regular season given that together they scored 4413 points and Kobe Bryant scored 147 fewer points than Kevin Durant.
To solve this, we can set up two equations based on the information provided:
We can substitute the second equation into the first to find Durant's score:
Now that we have Durant's score, we can use it to find Bryant's score:
Therefore, Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.
A radioactive isotope is unstable, and will decay by emitting a particle, transforming into another isotope. The assumption used to model such situations is that the amount of radioactive isotope decreases proportionally to the amount currently present.(a) Let N(t) designate the amount of the radioactive material present at time t and let N0 = N(0). Write and solve the IVP for radioactive decay of a radioactive material.(b) The half-life of a radioactive material is the time required for it to reach one-half of the original amount. What is the half-life of a material that in one day decays from 12 mg to 9 mg?
Answer:
Part a: The equation is [tex]N=N_{0}e^{-kt}[/tex]
Part b: The half life of the material is 2.4 days.
Step-by-step explanation:
Part a
The relation is given as
[tex]\frac{dN}{dt}=-kN[/tex]
Rearranging the equation gives
[tex]\frac{dN}{N}=-kdt[/tex]
Integrating and simplifying the equation as
[tex]\int \frac{dN}{N}=\int (-kdt)\\ln N=-kt+lnC\\N=e^{-kt+lnC}\\N=Ce^{-kt}[/tex]
This is the equation of radioactive decay of the radioactive material. For estimation of C consider following IVP where t=0,N=N_o so the equation becomes
[tex]N=Ce^{-kt}\\N_0=Ce^{-k(0)}\\N_0=Ce^{0}\\N_0=C[/tex]
Now substituting the value of C in the equation gives
[tex]N=N_{0}e^{-kt}[/tex]
This is the relation of concentration of unstable radioactive material at a given time .
Part b
From the given data the equation becomes as
[tex]N=N_{0}e^{-kt}\\9=12\times e^{-k*1}\\e^{-k}=\frac{9}{12}[/tex]
Now half like is defined as the time when the quantity is exact half, i.e. N/N_o =0.5 so
[tex]N=N_{0}e^{-kt}\\6=12\times e^{-k*t}\\e^{-kt}=\frac{6}{12}\\{e^{-k}}^t}=0.5\\(0.75)^t=0.5\\So\, t\, is\, given\, as \\t=\frac{ln 0.5}{ln 0.75}\\t=2.4 \, days[/tex]
So the half life of the material is 2.4 days.
A single card is drawn at random from each of six well-shuffled decks of playing cards. Find the probability that all six cards drawn are different.
Answer:
0.74141
Step-by-step explanation:
There are 52 cards in total
since each card is different,
the probability = number of favorable cards / total number of outcomes
P(C₁) = number of favorable cards / total number of outcomes
= [tex]\frac{52}{52}[/tex]
P(C₂, C₁) = number of favorable cards / total number of outcomes
= [tex]\frac{51}{52}[/tex]
P(C₃,C₁ ∩ C₂) = [tex]\frac{50}{52}[/tex]
P(C₄,C₁ ∩ C₂ ∩ C₃) = [tex]\frac{49}{52}[/tex]
P(C₅,C₁ ∩ C₂ ∩ C₃ ∩ C₄) = [tex]\frac{48}{52}[/tex]
P(C₆,C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅) = [tex]\frac{47}{52}[/tex]
General multiplication rule
P(A) =P(C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅ ∩ C₆)
= [tex]\frac{52}{52}. \frac{51}{52}. \frac{50}{52}. \frac{49}{52}. \frac{48}{52} .\frac{47}{52}[/tex]
= 8,808,975 / 11,881,376
= 0.74141
Given the following data, find the weight that represents the 73rd percentile. Weights of Newborn Babies 8.2 6.6 5.6 6.4 7.9 7.1 6.5 6.0 7.8 8.0 6.8 8.8 9.3 7.7 8.8
To find the 73rd percentile of the given data, we need to arrange the weights in ascending order and calculate the rank. Then, we interpolate to find the weight at the desired percentile.
Explanation:To find the weight that represents the 73rd percentile, we need to follow these steps:
Arrange the weights in ascending order: 5.6, 6.0, 6.4, 6.5, 6.6, 6.8, 7.1, 7.7, 7.8, 7.9, 8.0, 8.2, 8.8, 8.8, 9.3Calculate the rank of the percentile by using the formula: rank = (percentile/100) * (n - 1) + 1, where n is the number of data points. In this case, n = 15.The rank of the 73rd percentile is (73/100) * (15 - 1) + 1 = 11.02.Since the rank is not a whole number, we need to interpolate to find the weight. The weight at rank 11 is 8.0 and the weight at rank 12 is 8.2.Use the formula for interpolation: weight = weight at lower rank + (rank - lower rank) * (weight at higher rank - weight at lower rank), which gives: weight = 8.0 + (11.02 - 11) * (8.2 - 8.0) = 8.02.Therefore, the weight that represents the 73rd percentile is 8.02.
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Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non-wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 727 that were judged defective, inspector B found 756 such joints, and 940 of the joints were judged defective by at least one of the inspectors.
Suppose that one of the 10,000 joints is randomly selected.
(a) What is the probability that the selected joint was judged to be defective by neither of the two inspectors? (Enter your answer to four decimal places.)
(b) What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A? (Enter your answer to four decimal places.)
Answer:
(a) The probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.
(b) The probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.
Step-by-step explanation:
The sample of joints randomly selected is, n = 10,000.
Number of joints judged defective by inspector A is, n (A) = 727.
The probability that a joint is judged defective by inspector A is:
[tex]P(A)=\frac{n(A)}{n} =\frac{727}{10000} =0.0727[/tex]
Number of joints judged defective by inspector B is, n (B) = 756.
The probability that a joint is judged defective by inspector B is:
[tex]P(B)=\frac{n(B)}{n} =\frac{756}{10000} =0.0756[/tex]
Number of joints judged defective by at least one of the inspectors is,
n (At least 1) = 940.
The probability that a joint is judged defective by at least one of the inspectors is:
[tex]P(At\ least\ 1)=\frac{n(At\ least\ 1)}{n} =\frac{940}{10000}=0.094[/tex]
(a)
Compute the probability that a selected joint was judged to be defective by neither of the two inspectors as follows:
P (At least 1) = 1 - P (Less than 1)
= 1 - P (None)
P (None) = 1 - P (At least 1)
[tex]=1-0.094\\=0.906[/tex]
Thus, the probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.
(b)
Compute the probability that a selected joint was judged to be defective by inspector B but not by inspector A as follows:
P (B but not A) = P (At least 1) - P (A)
[tex]=0.094-0.0727\\=0.0213[/tex]
Thus, the probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.
An airport official wants to prove that the proportion of delayed flights for Airline A (denoted as p1) is less than the proportion of delayed flights for Airline B (denoted as p2). Random samples for both airlines after a storm showed that 51 out of 200 flights for Airline A were delayed, while 60 out of 200 of Airline B's flights were delayed. The test statistic for this problem is -1.00. The p-value for the test statistic for this problem is:
A. p = 0.3413 B. p = 0.0668 C. p = 0.1587 D. p = 0.0228
Answer:
The p-value for the test statistic is 0.1587.
Step-by-step explanation:
As the sample size for both the airlines are large, according to the central limit theorem the sampling distribution of sample proportion follows a normal distribution.
The test statistic for the difference between two proportions is:
[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{P(1-P)\frac{1}{n_{1}}+\frac{1}{n_{2}} } }[/tex]
The test statistic value is, z = -1.00
The p-value of the test statistic is:
[tex]P (Z<-1.00)=1-P(Z<1.00)=1-0.8413=0.1587[/tex]
**Use the standard normal table for the probability.
Thus, the p-value for the test statistic is 0.1587.
What is the critical value at the 0.05 level of significance for a goodness-of-fit test if there are six categories? Select one: a. 3.841 b. 5.991 c. 7.815 d. 11.070
Answer:
The correct option is (d) 11.070
Step-by-step explanation:
The test statistic for Goodness of fit test for k observations is:
[tex]\chi^{2}=\sum\frac{(Observed-Expected)^{2}}{Expected}[/tex]
This statistic follows a Chi-square distribution with (k - 1) degrees of freedom and α level of significance.
Here α = 0.05 and degrees of freedom is, k - 1 = 6 - 1 = 5 d.f.
Use the chi-square table for the critical value.
[tex]\chi^{2}_{(5)}=11.070[/tex]
Thus, the correct option is (d).
Consider the points below. P(0, −4, 0), Q(5, 1, −3), R(5, 2, 1) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Correct: Your answer is correct. (b) Find the area of the triangle PQR.
a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].
b) The area of the triangle PQR is 32.09 square units.
Given points are P(0, -4, 0), Q(5, 1, -3), T(5, 2, 1)
a) [tex]PQ= < 5, 1, -3 > - < 0, -4, 0 >[/tex]
[tex]PQ= < 5-0, 1-(-4), -3-0 >[/tex]
[tex]PQ= < 5, 5, -3 >[/tex]
[tex]PR= < 5, 2, 1 > - < 0, -4, 0 >[/tex]
[tex]PR= < 5-0, 2-(-4), 1-0 >[/tex]
[tex]PR= < 5, 6, 1 >[/tex]
b) The cross product of PQ and PR is
[tex]PQ\times PR=\left[\begin{array}{ccc}i&j&k\\5&5&-3\\5&6&1\end{array}\right][/tex]
[tex]PQ\times PR=(5+18)i-(5+15)j+(30-25)k[/tex]
[tex]PQ\times PR=23i-20j+10k[/tex]
The magnitude of [tex]PQ\times PR[/tex] is
[tex]|PQ\times PR| =|23i-20j+10k|[/tex]
[tex]|PQ\times PR| =\sqrt{23^2+(-20)^2+10^2}[/tex]
[tex]|PQ\times PR| =\sqrt{529+400+100}[/tex]
[tex]|PQ\times PR| =\sqrt{1029}[/tex]
[tex]|PQ\times PR| =32.09[/tex]
Therefore,
a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].
b) The area of the triangle PQR is 32.09 square units.
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Final answer:
To find a nonzero vector orthogonal to the plane through points P, Q, and R, calculate the cross product of two vectors in the plane. The area of triangle PQR can be found by dividing the length of the cross product vector by 2.
Explanation:
To find a nonzero vector orthogonal to the plane through points P, Q, and R, we can calculate the cross product of two vectors in the plane. Let's take vectors PQ and PR. Using the formula for cross product, we have:
PQ x PR = (5 - 0, 1 - (-4), -3 - 0) = (5, 5, -3)
Therefore, the vector (5, 5, -3) is orthogonal to the plane through points P, Q, and R.
To find the area of triangle PQR, we can use the length of the cross product vector and divide it by 2. The length of the cross product vector PQ x PR is:
|PQ x PR| =√(5² + 5² + (-3)²) = sqrt(59)
Dividing by 2, we get the area of triangle PQR as:
Area = √(59) / 2
Suppose we toss a fair coin three times in a row. Let A be the event of exactly 2 tails. Let B be the event that the first 2 tosses are tails. Let C be the event that all three tosses are tails. What is the probability of the union of A, B, and C
Answer: 1/2
Step-by-step explanation:
Since it's a fair coin, the probability of tossing head, P(H) = probability of tossing tail, P(T) = 1/2.
Hence, for the event of A being exactly 2 tails, then we have a possibility of:
A= [ HTT, THT, TTH]
For the event of B being first two tosses resulting in tail, it becomes:
B= [TTH)
For event of C being that the three tosses are tail, it becomes:
C=[TTT]
Hence, union of Event A, B and C is given as:
AuBuC= [HTT, THT, TTH, TTT]
Prob [AuBuC] = [ (1/2 * 1/2 * 1/2)] * 4
P[AuBuC] = (1/8) *4
P(AuBuC) = 1/2.
A batch contains 31 bacteria cells. Assume that 12 of the cells are not capable of cellular replication. Six cells are selected at random, without replacement, to be checked for replication. Round your answers to four decimal places (e.g. 98.7654).What is the probability that all six cells of the selected cells are able to replicate
Answer:
0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
The combinations formula is important in this problem:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Desired outcomes
31 cells
19 are able to replicate.
We pick six.
The order is not important, that is, if it had been 2 cells, cell A and cell B would be the same outcome as cell B and cell A. So we use the combinations formula to find the number of desired outcomes.
It is a combination of 6 from 19(cells who are able to replicate).
[tex]D = C_{19,6} = \frac{19!}{6!(19 - 6)!} = 27132[/tex]
Total outcomes
31 cells
6 are picked.
[tex]T = C_{31,6} = \frac{31!}{6!(31 - 6)!} = 736281[/tex]
What is the probability that all six cells of the selected cells are able to replicate?
[tex]P = \frac{D}{T} = \frac{27132}{736281} = 0.0369[/tex]
0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.
Answer:
The answer would be 98.8
Step-by-step explanation:
Resin being you must round simply to the nearest hundreds
A bacteria culture starts with 100 bacteria and doubles in size every half hour.
(a) How many bacteria are there after 4 hours?
bacteria
(b) How many bacteria are there after t hours?
y = bacteria
(c) How many bacteria are there after 40 minutes? (Round your answer to the nearest whole number.)
bacteria
(d) Graph the population function.
Answer:
y=25600
y=100(2)^{2t}
y=252
Step-by-step explanation:
A bacteria culture starts with 100 bacteria and doubles in size every half hour.
[tex]y=100(2)^{2x}[/tex]
where x represents the number of hours
(a) t= 4 hours
Plug in 4 for t and find out the number of bacteria y
[tex]y=100(2)^{2 \cdot 4}\\y=25600[/tex]
25600 bacteria
[tex]y=100(2)^{2t}[/tex]
convert 40 seconds into hour
40 divide by 60 =2/3
[tex]y=100(2)^{2\frac{2}{3} }\\y=100(2)^{\frac{4}{3} }\\\\y=252[/tex]
The graph is attached below
Find the radian measure of an angle at the center of a circle with radius 70.0 cm that intercepts an arc length of 127 cm. The measure of the angle is nothing.g
Answer:
1.8413 is the radian measure of angle at the center of circle.
Step-by-step explanation:
We are given the following in the question:
Radius,r = 70.0 cm
Arc length,s = 127 cm
Formula:
[tex]\theta = \dfrac{s}{r}[/tex]
where [tex]\theta[/tex] is the angle measure in radians, s is the intercepted arc and r is the radius of the circle.
Putting the values, we get,
[tex]\theta = \dfrac{127}{70} = 1.8143[/tex]
Thus, 1.8413 is the radian measure of angle at the center of circle.
The radian measure of an angle in a circle, given an arc length of 127 cm and radius of 70 cm, is approximately 1.814. This is computed by dividing the arc length by the radius.
Explanation:In mathematics, the measure of an angle in radians in a circle is given by the ratio of the length of the arc that it subtends and the radius of the circle in which this occurs. This is represented by the formula: θ = s / r where s is the arc length and r is the radius.
Given the radius r = 70 cm and arc length s = 127 cm, you can substitute these values into the formula to get the radian measure of the angle. Therefore, by substitution we get: θ = 127 cm / 70 cm which simplifies to approximately 1.814 radians.
To note, a radian is the standard unit of angular measure, used in many areas of mathematics. An angle's measurement in radians is numerically equal to the length of a corresponding arc of a unit circle, hence why this method works.
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A bee with a velocity vector r ′ ( t ) r′(t) starts out at ( 3 , − 6 , 10 ) (3,−6,10) at t = 0 t=0 and flies around for 9 9 seconds. Where is the bee located at time t = 9 t=9 if ∫ 9 0 r ′ ( u ) d u = 0
The velocity vector of the bee is simply the rate of change of the position of the bee.
The bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.
The given parameters are:
[tex]\mathbf{r'(t) = (3,-6,10)}[/tex], when t = 0
A vector is represented as:
[tex]\mathbf{r'(t) = xi + yj + zk}[/tex]
From the question, we have:
[tex]\mathbf{\int\limits^9_0 {r'(u)} \, du = 0}[/tex]
Integrate
[tex]\mathbf{r(u)|\limits^9_0 = 0}[/tex]
Expand
[tex]\mathbf{r(9) - r(0) = 0}[/tex]
Rewrite as:
[tex]\mathbf{r(9) = r(0) }[/tex]
Recall that: [tex]\mathbf{r'(t) = (3,-6,10)}[/tex]
So, we have:
[tex]\mathbf{r(9) = 3i -6j + 10k }[/tex]
Also, we have:
[tex]\mathbf{xi + yj + zk = 3i -6j + 10k }[/tex]
By comparison;
[tex]\mathbf{x = 3}[/tex]
[tex]\mathbf{y = -6}[/tex]
[tex]\mathbf{z = 10}[/tex]
So, the bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.
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After 9 seconds, the bee returns to its initial position (3, -6, 10) because the integral of the velocity function from 0 to 9 seconds is zero, indicating that the net displacement of the bee in these 9 seconds is zero.
Explanation:The question is essentially asking where the bee is located based on its velocity vector, initial position, and the change in time. To answer this, we will use the concept of integrals from calculus. The integral of the velocity function, r′(t), over a period of time gives the displacement of the bee. However, it's given that ∫9 0 r′(u) du = 0. This tells us that the net displacement of the bee in that 9 seconds is zero.
Displacement refers to the change in position of an object, so if the net displacement in these 9 seconds is zero, then the bee is at the same position at t=9 as at t=0. In other words, after travelling around for 9 seconds, the bee returns to the same starting point. Therefore, at time t=9, the bee is located at the same initial position, that is, (3, -6, 10).
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A computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent.
Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by
p(x)={ cx for x=1,2,3,4, or 5
{ 0 otherwise
where c is a constant.
Find P(X = 2).
Answer:
Step-by-step explanation:
given that a computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent
Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by
p(x)={ cx for x=1,2,3,4, or 5
{ 0 otherwise
where c is a constant.
Let us write pdf as follows
x 1 2 3 4 5 Total
p c 2c 3c 4c 5c 15c
Since total probability = 1
we get
[tex]c=\frac{1}{15}[/tex]
P(X=2)=2c = [tex]\frac{2}{15}[/tex]
A researcher wants to obtain a sample of 30 preschool children consisting of 10 two-year-old children, 10 three-year-old, and 10 four-year-old children. Assuming that the children are obtained from local daycare centers, this researcher should use ____ sampling.
Answer:
Quota sampling
Step-by-step explanation:
Quota sampling is a type of sampling method in which a representative data is collected from a group. It is a non probability based sampling method, a type of stratified sampling. Quota sampling is required when researchers are working with limited time and are without a sampling frame. Quota sampling method helps to determine the characteristics of a subgroup under research. It is an highly accurate method of sampling.
This week, one of the topics is describing sampling distributions. In project 2 you will be using female heights to answer a variety of questions. As practice for this project, please address the bulleted items below for male heights. It will be necessary to know that the average height for men is assumed to be 70 inches with a standard deviation of 4 inches. To receive full credit, you also need to comment on two other posts from your classmates (you can just state whether or not you agree with their solution and if not, what you did differently).George Washington was 6 feet tall. Find the z-score for George Washington. Find the probability that a randomly selected individual will be as tall or taller than George Washington. Interpret both the z-score and the probability in a sentence.
Answer:
a) 0.5
b) 0.309
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 70 inches
Standard Deviation, σ = 4 inches
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Height of George Washington = 6 feet = [tex]6\times 12 = 72\text{ inches}[/tex]
x = 72
a) z-score
[tex]z_{score} = \displaystyle\frac{72-70}{4} = 0.5[/tex]
Interpretation:
This, mean that George Washington is 0.5 standard deviation more than the mean height of for men.
b) P(individual will be as tall or taller than George Washington)
[tex]P( x \geq 72) = P( z \geq 0.5)[/tex]
[tex]= 1 - P(z < 0.5)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \geq 72) = 1 - 0.691 = 0.309 = 30.9\%[/tex]
Interpretation:
0.309 is the probability that a randomly selected individual will be as tall or taller than George Washington.
An astronaut with a mass of 90 kg (including spacesuit and equipment) is drifting away from his spaceship at a speed of 0.20 m/s with respect to the spaceship. The astronaut is equipped only with a 0.50-kg wrench to help him get back to the ship.
A). With what speed must he throw the wrench for his body to acquire a speed of 0.10 m/s?
B). In what direction relative to the spaceship must he throw the wrench? Towards the spaceship or away from the spaceship.
Answer:
Part A:
[tex]v_w=53.9\ m/s[/tex]
Part B:
Wrench is thrown away from the spaceship
Step-by-step explanation:
This the problem related to conservation of momentum.
According to the conservation of momentum:
Initial Momentum=Final Momentum
[tex]m_av_a=m_wv_w+(m_a-m_v)v[/tex]
where:
[tex]m_a[/tex] is the mass of astronaut
[tex]m_w[/tex] is the mass of wrench
[tex]m_a-m_w[/tex] is the mass when wrench is thrown
[tex]v_a[/tex] is the speed of astronaut
[tex]v_w[/tex] is the speed of wrench
v is the speed acquired
Part A:
(+ve sign for away from ship),( -ve sign for towards ship)
v= -0.10 m/s
[tex]90*0.2=0.5*v_w+(90-0.5)(-0.1)\\v_w=53.9\ m/s[/tex]
Part B:
Since velocity is +ve as calculated above and according to conditions:
+ve sign for away from ship -ve sign for towards shipWrench is thrown away from the spaceship
In an isolated environment, a disease spreads at a rate proportional to the product of the infected and non-infected populations. Let I(t) denote the number of infected individuals. Suppose that the total population is 2000, the proportionality constant is 0.0002, and that 1% of the population is infected at time t=0. Write down the intial value problem and the solution I(t).
Answer:
Expression: N = C·L·l(t)· T + 20
The initial value problem and solution are expressed as a first order differential equation.
Step-by-step explanation:
First, gather the information:
total population, N = 2 000
Proportionality constant, C = 0.0002
l(t) number of infected individuals = l(t)
healthy individuals = L
The equation is given as follows:
N = C·L·l(t)
However, there is a change with time, so the expression will be:
[tex]\frac{dN}{dt}[/tex] = C·L·l(t)
multiplying both sides by dt gives:
dN = C·L·l(t)
Integrating both sides gives:
[tex]\int\limits^a_b {dN} \, dt[/tex] = [tex]\int\limits^a_b {CLl(t)} \, dt[/tex]
N = C·L·l(t)· T + K
initial conditions:
T= 0, N₀ = (0.01 ₓ 2 000) = 20
to find K, plug in the values:
N₀ = K
20 = K
At any time T, the expression will be:
N = C·L·l(t)· T + 20 Ans
To verify and compare data across variables, data can be broken down and studied by subgroups in a process called _____. A. convenience sampling B. stratification C. cross tabulation D. non probability sampling E. back-translation
Answer:
B. stratification
Step-by-step explanation:
Stratification of data is a process in which data is agrupated into some groups or strats depending on its value.
This action is helpful when the person has a lot of data that can be clasified on several variables that will make their analysis easier to manage.
Answer:convenience sampling
Step-by-step explanation:
This is the type of sampling where the first available primary data source will be used without additional requirements.
Find the balance of $2,500 deposited at 3% compounded annually for 3 years
Answer:
The balance will be $2,731.82.
Step-by-step explanation:
This problem can be solved by the following formula:
[tex]A = P(1 + r)^{t}[/tex]
In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.
In this problem, we have that:
[tex]P = 2500, r = 0.03 t = 3[/tex]
We want to find A
So
[tex]A = P(1 + r)^{t}[/tex]
[tex]A = 2500(1 + 0.03)^{3}[/tex]
[tex]A = 2,731.82[/tex]
The balance will be $2,731.82.
Answer:
2731.8
Step-by-step explanation:
2500 deposited at 3 percent interest annually equation form
2500(1+(3/100))
for 3 years power the bract by 3
2500(1+(3/100))^3=2731.81
aka 2732
A test has the capability to detect a defect with 98% accuracy. It also tends to indicate a defect when there is none about 1% of the time. About 1 in 2000 test subjects has a defect. What is the probably that a subject chosen at random and is indicated to have a defect, really does have a defect? (0.25 points)
Answer:
OUR ANSWER IS =0.0467
Step-by-step explanation:
P(detect/defect)=0.98
P(detect/not defect)=0.01
P(defect)=1/2000=0.0005
P(not defect)=1-0.0005=0.9995
P(defect/detect)=P(detect/defect)*P(defect)/[P(detect/defect)*P(defect)+P(detect/not defect)*P(not defect)]
=0.98*0.0005/[0.98*0.0005+0.01*0.9995]
=0.00049/[0.00049+0.009995]
=0.00049/0.010485
=0.0467
For the month of May in a certain city, the probability that the weather on a given day is cloudy is 0.77. Also in the month of May in the same city, the probability that the weather on a given day is cloudy and snowy is 0.27. What is the probability that a randomly selected day in May will be snowy if it is cloudy?The probability is approximately nothing. (Round to three decimal places as needed.)
Answer:
0.3506 is the required probability.
Step-by-step explanation:
We are given the following in the question:
A: Weather on a given day is cloudy
B: Weather on a given day is snowy
[tex]P(A) = 0.77\\P(A\cap B) = 0.27[/tex]
We have to find the probability that the randomly selected day in May will be snowy if it is cloudy.
That is we have to evaluate P(B|A)
[tex]P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\\\P(B|A) = \dfrac{0.27}{0.77}\\\\P(B|A) = 0.3506[/tex]
Thus, 0.3506 is the probability hat a randomly selected day in May will be snowy if it is cloudy.
U.S. craft-beer breweries (breweries that make fewer than 6 million barrels annually and are less than 25% owned by big breweries) have been doing a booming business. The number of these small breweries from 2008 through 2012 can be modeled by using a quadratic function of the form f(t) = at² + bt + c where a, b, and c are constants and t is measured in years, with t = 0 corresponding to 2008.
(a) Find a, b, and c if f(0) = 1547, f(2) = 1802, and f(4) = 2403.
(b) Use the model obtained in part (a) to estimate the number of craft-beer breweries in 2016, assuming that the trend continued. (Round your answer to the nearest integer.)
a) The values of a, b and c are [tex]\frac{173}{4}, 41[/tex] and [tex]1547[/tex] respectively.
b) The number of craft beer breweries in 2016 will be 4643.
The quadratic equation holds an important significance in mathematics and is used in many situations, like calculating area or finding profit. When two linear expressions get multiplied, the resultant expression is a quadratic expression.
The number of craft beer breweries is modeled by a quadratic equation [tex]f(t)=at^2+bt+c[/tex].
In 2008, t=0.
a) Here, it is given that [tex]f(0)=1547[/tex], [tex]f(2)=1802[/tex] and [tex]f(4)=2403[/tex].
Now, use these values to find a, b and c as follows:
[tex]f(0)=a(0)^2+b(0)+c[/tex]
[tex]1547=c[/tex]
[tex]f(2)=a(2)^2+2b+c[/tex]
[tex]1802=4a+2b+1547[/tex]
[tex]255=4a+2b[/tex]
[tex]f(4)=a(4)^2+b(4)+c[/tex]
[tex]2403=16a+4b+1547[/tex]
[tex]856=16a+4b[/tex]
[tex]214=4a+b[/tex]
Now, setting the equations [tex]4a+2b=255[/tex] and [tex]4a+b=214[/tex] equal to each other:
[tex]4a=255-2b[/tex]
[tex]4a=214-b[/tex]
[tex]255-2b=214-b[/tex]
[tex]b=255-214[/tex]
[tex]b=41[/tex]
Now plugging the value of b into the equation [tex]4a+b=214[/tex], we get
[tex]4a+b=214[/tex]
[tex]4a=214-b[/tex]
[tex]a=\frac{214-b}{4}[/tex]
[tex]a=\frac{214-41}{4}[/tex]
[tex]a=\frac{173}{4}[/tex]
So, we have [tex]a=\frac{173}{4}[/tex], [tex]b=41[/tex] and [tex]c=1547[/tex]
b) Using the values calculated in part (a), the quadratic equation becomes
[tex]f(t)=\frac{173}{4}t^2+41t+1547[/tex]
Now, in 2016, the value of t=8
So, the number of craft beer breweries in 2016 will be calculated by [tex]f(8)[/tex]:
[tex]f(8)=\frac{173}{4}(8)^2+41(8)+1547[/tex]
[tex]= 2768+328+1547[/tex]
[tex]= 4643[/tex]
Therefore, the number of craft beer breweries in 2016 will be 4643.
Learn more about the quadratic function here:
https://brainly.com/question/29775037.
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We first find the constants by substituting in the provided points into the quadratic equation to create a series of equations. Solving them gives us a = 61.5, b = 47, and c = 1547. To predict the number of breweries in 2016, we substitute t = 8 into the function to get approximately 3455 breweries.
Explanation:To find the constants a, b, and c, we will substitute the given points (0, 1547), (2, 1802), and (4, 2403) into the equation f(t) = at² + bt + c. This will give us three equations:
1547 = a(0)² + b(0) + c 1802 = a(2)² + b(2) + c 2403 = a(4)² + b(4) + cSolving these gives us a = 61.5, b = 47, and c = 1547. Therefore the quadratic function is: f(t) = 61.5t² + 47t + 1547.
For (b), we want to find the number of breweries in 2016, which is 8 years from 2008. Substituting t = 8 into our quadratic function gives us f(8) = 61.5(8²) + 47(8) + 1547 which is approximately 3455. So, the model estimates there will be around 3455 craft-beer breweries in 2016.
Learn more about quadratic function here:https://brainly.com/question/35505962
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How do I solve this using the substitution method 3x+4y=0 2x+5y=7
Answer:
x=4
y=-3
Step-by-step explanation:
3x+4y=0
2x+5y=7
Change 3x+4y=0 into y= -3/4x
so substitute for y
2x-15/4x=7
-1 3/4x = 7
x=-4
substitue x for y and get
-3/4 * 4
y=-3
Answer: x = - 4
y = 3
Step-by-step explanation:
The given system of simultaneous equations is expressed as
3x+4y=0 - - - - - - - - - - - - -1
2x+5y=7- - - - - - - - - - - - - - - 2
From equation 1, we would make x the subject of the formula. Firstly, we would subtract 4y from the Left hand side and the right hand side of the equation. It becomes
3x +4y - 4y = 0 - 4y
3x = - 4y
We would divide the Left hand side and the right hand side of the equation by 3. It becomes
3x/3 = - 4y/3
x = - 4y/3
Substituting x = - 4y/3 into equation 2, it becomes
2 × - 4y/3 +5y = 7
- 8y/3 + 5y = 7
(- 8y + 15y)/3 = 7
7y/3 = 7
Cross multiplying, it becomes
7y = 21
Dividing the left hand side and the right hand side of the equation by 7, it becomes
7y/7 = 21/7
y = 3
Substituting y = 3 into x = - 4y/3, it becomes
x = - 4× 3/3
x = - 4
A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random. (a) When he flips the coin, what is the probability that it will show heads? (b) The coin shows heads. Now what is the probability that it is the fair coin?
Answer:
a) probability of choosing heads= 1/2 (50%)
b) probability of choosing the fair coin knowing that it showed heads is= 1/3 (33.33%)
Step-by-step explanation:
Since the unfair coin can have 2 heads or 2 tails , and assuming both are equally possible . then
probability of choosing the fair coin (named A)= 1/2
probability of choosing an unfair coin with 2 heads (named B)= (1-1/2)*1/2= 1/4
probability of choosing an unfair coin with 2 tails (named C)= (1-1/2)*(1-1/2)= 1/4
then
probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B + probability of choosing C * probability of getting heads from C =
1/2*1/2 + 1/4*1 + 1/4*0 = 2/4 = 1/2
the probability of choosing the fair coin knowing that it showed heads is
P(A/B) = P(A∩B)/P(B)
denoting event A= the coin is fair and event B= the result is heads
P(A∩B) = 1/2*1/2 = 1/4
but since we know now that that the unfair coin is not possible , the probability of choosing heads is altered:
P(B)=probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B = 1/2*1/2+1/2*1 = 3/4
then
P(A/B) = P(A∩B)/P(B) = (1/4)/(3/4) = 1/3
then the probability is 1/3
Suppose there is a correlation of 0.87 between the length of time a person is in prison and the amount of aggression the person displays on a psychological inventory. This means that spending a longer amount of time in prison causes people to become more aggressive. a. Trueb. False
Answer:
b. False
See explanation below
Step-by-step explanation:
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case we know that r =0.87. But we can't conclude that the linear correlation coefficient represent a cause-effect relationship, since the linear correlation coefficient measures the linear dependency between two variables, but we can have other types of association between the variables and that's not measured by the Pearson correlation coeffcient.
So for this case the answer woudl be
b. False
Final answer:
The claim that a correlation of 0.87 between time in prison and aggression implies causation is false because correlation does not equal causation. There may be other factors influencing aggression, and the correlation observed might be spurious.
Explanation:
The statement that a correlation of 0.87 between the length of time a person is in prison and the amount of aggression displayed means that spending a longer time in prison causes people to become more aggressive is false.
Correlation does not imply causation. Just because two variables are correlated does not mean that one variable causes the change in the other. There could be other factors or variables that contribute to the aggression, which are not accounted for in simply observing the correlation.
Negative experiences do increase aggression, as people are more prone to aggress when experiencing negative emotions such as frustration, pain, or being in a bad mood. In a prison context, the aggression could also stem from the social structure of prison life and not merely the length of time served.
The Stanford prison experiment by Philip Zimbardo is a notable study highlighting how quickly people can turn to aggressive behaviors based on the roles they are assigned within a prison environment.
Which equation shows a true relationship between the angle, arc length, and area of the sector shown?
Answer:c
Step-by-step explanation:because the arc length is x that is variable and multiplying it by the the formula gives the right answers.
Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight (S). Consider observing the direction for each of three successive vehicles. (Enter your answers in set notation. Enter EMPTY or ∅ for the empty set.)
(a) List all outcomes in the event A that all three vehicles go in the same direction.
(b) List all outcomes in the event B that all three vehicles take different directions.
(c) List all outcomes in the event C that exactly two of the three vehicles turn right.
(d) List all outcomes in the event D that exactly two vehicles go in the same direction.
(e) List outcomes in D'.
Answer:
a)
A={RRR,SSS,LLL}
b)
B={RSL,SLR,LRS,RLS,SRL,LSR}
c)
C={RRL,RLR,LRR,RRS,SRR,RSR}
d)
D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
e)
D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}
Step-by-step explanation:
Sample space=S= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS}
a)
Let A be the event that all three vehicles go in same direction. It means that the all three vehicles go to right or left or straight. Thus, event A can be represented as
A={RRR,SSS,LLL}
b)
Let B be the event that all three vehicles go in different direction. It means that the three vehicles can go to
1. Right, Straight, Left
2. Straight, Left, Right
3. Left, Right, Straight
4. Right, Left, Straight
5. Straight, Right, Left
6. Left, Straight, Right
Thus, event B can be represented as
B={RSL,SLR,LRS,RLS,SRL,LSR}
c)
Let C be the event that exactly two of three vehicles turn right. It means that the three vehicles can go to
1. Right, Right , Left
2. Right , Left, Right
3. Left, Right, Right
4. Right, Right , Straight
5. Straight, Right,Right
6. Right , Straight, Right
Thus, event C can be represented as
C={RRL,RLR,LRR,RRS,SRR,RSR}
d)
Let D be the event that exactly two of three vehicles go in same direction. It means that the three vehicles can go to
1. Right, Right, Left
2. Right, Right, Straight
3. Right, Left, Right
4. Right, Left, Left
5. Right, Straight, Right
6. Right, Straight, Straight
7. Left, Right, Right
8. Left, Right, Left
9. Left, Left, Right
10. Left, Left, Straight
11. Left, Straight, Left
12. Left, Straight, Straight
13. Straight, Right, Right
14. Straight, Right, Straight
15. Straight, Left, Left
16. Straight, Left, Straight
17. Straight, Straight, Right
18.Straight,Straight, Left
Thus, event D can be represented as
D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
e)
D'=S-D= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS} -{RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}
Final answer:
For a freeway exit where vehicles can go right, left, or straight, we determine the combination of directions for various events: Event A with all vehicles going the same direction has 3 outcomes; Event B with all vehicles taking different directions has 6 outcomes; Event C with two vehicles turning right has 6 outcomes; Event D with two vehicles going in the same direction has several outcomes depending on combinations; Event D' coincides with Event B.
Explanation:
When analyzing the outcomes of three successive vehicles taking different actions at a freeway exit, we can approach the given scenarios methodically:
(a) Event A: where all three vehicles go in the same direction. The outcomes are {RRR}, {LLL}, and {SSS}.To further explain, for (a), since there are three options for direction and all must be the same, we have exactly three outcomes when they are all identical. For (b), all vehicles must take a unique direction, and since there are three vehicles and three directions, we have 3! (factorial) permutations of the outcome, resulting in 6 possibilities. For (c), two vehicles must turn right and the third vehicle must take either of the two remaining directions, which can be arranged in 3 choose 2 (3C2 = 3) ways for each of the two non-right directions, making 6 outcomes total. For (d), it's similar to (c), but we include all same-direction pairs, resulting in 3C2 combinations for each pair of directions. Lastly, (e) is simply the case where no two outcomes are the same, hence it is the same as (b).
In a normally distributed data set of how long customers stay in your store, the mean is 50.3 minutes and the standard deviation is 3.6 minutes.
Within what range would you expect 95% of your customers to stay in your store?
Answer:
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 50.3 minutes
Standard deviation = 3.6 minutes
Within what range would you expect 95% of your customers to stay in your store?
Within 2 standard deviations of the mean.
So from
50.3 - 2*3.6 = 43.1 minutes
To
50.3 + 2*3.6 = 57.5 minutes
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
In order to determine whether or not a variable type of process is in control (whereoutput can be measured, not just classified in a binary fashion—i.e., good or bad), you must use which type of chart(s)? (Check all that apply.)A) p-chartB) c-chartC) x-bar chartD) fishbone diagramE) R-chart
Answer:
i dont know
Step-by-step explanation:
just working
An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.
What is the probability of:
(A) a white chip on the first and a red on the second?
(B) two green chips being drawn?
(C) a red chip on the second, given that a white chip was drawn on the first?
Answer:
(A) 0.04
(B) 0.25
(C) 0.40
Step-by-step explanation:
Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.
Given:
R = 8, G = 10 and W = 2.
Total number of chips = 8 + 10 + 2 = 20
[tex]P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}[/tex]
As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.
(A)
Compute the probability of selecting a white chip on the first and a red on the second as follows:
[tex]P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04[/tex]
Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.
(B)
Compute the probability of selecting 2 green chips:
[tex]P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25[/tex]
Thus, the probability of selecting 2 green chips is 0.25.
(C)
Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:
[tex]P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40[/tex]
Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.
The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is 0.04. The probability of drawing two green chips is 0.25. The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is 4.
Explanation:To find the probability of different events, we need to use the concept of probability. Probability is the likelihood of an event occurring. In this case, we have an urn with 8 red chips, 10 green chips, and 2 white chips.
(A) The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is:
P(white first and red second) = P(white first) * P(red second) = (2/20) * (8/20) = 16/400 = 0.04.
(B) The probability of drawing two green chips is:
P(green first and green second) = P(green first) * P(green second) = (10/20) * (10/20) = 100/400 = 0.25.
(C) The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is:
P(red second | white first) = P(red second and white first) / P(white first) = (8/20) / (2/20) = 8/2 = 4.