Answer:
Step-by-step explanation:
If you pay cash, the total amount that you will pay for the television is $349
If you pay $75 down, the balance would be paid in 18 monthly payments of 22.50 which is the installment price of the TV. Total amount paid in 18 months would be
22.5 × 18 = $405
Total cost of the TV when you pay in installments would be
405 + 75 = $480
Difference between the installment price and the cash price would be
480 - 349 = $131
The percent by which the installment price would be greater than the cash price is
131/349 × 100 = 37.5%
The blood type of successive children born to the same parents are independent and have fixedprobabilities that depend on the genetic makeup of the parents. Each child born to a particularset of parents has probability 0.15 of having blood type O. If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?
Answer:
There is a 9.75% probability that exactly 2 of them have type O blood.
Step-by-step explanation:
For each children, there is only two possible outcomes. Either they have blood type O, or they do not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
4 children, so [tex]n = 4[/tex]
Probability 0.15 of having blood type O, so [tex]p = 0.15[/tex]
If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{4,2}.(0.15)^{2}.(0.85)^{2} = 0.0975[/tex]
There is a 9.75% probability that exactly 2 of them have type O blood.
Ben swims 50,000 yards per week in his practices. Given this amount of training, he will swim the 100-yard butterfly in 51.5 seconds and place 10th in a big upcoming meet. Ben's coach calculates that if Ben increases his practice to 60,000 yards per week, his time will decrease to 46.4 seconds and he will place 7th in the meet. If Ben practices 70,000 yards per week, his time will be 44.6 seconds and he will win the meet. In terms of Ben's time in the big meet, what is his marginal productivity of the number of yards he practices?
Answer:
MArginal productivity: [tex]\frac{dt}{dL}=-0.0002[/tex]
We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.
Step-by-step explanation:
The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.
In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.
The marginal productivity can be expressed as:
[tex]\frac{dt}{dL}[/tex]
where dt is the variation in time and dL is the variation in training yards.
We can not derive the function because it is not defined, but we can approximate with the last two points given:
[tex]\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002[/tex]
Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.
This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.
The marginal productivity of the number of yards Ben practices is 0.00051 seconds per yard.
Explanation:The marginal productivity of the number of yards Ben practices can be calculated by dividing the change in his performance (measured in seconds) by the change in the number of yards he practices per week.
To find the marginal productivity between 50,000 and 60,000 yards per week, subtract Ben's initial time of 51.5 seconds from his new time of 46.4 seconds. This gives us a change in time of 5.1 seconds. Then divide the change in time by the change in yards, which is 10,000 yards (60,000 - 50,000).
Marginal productivity = Change in time / Change in yards = 5.1 seconds / 10,000 yards = 0.00051 seconds per yard.
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In Pennsylvania the average IQ score is 101.5. The variable is normally distributed, and the population standard deviation is 15. A school superintendent claims that the students in her school district have an IQ higher than the average of 101.5. She selects a random sample of 30 students and finds the mean of the test scores is 106.4. Test the claim at ???? = 0.05.
Answer:
We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 101.5
Sample mean, [tex]\bar{x}[/tex] = 106.4
Sample size, n = 30
Alpha, α = 0.05
Population standard deviation, σ = 15
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 101.5\\H_A: \mu > 101.5[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.96[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5
Of the 50 students in an undergraduate statistics class, 60% send email and/or text messages during any given lecture. They have a 50/50 chance of being caught by their professor, who is not amused by such unprofessional conduct. The eagle-eyed professor never charges an innocent student. What is the probability a student sends an email and/or text message during lecture and gets caught?
Answer:
0.3 or 30%
Step-by-step explanation:
Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :
[tex]P = 0.5*0.6 = 0.3[/tex]
The probability is 0.3 or 30%.
An air-water vapor mixture gas enters an insulated mixing chamber at 300 K, 1 atm, 60 % relative humidity with a mass flow rate of 1 kg/s. A second steam of moist air enters at 380 K, 1 bar, 80% relative humidity at a mass flow rate of 0.5 kg/s. Assume the mixing takes place completely and any condensed liquid or vapor mixture exit at 1 bar. Determine: (a) The humidity ratio for both entering streams. (b) The humidity ratio and the temperature of the mixed stream exiting the control volume. (c) The amount of liquid condensate that leaves the control volume. (d) Challenge: Determine the rate of entropy production, in kW/K.
Answer:
i don't know sorry hope this helps
Step-by-step explanation:
Find equations of both the tangent lines to the ellipse x2 + 9y2 = 81 that pass through the point (27, 3). y = (smaller slope) y = (larger slope)
Answer:
The tangent line equations are:
- The one that has the smaller slope [tex]\bold{y =3}[/tex]
- The one that has the larger slope [tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
Step-by-step explanation:
In order to find the tangent lines we need to find first the first derivative since the first derivative evaluated at a point give us the slope of the tangent line.
Working with implicit differentiation.
In order to find the derivative we can work with implicit differentiation and we get
[tex]2x+18y \cfrac{dy}{dx}= 0[/tex]
Notice that only when we are finding the derivative of an expression that has y we multiply by dy/dx due chain rule.
Solving for the first derivative we get
[tex]18y\cfrac{dy}{dx}=-2x[/tex]
[tex]\cfrac{dy}{dx}=-\cfrac{x}{9y}[/tex]
So the equation of the first derivative at the point (x,y) give us the first equation for the slope.
[tex]m=-\cfrac{x}{9y}[/tex]
Slope of a line using definition.
We can also find the slope of the line that will pass the point (x,y) trough the point (27,3) using the definition:
[tex]m = \cfrac{y_2-y_1}{x_2-x_1}[/tex]
Replacing the point we get
[tex]m=\cfrac{y-3}{x-27}[/tex]
That is another equation for the slope.
Finding the value of the slopes.
We can now set both slope equations equal to each other to find the point (x,y) where they intersect and the value of the slopes.
[tex]-\cfrac{x}{9y}= \cfrac{y-3}{x-27}[/tex]
We can work with cross multiplication to get
[tex]-x(x-27)=9y(y-3)[/tex]
And we can distribute and simplify
[tex]-x^2+27x=9y^2-27y[/tex]
Moving everything to the right side
[tex]0=x^2-27x+9y^2-27y\\\\0=x^2+9y^2-27x-27y[/tex]
At this point we can use the ellipse equation so we can replace [tex]x^2+9y^2[/tex] with 81, that will give us
[tex]0=81-27x-27y[/tex]
Then we can divide by 27 and solve for y.
[tex]0=3-x-y\\y= 3-x[/tex]
At this point we can replace on the ellipse equation.
[tex]x^2+9(3-x)^2=81[/tex]
And we can distribute and simplify.
[tex]x^2+9(9-6x+x^2)=81\\x^2+81-54x+9x^2=81\\x^2-54x+9x^2=0\\10x^2-54x=0\\[/tex]
So then we can factor and solve for x.
[tex]2x(5x-27)=0[/tex]
Setting each factor and solve for x we get
[tex]x= 0, \cfrac{27}{5}[/tex]
At x = 0, we can find the y value.
[tex]y= 3-x\\y= 3-0\\y= 3[/tex]
And the slope
[tex]m = -\cfrac{x}{9y}\\m = -\cfrac{0}{9y}\\m = 0[/tex]
So the line equation is
[tex]y-3=0(x-0)\\y=3[/tex]
For the second point x = 27/5 we have:
[tex]y= 3-\cfrac{27}{5}\\y=-\cfrac{12}{5}[/tex]
So slope is:
[tex]m = -\cfrac{\cfrac{27}{5}}{9\left(-\cfrac{12}{5}\right)}[/tex]
[tex]m = \cfrac 14[/tex]
So the line equation is
[tex]y-\left(-\cfrac{12}{5}\right)= \cfrac 14 \left(x -\cfrac{27}{5}\right)\\y = \cfrac 14 x - \cfrac{15}{4}[/tex]
Thus the equations of tangent lines are:
[tex]\bold{y =3}[/tex]
and
[tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]
To find the equations of the tangent lines to the ellipse that passes through a specific point, differentiate the equation of the ellipse to get a formula for the slope at any point on the ellipse. Set this equal to the slope of the line from the point on the ellipse to the specific point, and solve. The resultant solutions give the points of tangency, and the lines through these points and the specific point are the desired tangent lines.
Explanation:The subject of the question is about finding the equations of the tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through a specific point (27, 3). We first differentiate the equation of the ellipse to get a general formula for the slope of the tangent line at any point on the ellipse. The differentiation of x^2 + 9y^2 = 81 with respect to x gives 2x + 18yy' = 0, so y' = -x/ (9y). Now, we put the slopes y' equal to the slopes of line from (x, y) to (27, 3), and solve the resulting system of equations. The solutions give the points of tangency, and the lines through these points and (27, 3) are the desired tangent lines. The exact computations involve solving a quadratic equation and it'll give two slopes for the tangent line that passes through the point (27,3).
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An analyst is forecasting net income for Excellence Corporation for the next fiscal year. Her low-end estimate of net income is $250,000, and her high-end estimate is $350,000. Prior research allows her to assume that net income follows a continuous uniform distribution. The probability that net income will be greater than or equal to $337,500 is
Answer:
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Step-by-step explanation:
probability distribution fucntion is given as
[tex] F_x = P(X \leq x)[/tex]
[tex] = \frac{x -a}{b -a} a<x< b[/tex]
where a indicate lower end estimate = $250,000
b indicate high end estimate = $350,000
probability greater than $337500
[tex] P(X \geq 337,500) = 1- P(X < 337500)[/tex]
[tex] = 1 - \frac{x -a}{b -a}[/tex]
[tex] = 1 - \frac{337500 - 250000}{350000 - 250000}[/tex]
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Calculate the probability of net income being greater than or equal to $337,500 for Excellence Corporation following a continuous uniform distribution. Probability will be equal to 12.5%.
The probability that net income will be greater than or equal to $337,500 is 0.25. To calculate this probability for a continuous uniform distribution, we need to find the proportion of the area under the distribution curve that corresponds to values greater than $337,500.
Step-by-step calculation:
Calculate the total range of net income: $350,000 - $250,000 = $100,000
Calculate the proportion of the area for values greater than $337,500: ($350,000 - $337,500) / $100,000 = 0.125
Since the distribution is continuous and uniform, the probability is equal to the proportion calculated: 0.125 = 12.5%
(a) Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x)=x/(10 x**2+1) sum_(n=0)^infinity (-1)**n 10**n x**(2n+1) sum_(n=0)^infinity (-1)**n (x**(2n+1))/(10**n) sum_(n=0)^infinity (-1)**n 10**n x**(2n) 10 sum_(n=0)^infinity (-1)**n x**(2n+1) sum_(n=0)^infinity (-1)**n 10**(n+1) x**(n+1) Correct: Your answer is correct. (b) Determine the interval of convergence.
One way to do this is to recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
so that
[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]
(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.
If other factors are held constant, what happens to a confidence interval if the sample variance increases? The range of t scores increases and the width of the interval increases. The range of t scores decreases and the width of the interval decreases. The standard error increases and the width of the interval increases. The standard error decreases and the width of the interval decreases.
Answer:
The standard error increases and the width of the interval increases
Step-by-step explanation:
The variance is equal to the square of the standard deviation. An increase in variance will increase the standard deviation.
The width of the confidence interval can be defined as:
[tex]W=UL-LL=2z\sigma/\sqrt{n}[/tex]
The width is proportional to the standard deviation, so when the variance increases, the standard deviation and also the width of the interval increases.
Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The interval is to have a margin of error of 3 minutes, and the amount spent has a Normal distribution with a standard deviation σ = 40 minutes. The number of observations required is closest to:
1180.
683.
482.
22.
Answer: 482
Step-by-step explanation:
Formula to find the sample size is given by :-
[tex]n= (\dfrac{z^*\times \sigma}{E})^2[/tex] (1)
, where z* = critical z-value (two tailed).
[tex]\sigma[/tex] = Population standard deviation and E = Margin of error.
As per given , we have
Margin of error : E= 3
[tex]\sigma=40[/tex]
Confidence level = 90%
Significance level =[tex]\alpha=1-0.90=0.10[/tex]
Using z-table , the critical value for 90% confidence=[tex]z^*=z_{\alpha/2}=z_{0.05}=1.645[/tex]
Required minimum sample size = [tex]n= (\dfrac{(1.645)\times (40)}{3})^2[/tex] [Substitute the values in formula (1)]
[tex]n=(21.9333333333)^2[/tex]
[tex]n=481.07111111\approx482[/tex] [ Round to the next integer]
Hence, the number of observations required is closest to 482.
You manage a call center that processes orders. One measure of efficiency in your department is the time spent on each call. You need to report to your superiors on the time-per-call in your call center each month. You’ve identified some of your most efficient workers, and you know that efficiency tends to decrease, in general on Fridays.
1. Identify the ethical issues involved in developing a sampling protocol:
a. There are none. It's a simple mathematical operation
b. In order to get a random and representative sample, you need to consider how efficiency fluctuates over the month, and sample call times on a frequency that includes every day of the week.
c. The analysis and reporting should consider the impact of extreme values in the data.
d. Both "b." and "c." above.
Answer:
The answer is option A.
Step-by-step explanation:
If the report is made once a month, the daily variations (including the tendency to decrease efficiency on Friday) will be masked within the monthly result.
It would only generate an ethical problem in the case that Fridays fall in different samples, but in the case of the monthly report there are usually 4 or 5 Fridays included in each sample.
In ΔDEF, the measure of ∠F=90°, FE = 39, ED = 89, and DF = 80. What ratio represents the sine of ∠D?
Answer:Sin of ∠D = 39/89
Step-by-step explanation:
The diagram of the right angle DEF is shown in the attached photo.
To determine the ratio representing the sine of ∠D, we would apply the sine trigonometric ratio. It is expressed as
Sin# = opposite side/hypotenuse
Looking at the triangle
Opposite side = 39
Hypotenuse = 89
# = ∠D
Therefore
Sin of ∠D = 39/89
The sine of ∠D in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. So, in ΔDEF, sine of ∠D is FE/ED = 39/89.
Explanation:In trigonometry, the sine of an angle in a right triangle is defined as the ratio of the length of the side that is opposite that angle, to the length of the hypotenuse. Considering this, the ratio representing the sine of ∠D in ΔDEF would be the length of side FE (opposite ∠D) divided by the length of side ED (the hypotenuse in ΔDEF because ∠F is a right angle and ED lies opposite to it).
Therefore, the ratio representing the sine of ∠D is FE/ED = 39/89.
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Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use α= .05 and test to see whether the consultant with more experience has the higher population mean service rating.
Consultant A: n = 16, x = 6.82, s = 0.64
Consultant B: n = 10, x = 6.25, s = 0.75
a. State the null and alternative hypotheses.
b. Compute the value of the test statistic.
c. What is the p-value?
d. What is your conclusion?
Answer:
See the explanation
Step-by-step explanation:
(a)
H0: Consultant with more experience has the higher population mean service rating.
H1: Consultant with more experience doesn't have the higher population mean service rating.
(b)
t = 1.9923 (see the attached image)
(c)
The degrees of freedom for the test statistic,
df = 16
The P-value of the one tailed t- test with 16 degrees of freedom is,
P−value = tdist(X,df,tails)
P-value = tdist(1.9923,16,1)
P-value = 0.032
(d)
Since, P-value 0.032 is less than the significance level 0.05, there is an enough evidence to reject the null hypothesis.
Hence, there is a sufficient evidence to conclude that Consultant with more experience doesn't have the higher population mean service rating.
Hope this helps!
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 16 samples is 7.8 ppm with a standard deviation of 0.8. Assume the population is normally distributed. A level of significance of 0.01 will be used. Make the decision to reject or fail to reject the null hypothesis.
Answer:
We accept H₀ with the information we have, we can say level of ozone is under the major limit
Step-by-step explanation:
Normal Distribution
population mean = μ₀ = 7.5 ppm
Sample size n = 16 df = n - 1 df = 15
Sample mean = μ = 7.8 ppm
Sample standard deviation = s = 0.8
We want to find out if ozono level, is above normal level that is bigger than 7.5
1.- Hypothesis Test
null hypothesis H₀ μ₀ = 7.5
alternative hypothesis Hₐ μ₀ > 7.5
2.-Significance level α = 0.01 we will develop one tail-test (right)
then for df = 15 and α = 0,01 from t -student table we get
t(c) = 2.624
3.-Compute t(s)
t(s) = ( μ - μ₀ ) / s /√n ⇒ t(s) = ( 7.8 - 7.5 )*4/0.8
t(s) = 0.3*4/0.8
t(s) = 1.5
4.-Compare t(s) and t(c)
t(s) < t(c) 1.5 < 2.64
Then t(s) is inside the acceptance region. We accept H₀
Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use technology to evaluate or approximate the integral. y:31n x, for 2sxs5 a. The integral that gives the arc length of the curve is L dx &The arc length of the curve is approximately (Round to three decimal places as needed.)
Answer:
a. [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]
b. [tex]\displaystyle AL = 4.10322[/tex]
General Formulas and Concepts:
Calculus
Differentiation
DerivativesDerivative NotationDerivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Integration
IntegralsDefinite IntegralsIntegration Constant CIntegration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
U-Substitution
Arc Length Formula [Rectangular]: [tex]\displaystyle AL = \int\limits^b_a {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify
y = 3ln(x)
Interval [2, 5]
Step 2: Find Arc Length
[Function] Differentiate [Logarithmic Differentiation]: [tex]\displaystyle \frac{dy}{dx} = \frac{3}{x}[/tex]Substitute in variables [Arc Length Formula - Rectangular]: [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ [\frac{3}{x}]^2}} \, dx[/tex][Integrand] Simplify: [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex][Integral] Evaluate: [tex]\displaystyle AL = 4.10322[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Applications of Integration
Book: College Calculus 10e
The arc length of the curve [tex]\( y = 3 \ln x \)[/tex] over the interval [2, 5] is approximately 12.092 units, obtained through numerical methods such as Simpson's rule or a CAS like Wolfram Alpha.
a. Writing and Simplifying the Integral:
The arc length L of the curve y = f(x) over the interval [a, b] is given by the following definite integral:
L = [tex]\int_a^b \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx[/tex]
First, find the derivative dy/dx of the given function:
y' = [tex]\dfrac{d}{dx} (3 \ln x) = \dfrac{3}{x}[/tex]
Now, plug f(x) = 3 \ln x and [tex]f'(x) = \dfrac{3}{x}[/tex] into the arc length formula for the interval [2, 5]:
L = [tex]\int_2^5 \sqrt{1 + \left( \dfrac{3}{x} \right)^2} dx[/tex]
L = [tex]\int_2^5 \sqrt{1 + \dfrac{9}{x^2}} dx[/tex]
b. Evaluating or Approximating the Integral:
This integral cannot be solved analytically using standard techniques. Therefore, we can use numerical methods like Simpson's rule or a CAS (Computer Algebra System) to approximate the value.
Using Simpson's rule with 10 subintervals, we get an approximate arc length of:
L ≈ 12.092
Alternatively, using a CAS like Wolfram Alpha, we can obtain a more precise numerical value:
L ≈ 12.091971077
Therefore, the arc length of the curve is approximately 12.092 units.
The endpoints of the side of a square lie on the circle and the opposite side of the square tangent to the circle at its midpoint. Find the ratio of the side and the radius.
Answer:
[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]
Step-by-step explanation:
Circle and Square
We have a geometric construction as shown in the image below. We can see that
[tex]\displaystyle r+h=L[/tex]
Or, equivalently
[tex]\displaystyle h=L-r[/tex]
The triangle formed by r,h and L/2 is right, because the opposite side of the square is tangent to the circle at its midpoint. This means we can use Pythagoras's theorem:
[tex]\displaystyle r^2=h^2+\left(\frac{L}{2}\right)^2[/tex]
Replacing h
[tex]\displaystyle r^2=(L-r)^2+\left(\frac{L}{2}\right)^2[/tex]
Expanding squares
[tex]\displaystyle r^2=L^2-2Lr+r^2+\frac{L^2}{4}[/tex]
Simplifying
[tex]\displaystyle 2Lr=L^2+\frac{L^2}{4}[/tex]
Multiplying by 4
[tex]\displaystyle 8r=4L+L[/tex]
Joining terms
[tex]\displaystyle 8r=5L[/tex]
Solving for the ratio L/R as required
[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]
The ratio of the side and the radius is 8:5
calculation;Since
r + h = L
So we can write h = L - r
Now we applied the Pythagoras theorem
[tex]r^2 = h^2 +( \frac{L}{2}) ^2\\\\r^2 = (L-r)^2 + ( \frac{L^}{2}) ^2\\\\r^2 = L^2- 2Lr+ r^2 + \frac{L^2}{4}\\\\ 2Lr = L^2 \frac{L^2}{4}\\\\[/tex]
Now
8r = 4L + L
8r = 5L
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At a nationwide travel agency, newly hired employees are classified as beginners (B). Every six months the performance of each agent is reviewed. Past records indicate that transitions through the ranks to intermediate (I) and qualified (Q) are according to the following Markov chain, where F indicates workers that were fired BIQ F B 45.4 0.15 Q 0 0 1 0 F 0 0 0 1 (a) What fraction eventually become qualified? (b) What is the expected time until a beginner is fired or becomes qualified?
Answer
The answer and procedures of the exercise are attached in the following archives.
The information complete of the exercise is attached in the sheet.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Euro Coin. Statistics students at the Akademia Podlaka conducted an experiment to test the hypothesis that the one-Euro coin is biased (i.e., not equally likely to land heads up or tails up). Belgian-minted one-Euro coins were spun on a smooth surface, and 140 out of 250 coins landed heads up. Does this result support the claim that one-Euro coins are biased
Answer:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [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 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
Step-by-step explanation:
1) Data given and notation
n=250 represent the random sample taken
X=140 represent the number of heads obtained
[tex]\hat p=\frac{140}{250}=0.56[/tex] estimated proportion of heads
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that that one-Euro coins are biased, so the correct system of hypothesis are:
Null hypothesis:[tex]p=0.5[/tex]
Alternative hypothesis:[tex]p \neq 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, 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].
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
[tex]np_o =250*0.5=125>10[/tex]
[tex]n(1-p_o)=250*(1-0.5)=125>10[/tex]
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
4) 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 next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [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 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households in City 1 is more than that of households in City 2. She surveys 35 households in City 1 and obtains an average weekly food expenditure of $164. A sample of 30 households in City 2 yields an average weekly expenditure of $159. Historical data reveals that the population standard deviation for City 1 and City 2 are $12.50 and $9.25, respectively.Col1 City 1 x1(bar)=164 σ(1)=12.5 n(1)=35Col2 City 2 x2 (bar) =159 σ(2) =9.25 n2=30Let μ(1) be the mean weekly food expenditure for City 1 and μ(2) be that for City 2.1. To test the economist’s claim, the competing hypotheses should be formulated asSelect one:a. H0:μ1-μ2>0 versus Ha:μ1-μ2≤0b. H0:μ1-μ2≥0 versus Ha:μ1-μ2<02.The standard error of x(1)bar- x(2) bar isSelect one:a. 0.82b. 2.70c. 12.5d. 9.253.The value of the test statistics isSelect one:a. 0.40b. 1.85c. 0.54d. 27.784. The p-value of the test is
Select one:
a. 0.34
b. 0.03
c. 0.29
d. 0.08
5.At α=0.05,
Select one:
a. We can reject H(0) in favor of H(a)
b. We cannot reject H(0)
c. We can conclude that average weekly food expenditures in City 1 is less than that of City 2
Answer:
Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]
[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]
b) 2.70
[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]
b. 1.85
[tex]p_v =P(Z>1.85)=0.032[/tex]
b. 0.03
a. We can reject H(0) in favor of H(a)
Step-by-step explanation:
Data given and notation
[tex]\bar X_{1}=164[/tex] represent the mean for the sample 1
[tex]\bar X_{2}=159[/tex] represent the mean for the sample 2
[tex]\sigma_{1}=12.5[/tex] represent the population standard deviation for the sample 1
[tex]s_{2}=9.25[/tex] represent the population standard deviation for the sample B2
[tex]n_{1}=35[/tex] sample size selected 1
[tex]n_{2}=30[/tex] sample size selected 2
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]
We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Standard error
The standard error on this case is given by:
[tex]SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]
Replacing the values that we have we got:
[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]
b. 2.70
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]
P-value
Since is a one side right tailed test the p value would be:
[tex]p_v =P(Z>1.85)=0.032[/tex]
b. 0.03
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
a. We can reject H(0) in favor of H(a)
Suppose an ecologist collects data on the weights and basal trunk diameters of a randomly selected sample of felled trees. The ecologist fits a simple linear regression model, predicting tree weight from the trunk's cross-sectional area. After ensuring that all assumptions of the linear model are met, the ecologist computes a 95% confidence interval for the predicted mean weight of trees with a trunk cross-sectional area of 452 in?.
Select the correct interpretation of this confidence interval.
a. The weight of another tree with a trunk cross-sectional area of 452 in- sampled at random from the same population will fall in the ecologist's interval with 95% probability.
b.The true mean weight for trees in the population with a trunk cross-sectional area of 452 in- will fall within the ecologist's interval with 95% probability.
c.The weight of 95% of trees in the population with a trunk cross-sectional area of 452 inwill fall within the ecologist's interval.
d. The ecologist's interval has a 95% chance of containing the true sample mean weight of trees with a trunk cross- sectional area of 452 in².
e. There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in?.
Answer:
e) There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in
Step-by-step explanation:
Given that an ecologist collects data on the weights and basal trunk diameters of a randomly selected sample of felled trees. The ecologist fits a simple linear regression model, predicting tree weight from the trunk's cross-sectional area. After ensuring that all assumptions of the linear model are met, the ecologist computes a 95% confidence interval for the predicted mean weight of trees with a trunk cross-sectional area of 452 inches
We can interpret the confidence interval as
we are 95% confident that for samples of trees largely drawn with 452 inches the mean of these trees would fall within this interval.
e) There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in
A farmer uses two types of fertilizers. A 50-lb bag of Fertilizer A contains 8 lb of nitrogen, 2 lb of phosphorus, and 4 lb of potassium. A 50-lb bag of Fertilizer B contains 5 lb each of nitrogen, phosphorus, and potassium. The minimum requirements for a field are 440 lb of nitrogen, 260 lb of phosphorus, and 360 lb of potassium. If a 50-lb bag of Fertilizer A costs $80 and a 50-lb bag of Fertilizer B costs $30, find the amount of each type of fertilizer the farmer should use to minimize his cost C in dollars while still meeting the minimum requirements. (Let x represent the number of bags of Fertilizer A and y represent the number of bags of Fertilizer B.)
Answer:
The amount fertilizer A the farmer should use while still meeting the minimum requirement is approximately seventy-six 50-lb bags which would cost $6080.
The amount of fertilizer B the farmer should use while still meeting the minimum requirement is approximately seventy-one 50- lb bags which would cost $2120
Step-by-step explanation:
Fertilizer A
Let x represent the number of 50- lb bags needed for the minimum requirement
Cost of I 50-lb bag = $80
Cost of x 50-lb bag = $80x
1 50-lb bag contains 14lb nutrient ( 8lb Nitrogen+ 2lb Phosphorus + 4lb Potassium)
x 50-lb bag contains 1060lb nutrient (440lb N + 260lb P + 360lb K) which is the minimum requirement
x = 1060lb ÷ 14lb = 75.71
x = 76 ( to the nearest whole number)
76 50-lb bags would cost 76×$80 = $6080
Fertilizer B
Cost of a 50- lb bag is $30
Let y be the number of bags required
Cost of y bag is $30y
y = 1060÷ 15= 71
Cost = 71× $30 = $2130
Answer:
0 fertilizer A
88 fertilizer B
Step-by-step explanation:
Shown in the attachment
The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random sample of 50 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between $71000 and $73500?A 0.079B. 0.500C. 0.487D. 0.306
Answer: D. 0.306
Step-by-step explanation:
Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as
z = (x - u)/s
Where
x = annual salary for intermediate level executives
u = mean annual salary
s = standard deviation
From the information given,
u = $74000
s = $2500
We want to find the probability that the mean annual salary of the sample is between $71000 and $73500. It is expressed as
P(71000 lesser than or equal to x lesser than or equal to 73500)
For x = 71000,
z = (71000 - 74000)/2500 = - 1.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.1151
For x = 73500,
z = (73500 - 74000)/2500 = - 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.4207
P(71000 lesser than or equal to x lesser than or equal to 73500) is
0.4207 - 0.1151 = 0.306
In how many ways can 9 hearts be chosen if 12 cards are chosen from a well-shuffled deck of 52 playing cards? a) 715 b) 220 c) 108 d) 6,534,385 e) 117 f) None of the above.
Answer: d) 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
Step-by-step explanation:
Given;
Number of cards to be chosen = 12
Number of hearts to be chosen = 9
Number of non-hearts to be selected = 3
Total number of cards = 52
Number of hearts total = 13
Number of non-hearts total= 39
The number of ways of selecting hearts can be given by the illustration below.
Number of ways of selecting hearts × number of ways of selecting non-hearts
N = 13C9 × 39C3 (it's combination since order is not important)
N = 13!/(9! ×4!) × 39!/(36! × 3!)
N = 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
The number of ways to choose 9 hearts from 12 cards is 220, option (b) is correct.
To solve this problem, we'll use the combination formula, which is used when selecting objects from a larger set without regard to the order. The formula for combinations is:
[tex]\[ \text{C}(n, k) = \frac{n!}{k!(n - k)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and (!) denotes factorial.
Given that we need to choose 12 cards from a deck of 52, and we want to know the number of ways to choose 9 hearts, which is [tex]\( k = 9 \)[/tex] hearts out of a total of [tex]\( n = 12 \)[/tex] cards.
Step 1:
Calculate the number of ways to choose 9 hearts from the 12 chosen cards:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!(12 - 9)!} \][/tex]
Step 2:
Simplify the expression:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!3!} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{1320}{6} \][/tex]
[tex]\[ \text{C}(12, 9) = 220 \][/tex]
Therefore, there are 220 ways to choose 9 hearts from 12 cards. So, the correct answer is option (b).
In the Country A legal system, a defendant is presumed innocent until proven guilty. Consider a null hypothesis, Upper H 0, that the defendant is innocent, and an alternative hypothesis, Upper H 1, that the defendant is guilty. A jury has two possible decisions: Convict the defendant (i.e., reject the null hypothesis) or do not convict the defendant (i.e., do not reject the null hypothesis). Explain the meaning of the risks of committing either a Type I or Type II error in this example.
(A) A Type I error would be incorrectly convicting the defendant when he is guilty. A Type II error would be incorrectly failing to convict the defendant when he is innocent.
(B) A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
(D) A Type I error would be incorrectly failing to convict the defendant when he is innocent. A Type II error would be incorrectly convicting the defendant when he is guilty.
Answer:
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
Step-by-step explanation:
Type I error is rejecting the true null hypothesis and type II error is not rejecting the false null hypothesis. Hence in this scenario, it will be:
A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
Option C is correct.
find 72 degrees and 40 degrees from the picture
Answer:a) 5 sides
b) 9 sides
Step-by-step explanation:
a) Since the polygon is a regular polygon, all the interior angles are equal. If one exterior angle is 72 degrees, the interior angle would be
180 - 72 = 108 degrees(sum of the angles on a straight line is 180 degrees)
The sum of the interior angles of a polygon is expressed as
180(n - 2)
Where n represents the number of sides that the polygon has. Since the interior angles are equal, then
108n = 180(n - 2)
108n = 180n - 360
180n - 108n = 360
72n = 360
n = 360/72 = 5
b) If one exterior angle is 40 degrees, the interior angle would be
180 - 40 = 140 degrees
Since the interior angles are equal, then
140n = 180(n - 2)
140n = 180n - 360
180n - 140n = 360
40n = 360
n = 360/40 = 9
the Earth has an average distance from the Sun of 1.5* 10 to the 8th power kilometers and the Earth has an approximate distance from the Moon of 4.0 x 10 to the 5th km how many times greater is Earth's average distance from the Sun than is from the Moon
Answer:
Earth's average distance from the Sun is [tex]375[/tex] times more than is from the Moon.
Step-by-step explanation:
Given the average distance of Earth from the Sun is [tex]1.5\times10^{8}[/tex] kilometers.
Also, the approximate distance of Earth from the Moon is [tex]4\times10^{5}[/tex] kilometers.
For this, we will divide the average distance of Earth from the Sun by approximate distance of Earth from the Moon.
So,
[tex]\frac{1.5\times10^{8}}{4\times10^{5}}=375\ km[/tex]
20)
The projected value of an investment is modeled by the exponential function V(t) = 30,000(1.125) where V(t) is the total value
after t years. What does 1.125 represent in the function?
The growth factor of the investment
The initial value of the investment
The projected value of the investment
D)
The value of the investment after t years:
The 1.125 in the exponential function V(t) = 30,000(1.125)^t represents the growth factor of the investment, indicating an annual growth of 12.5%.
Explanation:In the exponential function V(t) = 30,000(1.125)^t, the 1.125 represents the growth factor of the investment. This means each year, the value of the investment grows by 12.5%. It's the rate at which the initial value of the investment, 30,000 in this case, increases on an annual basis. Therefore, the exponential function shows us how the value of the investment changes over time with this constant growth factored in.
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Choose an American household at random, and let the random variable X be the number of cars, including SUVs and light trucks, the residents own. The table gives the the probability model if we ignore the few households that own more than 8 cars.Number of cars X 0 1 2 3 4 5 6 7 8Probability 0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001A housing company builds houses with two?car garages. What percent of households have more cars than the garage can hold?A- 22.7%B- 41.0%C- 59.0%D- 14.4%
Answer:
Option A.
Step-by-step explanation:
The given probability table is
Number of cars X: 0 1 2 3 4 5 6 7 8
Probability :0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001
It is given that a housing company builds houses with two car garages.
We need to find the percent of households have more cars than the garage can hold.
[tex]P(X>2)=1-P(X\leq 2)[/tex]
[tex]P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]
Substitute the probability values from the given table.
[tex]P(X>2)=1-[0.087+0.323+0.363][/tex]
[tex]P(X>2)=1-0.773[/tex]
[tex]P(X>2)=0.227[/tex]
It means 22.7% of households have more cars than the garage can hold.
Therefore, the correct option is A.
The percent of households that have more cars (including SUVs and light trucks) than a two-car garage can accommodate is approximately 22.7%.
Explanation:To answer the student's question, let's examine the probability model provided. The random variable X represents the number of cars, including SUVs and light trucks, in a randomly selected American household.
If a housing company builds houses with two-car garages, we want to determine what percent of households have more cars than the garage can hold. To find this, we would add up the probabilities for the households that own more than 2 cars.
The probabilities for owning 3, 4, 5, 6, 7 or 8 cars are 0.144, 0.053, 0.019, 0.007, 0.002, and 0.001, respectively. Adding all these probabilities gives us:
0.144 + 0.053 + 0.019 + 0.007 + 0.002 + 0.001 = 0.226
To convert this to a percentage, we multiply by 100, which gives us 22.6%. Therefore, approximately 22.7% of households have more cars than the garage can hold, which corresponds to answer choice A.
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The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with an average of 2.83 rescues every eight hours. What is the probability that the squad will have at most 2 calls in an hour? Round answer to 4 decimal places.
Answer:
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that:
2.83 rescues every eight hours.
What is the probability that the squad will have at most 2 calls in an hour?
This is
[tex]P = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
We have 2.83 rescues every 8 hours. So for an hour, we have [tex]\mu = \frac{2.83}{8} = 0.354[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019[/tex]
[tex]P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485[/tex]
[tex]P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440[/tex]
So
[tex]P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944[/tex]
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Final answer:
The probability that the rescue squad will receive at most 2 calls in an hour is 0.9951
when the average is 2.83 calls every eight hours.
Explanation:
The number of rescue calls that a rescue squad receives is given to be a Poisson distribution with an average of 2.83 rescues every eight hours. To calculate the probability of receiving at most 2 calls in one hour, we first find the average number of rescues per hour by dividing the given average by eight, since there are eight hours in the period mentioned. This gives us an average rate ([tex]λ[/tex]) of 2.83 rescues / 8 hours = 0.35375 rescues per hour. The Poisson probability formula is:
[tex]P(X=k) = (e^{-λ} * λ^k) / k![/tex]
To find the probability of at most 2 calls, we sum the probabilities of 0, 1, and 2 calls:
[tex]P(X=0) = e^{-0.35375} * (0.353750)^0 / 0! = 0.7026 \\\\P(X=1) = e^{-0.35375} * (0.353751)^1 / 1! = 0.2485 \\\\P(X=2) = e^{-0.35375} * (0.353752)^2 / 2! = 0.0440[/tex]
Thus, the desired probability is the sum of these probabilities:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) ≈ 0.7026 + 0.2485 + 0.0440 ≈ 0.9951
After rounding to four decimal places we have that the probability that the squad will have at most 2 calls in an hour is 0.9951.
A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.
Answer:
The three statements are true.
Step-by-step explanation:
The question is incomplete:
A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.
A. There are 54 clients who own dogs.
B. There are 34 clients who own dogs but not cats.
C. There are 16 clients who own cats but not dogs.
A. There are 54 clients who own dogs.
TRUE. Of the 70 clients, only 36 own cats. There are left 34 clients that own only dogs. If we add the 20 clients that own both cats and dogs, we have 34+20=54 clients who own dogs.
B. There are 34 clients who own dogs but not cats.
TRUE. Of the 70 clients, only 36 own cats. Then, there are left 70-36=34 clients that own only dogs.
C. There are 16 clients who own cats but not dogs.
TRUE. Out of the 36 clients that own cats (only of with dogs), there are 20 that own both. Therefore, there are 36-20=16 clients that own only cats.