Answer:
a) A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
b) [tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
c) [tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Part a
Let X the random variable that represent the diameter of a brand of ping pong of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1.32,0.08)[/tex]
Where [tex]\mu=1.32[/tex] and [tex]\sigma=0.08[/tex]
And we select a sample of size n =4 and we want to find the distribution for [tex]\bar X[/tex]
Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
So the best option for this case would be:
A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
Part b
[tex] P(\bar X<1.28)[/tex]
We can use the z score given by:
[tex]z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
Part c
[tex] P(1.28< \bar X<1.34)[/tex]
[tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
The sampling distribution of the mean diameter of the ping-pong balls, given that the population diameter is normally distributed, will also be normally distributed. The expected mean is 1.32 inches and the standard deviation will be 0.04 inches.
Explanation:The correct answer to this question should be A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. This is grounded on the Central Limit Theorem which states that the distribution of sample means tends to be normal regardless of the shape of the population distribution, especially when the sample size is large.
In this case, though our sample size (4) is fairly small due, because the distribution of the population is specified as normal, the sampling distribution will remain normal. The important numbers to know would be the expected value (mean which is the same as population mean i.e., 1.32 inches) and the standard deviation (which is the population standard deviation divided by square root of sample size i.e., 0.08/sqrt(4)).
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A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represent the three different kinds of defects.
Suppose further that for each unit produced P(D) = .07 P(D) = .12 P(Ds) = .05 P(D, U Ds) = .14 P(Din D2nDs) = .01
(a) What is the probability that a unit does not have a type 1 defect?
(b) What is the probability that a unit has both a type 2 and 3 defect?
(c) What is the probability that a unit has both a type 2 and 3 defect, but not a type 1 defect?
(d) What is the probability that a unit has at most two defects?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) 0.88
b) 0.02
c) 0.01
d) 0.99
Step-by-step explanation:
Step one: State the given parameters
[tex]P(D_{1} ) = 0.12[/tex] [tex]P(D_{2} ) = 0.07[/tex]
[tex]P(D_{3} ) = 0.05[/tex] [tex]P (D_{1} U D_{2} ) = 0.13[/tex]
[tex]P(D_{1}n D_{2}n D_{3}) = 0.01[/tex] [tex]P(D_{1} U D_{3}) = 0.14[/tex]
Step 2 : Obtain the probability that a unit does not have a type 1 defect
[tex]P(\frac{}{D_{1} })[/tex] = [tex]1 -P(D_{1} )[/tex]
= [tex]1 - 0.12[/tex]
= 0.88
Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?
The probability of the unit having both type 2 and type 3 defect is denoted as [tex]P(D_{2} n D_{3} )[/tex]
This is calculated as
[tex]P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13[/tex]
= 0.02
Therefore P(D_{2} n D_{3} ) = 0.02
Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect
Let [tex]P(\frac{}{D_{1}} n D_{2} n D_{3} )[/tex] denote the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.
This can be calculated as follows :
[tex]P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})[/tex]
= 0.02 - 0.01
= 0.01
Step 4 : Obtain the probability that a unit has at most two defects
P(at most 2 defects) = 1 - P(all three defects)
= [tex]1- P(D_{1} n D_{2}nD_{3})[/tex]
= 1 - 0.01
= 0.99
What is the volume of a cylinder with a height of 2 feet and a radius of 6 feet? Use 3.14 for pi. Enter your answer in the box. ft³
Answer:
[tex]V=226.08\ ft^3[/tex]
Step-by-step explanation:
we know that
The volume of a cylinder is equal to
[tex]V=\pi r^{2} h[/tex]
where
r is the radius of the base of the cylinder
h is the height of the cylinder
we have
[tex]r=6\ ft\\h=2\ ft\\\pi=3.14[/tex]
substitute the given values in the formula
[tex]V=(3.14)(6)^{2}(2)\\ V=226.08\ ft^3[/tex]
How many three-digit phone prefixes that are used to represent a particular geographic area are possible that have no 0 or 1 in the first or second digits?
There are 640 possible three-digit phone prefixes that do not have 0 or 1 in the first or second digits.
The number of three-digit phone prefixes are possible that have no 0 or 1 in the first or second digits.
Since each digit in a phone number can be a number from 0-9, but the first two digits cannot be 0 or 1, we have 8 choices (2-9) for the first digit, 8 choices (2-9) for the second digit, and 10 choices (0-9) for the third digit because the third digit has no such restriction.
Therefore, the total number of possible phone prefixes is calculated by multiplying the number of choices for each digit,
8 (choices for the first digit) × 8 (choices for the second digit) × 10 (choices for the third digit) = 640 possible three-digit phone prefixes.
(19.-2).(-11, 10) find the slope
Answer:
[tex]\large\boxed{\large\boxed{slope=-2/5}}[/tex]
Explanation:
The problem is: given the points (19,−2) and (−11,10) find the slope of the line that joins them.
The slope of a line is the change in the y-coordinate over the change of the x-coordinate:
slope = rise / run = Δy / ΔxThus:
[tex]slope=[10-(-2)]/[-11-19]\\\\slope=12/(-30)\\\\slope=-12/30[/tex]
Simplify, dividing both numerator and denominator by 6:
[tex]slope=-2/5\leftarrow answer[/tex]
Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
yy'' + (y')^ 2 = 0, t > 0. (3)
Then show that for any nonzero constants c1 and c2, c1 + c2t^1/2 is not a solution of this equation.
Answer: it is verified that:
* y1 and y2 are solutions to the differential equation,
* c1 + c2t^(1/2) is not a solution.
Step-by-step explanation:
Given the differential equation
yy'' + (y')² = 0
To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.
Now,
y1 = 1
y1' = 0
y'' = 0
So,
y1y1'' + (y1')² = (1)(0) + (0)² = 0
Hence, y1 is a solution.
y2 = t^(1/2)
y2' = (1/2)t^(-1/2)
y2'' = (-1/4)t^(-3/2)
So,
y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0
Hence, y2 is a solution.
Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.
Let us differentiate this twice, and verify if it satisfies the differential equation.
y = c1 + c2t^(1/2)
y' = (1/2)c2t^(-1/2)
y'' = (-1/4)c2t(-3/2)
yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²
= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)
= (-1/4)c1c2t(-3/2)
≠ 0
This clearly doesn't satisfy the differential equation, hence, it is not a solution.
Final answer:
The provided solutions y₁(t) = 1 and y₂(t) = [tex]t^{(1/2)[/tex] satisfy the given differential equation, verified through substitution and simplification. However, a linear combination of the form [tex]c_1 + c_2t^{(1/2)[/tex] is not a solution as it does not satisfy the original equation when its derivatives are substituted.
Explanation:
We are given the second-order differential equation yy'' + (y')² = 0, where y = y(t) is a function of t, and we are asked to verify solutions and understand properties of certain types of solutions to this equation.
To verify that y₁(t) = 1 is a solution, we calculate the derivatives: y₁' = 0 and y₁'' = 0. Substituting these into the differential equation yields (1)(0)+(0)² = 0, which holds true, confirming that y₁(t) = 1 is indeed a solution.
Next, to verify y₂(t) = [tex]t^{(1/2)[/tex], we find y₂' = [tex](1/2)t^{(-1/2)[/tex] and [tex]y_2'' = -(1/4)t^{(-3/2)[/tex]. Substituting these values gives [tex](t^{(1/2)})(-(1/4)t^{(-3/2)}) + ((1/2)t^{(-1/2)})^2 = 0[/tex], which simplifies to 0, showing that y₂(t) is also a solution.
For the linear combination of solutions, we consider [tex]y(t) = c_1 + c_2t^{(1/2)[/tex]. Derivatives are [tex]y' = c_2(1/2)t^{(-1/2)[/tex] and [tex]y'' = -c_2(1/4)t^{(-3/2)[/tex]. Substituting into the given differential equation does not yield zero, thus [tex]c_1 + c_2t^{(1/2)[/tex] is not a solution.
What is the difference in mass between a nickel that weighs 4.7 g and a nickel that weighs 4.874 g ?
Nickel a: 4.874
Nickel b: 4.7
The difference will be:
Higher- lower
4.874 - 4.7
0.174
So the difference is 0.174g
The difference in mass between the two nickels is 0.174 grams.
Given that:
Weight of first nickel, n = 4.7 g
Weight of second nickel, N = 4.7 g
To find the difference in mass between the two nickels, subtract the weight of one nickel from the weight of the other nickel:
Difference in mass = Weight of the second nickel - Weight of the first nickel
Let's calculate the difference:
Weight of the second nickel = 4.874 g
Weight of the first nickel = 4.7 g
Difference in mass = 4.874 g - 4.7 g
Difference in mass = 0.174 g
Hence, the difference in mass between the two nickels is 0.174 grams.
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Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that the time spent per session is normally distributed. Complete parts (a) through (d). a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? . 259 (Round to three decimal places as needed.) b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes? . 297 (Round to three decimal places as needed.) c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? . 68 (Round to three decimal places as needed.)
Answer:
a) 0.259
b) 0.297
c) 0.497
Step-by-step explanation:
To solve this problem, it is important to know the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 11, \sigma = 3[/tex]
a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that [tex]n = 25, s = \frac{3}{\sqrt{25}} = 0.6[/tex]
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{11.2 - 11}{0.6}[/tex]
[tex]Z = 0.33[/tex]
[tex]Z = 0.33[/tex] has a pvalue of 0.6293.
X = 10.8
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10.8 - 11}{0.6}[/tex]
[tex]Z = -0.33[/tex]
[tex]Z = -0.33[/tex] has a pvalue of 0.3707.
0.6293 - 0.3707 = 0.2586
0.259 probability, rounded to three decimal places.
b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?
Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So
X = 11
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{11 - 11}{0.6}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5.
X = 10.5
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10.5 - 11}{0.6}[/tex]
[tex]Z = -0.83[/tex]
[tex]Z = -0.83[/tex] has a pvalue of 0.2033.
0.5 - 0.2033 = 0.2967
0.297, rounded to three decimal places.
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that [tex]n = 100, s = \frac{3}{\sqrt{100}} = 0.3[/tex]
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{11.2 - 11}{0.3}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a pvalue of 0.7486.
X = 10.8
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10.8 - 11}{0.3}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514.
0.7486 - 0.2514 = 0.4972
0.497, rounded to three decimal places.
To find the probability that the sample mean is between eight minutes and 8.5 minutes, calculate the z-scores for both values and find the area under the standard normal distribution curve between these z-scores.
Explanation:To find the probability that the sample mean is between eight minutes and 8.5 minutes, we need to calculate the z-scores for both values and then find the area under the standard normal distribution curve between these two z-scores.
The formula to calculate the z-score is: z = (x - mu) / (sigma / sqrt(n))
where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.
Using the given information, we can calculate the z-scores as follows:
z1 = (8 - 11) / (3 / sqrt(25))
z2 = (8.5 - 11) / (3 / sqrt(25))
Next, we use a standard normal distribution table or a calculator to find the area between these two z-scores, which represents the probability that the sample mean is between eight minutes and 8.5 minutes.
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PLEASE SHOW WORK
Area of the triangle: 1 1/3 yards and 6 yards
[tex]\boxed{A=4yd^2}[/tex]
Explanation:I'll assume the dimensions are:
[tex]base \ (b)=1 \frac{1}{3}yards \\ \\ height \ (h)=6yards[/tex]
First of all, let's convert mixed fraction into improper fraction:
[tex]Add \ whole \ part \ and \ fractional \ part: \\ \\ 1\frac{1}{3}=1+\frac{1}{3} \\ \\ \\ Simplifying: \\ \\ 1\frac{1}{3}=\frac{3+1}{3} =\frac{4}{3}[/tex]
The formula for the area of a triangle is:
[tex]A=\frac{1}{2}b\times h \\ \\ A=\frac{1}{2}(\frac{4}{3})(6) \\ \\ \boxed{A=4yd^2}[/tex]
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A particular electronic component is produced at two plants for an electronics manufacturer. Plant A produces 60% of the components used and the remainder are produced by plant B. The proportion of defective components produced at plant A is 1% and the proportion of defective components produced at plant B is 2%.If a component received by the manufacturer is defective, the probability that it was produced at plant A isA. 3/7.
B. 2/7.
C. 4/7.D. 1/7.
Answer:
A) 3/7
Step-by-step explanation:
We start by calculating the following probabilities:
P(produced by A) = 0.6
P(produced by A and defective) = P(A ∩ def) = 0.6*0.01 = 0.006
P(produced by A and not defective) = P(A ∩ not def) = 0.6*0.99 = 0.594
P(produced by B and defective) = P(B ∩ def) = 0.4*0.02 = 0.008
P(produced by B and not defective) = P(B ∩ not def) = 0.4*0.98 = 0.392
The probability that it was produced by A given that it is defective is:
P(A|def) = P(A ∩ def) / P(def) = P(A ∩ def) / (P(A ∩ def)+P(B ∩ def)) = 0.006 / (0.006+0.008) = 6/14 = 3/7
Final answer:
The probability that a defective electronic component was produced by Plant A is 3/7. This was found using conditional probability and the information on production percentages and defect rates from both plants.
Explanation:
The question involves applying the concept of conditional probability to determine the probability that a defective electronic component was produced by Plant A. We need to calculate this using Bayes' theorem with the given probabilities for production and defect rates at both plants.
Calculate the probability of a component being defective, considering both plants.Compute the conditional probability that a defective component comes from Plant A.To answer the question: The probability of a component being defective from either Plant A or Plant B is calculated as follows:
P(Defective) = P(Defective | A)P(A) + P(Defective | B)P(B)
= (0.01)(0.60) + (0.02)(0.40)
= 0.006 + 0.008
= 0.014
Next, the probability that the component was produced at Plant A given that it is defective is:
P(A | Defective) = P(Defective | A)P(A) / P(Defective)
= (0.01)(0.60) / 0.014
= 0.006 / 0.014
= 3/7
Therefore, the correct answer is A. 3/7.
Steve Goodman, production foreman for the Florida Gold Fruit Company, estimates that the average sale of oranges is 4,700 and the standard deviation is 500 oranges. Sales follow a normal distribution. What is the probability that sales will be less than 4,300 oranges?
The probability that sales will be less than 4,300 oranges is 21.19%
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score, \mu=mean, \sigma=standard\ deviation\\\\Given\ that:\\\mu=4700,\sigma=500\\\\For\ x=4300:\\\\z=\frac{4300-4700}{500} =-0.8[/tex]
From the normal distribution table:
P(x < 4300) = P(z < -0.8) = 0.2119 = 21.19%
The probability that sales will be less than 4,300 oranges is 21.19%
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The probability that sales will be less than 4,300 oranges is approximately 21.23%.
Explanation:To find the probability that sales will be less than 4,300 oranges, we need to standardize the value using the z-score formula.
The z-score formula is given by z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. In this case, x = 4,300, μ = 4,700, and σ = 500.
Substituting these values into the formula, we get z = (4,300 - 4,700) / 500 = -0.8. We can then use a z-table or a calculator to find the probability associated with a z-score of -0.8, which is approximately 0.2123 or 21.23%.
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Assume that a procedure yields a binomial distribution with a trial repeated n times. Using the binomial probability formula, what is the probability of x successes given the probability p of success on a single trial? Round your answer to three decimal places.
Answer:
[tex]P(X=5)=(30C5)(0.2)^5 (1-0.2)^{30-5}=0.172[/tex]
Step-by-step explanation:
Assuming this complete question :"Assume that a procedure yields a binomial distribution with a trial repeated n times. Using the binomial probability formula, what is the probability of x successes given the probability p of success on a single trial? Round your answer to three decimal places.
n=30, x= 5, p=1/5"
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we know that:
[tex]X \sim Binom(n=30, p=0.2)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And for this case if we find the probability for x=5 we got:
[tex]P(X=5)=(30C5)(0.2)^5 (1-0.2)^{30-5}=0.172[/tex]
In expanded notation, the hexadecimal 74AF16 is (7*4096) + (4*256) + (A*16) + (F*1). When converting from hexadecimal to decimal, what value is assigned to F?
Answer:
F is assigned the value of 15
[tex]74AF_{16} = 29871_{10}[/tex]
Step-by-step explanation:
Hexadecimal number system is base 16 and it contain the following numbers:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
A has a value of 10
B has a value of 11
C has a value of 12
D has a value of 13
E has a value of 14
F has a value of 15
By completing the expanded notation:
[tex](7*4096) + (4*256) + (A * 16) + (F *1)\\= (7*4096) + (4*256) + (10 * 16) + (15 *1)\\= 28672 + 1024 + 160 + 15\\= 29871[/tex]
In hexadecimal notation, the letter 'F' corresponds to the decimal value 15. Therefore, when converting from hexadecimal to decimal, you would assign the value 15 to 'F'.
Explanation:In hexadecimal notation, the letters A through F correspond to the decimal values 10 through 15, respectively. When converting from hexadecimal to decimal, you would replace the hexadecimal digit 'F' with its decimal equivalent. Therefore, in the hexadecimal system, the letter 'F' signifies the decimal number 15. To calculate the value of 74AF16 in decimal, you replace 'F' with 15 and compute the expression (7*4096) + (4*256) + (10*16) + (15*1).
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Pretend today is your birthday, and you're hoping for some money. But your grandma is a finance professor and likes making things difficult for you. She tells you that she'll either give you $1,500 today, or give you $550 each year at the end of the year for the next 3 years. If the applicable discount rate is 6%, should you take the $1,500?
Answer:
no, the 550 each year is best offer
Step-by-step explanation:
Answer:
I’ll just stay with the 550 a year
Step-by-step explanation:
Suppose a test for a virus has a false-positive rate of 0.009 and a false-negative rate of 0.002. Assume that 1.5% of the population has the virus. (a) What is the chance someone from this population will test positive? (Enter exact answer.) (b) If someone tests positive, what is the chance he actually has the virus? (Answer correct to four decimal places.)
Answer:
(a) 0.023835
(b) 0.6281
Step-by-step explanation:
(a) The chance someone from this population will test positive is given by the percentage of people who have the virus multiplied by the change of testing positive (1 - false-negative rate) added to the percentage of people who do not have the virus multiplied by the change of testing positive (false-positive rate)
[tex]P(+) = 0.015*(1-0.002)+(1-0.015)*0.009\\P(+) = 0.023835[/tex]
(b) The probability that someone actually has the virus given that they have tested positive is determined as the probability of having the virus and testing positive divided by the probability of testing positive:
[tex]P(V|+) = \frac{ 0.015*(1-0.002)}{0.023835}\\P(V|+) = 0.6281[/tex]
Let A = {•, □, ⊗} and B = {□, ⊖, •}. (a) List the elements of A×B and B ×A. The parentheses and comma in an ordered pair are not necessary in cases such as this where the elements of each set are individual symbols.
Answer:
Elements of AxBb and BxA have been listed in the attached file
Step-by-step explanation:
The concept applied is that of binary operation and generally using the rule of combining more than one operations sign in either communitative or associative property as shown in the attachment.
You spend $40 for 8 hamburgers and 4 hotdogs at a ballgame. The next game you spend $32 for 3 hamburgers and 10 hotdogs. Write a system of linear equations to represent this scenario.
Answer: the system of linear equations to represent this scenario are
8x + 4y = 40
3x + 10y = 32
Step-by-step explanation:
Let x represent the cost of one hamburger.
Let y represent the cost of one hotdog.
You spend $40 for 8 hamburgers and 4 hotdogs at a ballgame. This means that
8x + 4y = 40 - - - - - - - - - - - - 1
The next game you spend $32 for 3 hamburgers and 10 hotdogs. This means that
3x + 10y = 32- - - - - - - - - - - - 2
Multiplying equation 1 by 3 and equation 2 by 8, it becomes
24x + 12y = 120
24x + 80y = 256
- 68y = - 136
y = - 136 /- 68
y = 2
Substituting y = 2 into equation 2, it becomes
3x + 10y = 32
3x + 10 × 2 = 32
3x = 32 - 20 = 12
x = 12/3 = 4
A forensic scientist uses the functions
G() = 2.56f+47.24 and H(t) = 2.74t+61.22
to find the height of a woman if the scientist is given the length of the woman's
femur bone for the length of the woman's tibia bone t in centimeters. Find the height of a woman whose femur measures 49 centimeters
The height of a woman whose femur measures 49 centimeters is
(Simplify your answer.)
Since we're given the femour's length, we'll have to use the first function.
If we substitute [tex]f=49[/tex] in the expression we have
[tex]g(f)=2.56f+47.24 \implies g(49)=2.56\cdot 49+47.24=125.44+47.24=172.68[/tex]
The height of the woman with a femur length of 49 cm is 172.68 cm.
What is a function?A function is a relationship between inputs where each input is related to exactly one output.
Example:
f(x) = 2x + 1
f(1) = 2 + 1 = 3
f(2) = 2 x 2 + 1 = 4 + 1 = 5
The outputs of the functions are 3 and 5
The inputs of the function are 1 and 2.
We have,
G(f) = 2.56f + 47.24
f = 49 cm
G(49) = 2.56 x 49 + 47.24
G(49) = 125.44 + 47.24
G(49) = 172.68 cm
Thus,
The height of the woman is 172.68 cm.
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A 11-inch candle is lit and burns at a constant rate of 1.3 inches per hour. Let t represent the number of hours since the candle was lit, and suppose f is a function such that f ( t ) represents the remaining length of the candle (in inches) t hours after it was lit. Write a function formula for f .
Answer:
f(t)= 11 in - 1.3 in/h *t
Step-by-step explanation:
defining the length of the candle as L , then since the candle burns at a constant rate , then
-dL/dt = 1.3 in/h = a
therefore
-∫dL = a∫dt
-L(t)=a*t + C , C=constant
at t=0 , the length of the candle is L₀= 11 in ,thus
-L₀=a*0 + C → C= -L₀
replacing the value of C
-L(t)=a*t - L₀
L(t) = L₀ - a*t = 11 in - 1.3 in/h *t
then
f(t)= 11 in - 1.3 in/h *t
A group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets. If they randomly decide who will watch the movie, what is the probability that there are at least 3 girls in the group that watch the movie? A. 0.018 B. 0.268 C. 0.536 D. 0.821
Answer:
D. 0.821
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The combinations formula is important to solve this question:
[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
The order is not important. For example, Elisa, Laura and Roze is the same outcome as Roze, Elisa and Laura. This is why we use the combinations formula.
At least 3 girls.
3 girls
3 girls from a set of 5 and 2 boys from a set of 3. So
[tex]C_{5,3}*C_{3,2} = 30[/tex]
4 girls
4 girls from a set of 5 and 1 boy from a set of 3. So
[tex]C_{5,4}*C_{3,1} = 15[/tex]
5 girls
5 girls from a set of 5
[tex]C_{5,5} = 1[/tex]
[tex]D = 30+15+1 = 46[/tex]
Total outcomes
5 from a set of 8. So
[tex]T = C_{8,5} = 56[/tex]
Probability
[tex]P = \frac{D}{T} = \frac{46}{56} = 0.821[/tex]
So the correct answer is:
D. 0.821
Answer:
I got answer choice D
Hope this helps :)
Step-by-step explanation:
On an assembly line that fills 8-ounce cans, a can will be rejected if its weight is less than 7.90 ounces. In a large sample, the mean and the standard deviation of the weight of a can is measured to be 8.05 and 0.05 OZ, respectively. (a) Calculate the percentage of the cans that is expected to be rejected on the basis of the given criterion. (b) If the filling equipment is adjusted so that the average weight becomes 8.10 OZ, but the standard deviation remains 0.05 OZ, calculate the rejection rate (% of cans being rejected) . (c) If the filling equipment is adjusted so that the average weight remains 8.05 OZ, but the standard deviation is reduced to 0.03 OZ, calculate the rejection rate.
The percentage of cans expected to be rejected based on given mean and standard deviation are calculated using the Z score and standard normal distribution table. By adjusting the mean and standard deviation, the rejection rates will change accordingly.
Explanation:This question is about calculating the expected rejection rate of cans based on different conditions using statistical concepts like mean and standard deviation.
(a) The Z score for 7.9 is (7.9 - 8.05) / 0.05 = -3. We use the standard normal distribution table to find the probability of a can having weight less than 7.9 ounces. That's almost 0.1% (0.001), so about 0.1% of cans are expected to be rejected.
(b) After adjusting the average weight to 8.1 oz, the Z score for 7.9 becomes (7.9 - 8.1) / 0.05 = -4. Again, find the probability in the standard normal distribution table, it is almost 0, so the rejection rate will drastically decrease.
(c) When the standard deviation is reduced to 0.03 but mean remains 8.05, the Z score becomes (7.9 - 8.05) / 0.03 = -5. The rejection rate will be extremely close to 0 as per standard normal distribution table reference.
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a newborn is treated for pulmonary valve stenosis; stretching of the valve opening is accomplished via a percutaneous balloon pulmonary valvuloplasty. what is the root operation?
Answer:
dilation
Step-by-step explanation:
When the pulmonary valve does not work properly, it can interfere with blood flow from the heart to the lungs, as well as force the heart to work harder to carry the blood that is needed to the rest of the body. Some children have heart conditions present at the time of birth and may require repair or replacement of the pulmonary valve, this option has a lower risk of infection, preserves the strength and functioning of the valve, and eliminates the need to take medication.
Final answer:
The root operation for treating pulmonary valve stenosis in newborns using balloon valvuloplasty is the valvuloplasty itself, which is a non-surgical procedure to widen the stenosed heart valve.
Explanation:
The root operation for treating newborn pulmonary valve stenosis with percutaneous balloon pulmonary valvuloplasty is the procedure known as valvuloplasty.
This procedure involves the insertion of a specialized catheter with a balloon at its tip into a blood vessel, usually via the leg, and navigating it to the valve.
The balloon is then inflated to widen the stenosed valve and allow better blood flow. Subsequently, the balloon is deflated and removed, completing the valvuloplasty.
Examine the diagram below and find the value of angle a and angle b
Answer:
50°, 105°
Step-by-step explanation:
a + 130 = 180 ( sum of angles on a straight line)
a = 180 - 130 = 50°
a + b + 45 = 180 ( sum of angles in a Δ)
30 + b + 45 = 180
75 + b = 180
b = 180 - 75 = 105°
Howard collected data from a random sample of 600 people in his department asking whether or not they use the company's healthcare . Based on the results, he reports that 48% of the people in his company use the company's healthcare. Why is this statistic misleading
Answer:
This statistic is misleading because Howard surveys only his department, and not membes of all the departments that the company has.
Step-by-step explanation:
This is a common statistics practice, when we want to study something from a population, we find a sample of this population.
However, the sample has to be representative
For example:
I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected Buffalo residents wheter they are Buffalo Bills fans, and expand this to the entire population of New York State residents. This is not representative of all New York State residents, just Buffalo residents.
In this problem, we have that:
Howard wants to know the proportion of employees of a company who use the company's healthcare. He asks only his department. However, a company as multiple departments, which leads to the statistics found in Howard's survey being misleading.
Final answer:
Howard's statistic that 48% of people in his company use the healthcare may be misleading due to potential issues like non-representative sampling, biased survey questions, response bias, and lack of current context.
Explanation:
The question about Howard reporting that 48% of the people in his company use the company's healthcare is misleading may have several underlying reasons. Firstly, the sample size and how the sample was obtained are critical factors that determine the reliability and representation of the poll. A random sample of 600 people is generally a good size, but if the sample is not representative of the entire company's demographics, the statistic could be misleading.
Another potential issue could be the phrasing of the survey question. Questions that are biased or leading can influence the way people respond and thus skew the results. In addition, respondents might have reasons to provide socially desirable answers rather than true ones, or they might misremember their actual usage of healthcare. This is an example of response bias, which can lead to inaccurate data.
Finally, it's vital to consider the context, such as whether the data is recent and if there have been any significant changes in the company's healthcare policy or overall employee demographics since the survey. All these factors can contribute to why a statistic might be considered misleading, as it may not accurately reflect the true situation.
We apply the Empirical Rule when the relative frequency distribution of the sample is not bell-shaped or symmetric Group of answer choices True False
Answer:
We can conclude that this statement is False. Because the Empirical Rule does not apply to data sets with severely asymmetric distributions, since by definition the use of the rule is satisfied just for symmetric distributions like the normal distribution.
And if the distribution is not bell shaped or symmetric then we can't use it.
Step-by-step explanation:
Previous concepts
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). "Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".
Solution to the problem
We can conclude that this statement is False. Because the Empirical Rule does not apply to data sets with severely asymmetric distributions, since by definition the use of the rule is satisfied just for symmetric distributions like the normal distribution.
And if the distribution is not bell shaped or symmetric then we can't use it.
The statement 'We apply the Empirical Rule when the relative frequency distribution of the sample is not bell-shaped or symmetric' is False. The Empirical Rule is applied when the distribution of the data is bell-shaped and symmetric.
The statement 'We apply the Empirical Rule when the relative frequency distribution of the sample is not bell-shaped or symmetric' is False.
The Empirical Rule is applied when the distribution of the data is bell-shaped and symmetric.
It states that approximately 68% of the data is within one standard deviation of the mean, 95% is within two standard deviations, and more than 99% is within three standard deviations.
Therefore, the Empirical Rule is not applied when the data is not bell-shaped or symmetric.
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A publication released the results of a study of the evolution of a certain mineral in the Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 55 and 1010 parts per million
a. Find E(x) and interpret its value
b. Compute P(2.875 x35)
c. Computn Plx<4.125)
Answer:
a) [tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]
b) [tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]
c) [tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]
Step-by-step explanation:
If we work with the limits defined from 5 to 10 then part b and c from this question not makes sense. If we work with the limits 1 and 6 all the parts for the question makes sense because if we work with 5 and 10 the only thing that we can find is the expected value [tex] E(A) = \frac{5+10}{2}= 7.5[/tex]
Assuming the following correct question : "A publication released the results of a study of the evolution of a certain mineral in the Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 1 and 6 parts per million"
Solution to the problem
Let A the random variable that represent " amount of the mineral x ". And we know that the distribution of A is given by:
[tex]A\sim Uniform(1 ,6)[/tex]
Part a
For this uniform distribution the expected value is given by [tex]E(X) =\frac{a+b}{2}[/tex] where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:
[tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]
Part b
For this case we can use the cumulative distribution function for the uniform distribution given by:
[tex] F(X=x)= \frac{x-a}{b-a} = \frac{x}{6-1} =\frac{x-1}{5} , 1 \leq X \leq 6[/tex]
And we want this probability:[tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]Part c
For this case we want this probability:
[tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]
In a normally distributed data set with a mean of 19 and a standard deviation of 2.6, what percentage of the data would be between 16.4 and 21.6?
Answer:
68.26% of the data would be between 16.4 and 21.6.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 19, \sigma = 2.6[/tex]
What percentage of the data would be between 16.4 and 21.6?
This is the pvalue of Z when X = 21.6 subtracted by the pvalue of Z when X = 16.4. So
X = 21.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.6 - 19}{2.6}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
X = 16.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16.4 - 19}{2.6}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
So 0.8413 - 0.1587 = 0.6826 = 68.26% of the data would be between 16.4 and 21.6.
Answer: Percentage = 0.6826 X 100 = 68.26%
Step-by-step explanation: Please find the attached document for the step by step explanation
Answer if you have a big brain 96 POINTS
Answer:
1) 2x+7
2) -3x+11
3) 0.75x-2
4) -2x+0
5) -1.5x+2
6) -4x+16
Step-by-step explanation:
1) y = mx + c
m = 2 when x=1 , y=9
9 = 2(1)+c
c = 7
y = 2x + 7
2) m = -3
When x=4, y= -1
-1 = -3(4) + c
c = -1+12 = 11
y = -3x + 11
3) m = 0.75
When x= -4, y= -5
-5 = 0.75(-4) + c
-5 = -3 + c
c = -2
y = 0.75x - 2
4) m = (y2-y1)/(x2-x1)
m = (2-(-6))/(-1-3) = 8/-4 = -2
y = -2x + c
When x= -1, y= 2
2 = -2(-1) + c
2 = 2 + c
c = 0
y = -2x + 0
5) m = (-10-(-4))/(8-4)
m = (-10+4)/4 = -6/4 = -1.5
y = -1.5x + c
When x= 4, y= -4
-4 = -1.5(4) + c
-4 = -6 + c
c = 2
y = -1.5x + 2
6) m = (-4-4)/(5-3) = -8/2 = -4
When x= 3, y= 4
4 = -4(3) + c
4 = -12 + c
c = 16
y = -4x + 16
Answer:
1) 2x+7
2) -3x+11
3) 0.75x-2
4) -2x+0
5) -1.5x+2
6) -4x+16
Step-by-step explanation:
In a box containing 25 cherries, 2 of them are rotten. Susan randomly picks cherries in the box. How many cherries should be picked so that the probability of having exactly 2 rotten cherries among them equals 1/20?
Answer:
Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20
Step-by-step explanation:
assuming that each cherry is equally probable to be chosen , since each cherry is independent from the others and sampling is done without replacement , the random variable X= number of cherries that are rotten from the picked ones follows a hyper geometrical distribution , where
P(X=k)= C(M,k) * C(N-M, n-k) / C(N,n)
where
N= population size = 25
n= number of picks
M = total number of rotten cherries =2
k = number of rotten cherries picked =2
C( ) = combination
then
1/20=C(2,2)*C(25-2,n-2)/C(25,n) = 1 * (23!/(n-2)!*(25-n)! / (25!/(n!*(25-n)!
1/20 = n!/(n-2)! * 1/(24*25)
24*25/20 = n*(n-1)
n²-n-30 =0
n= (1 +√(1+4*1*30))/2 = 12/2= 6
n=6
then Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20
To find the number of cherries Susan should pick, use the combination formula and probability calculation. After setting up an equation with probability equal to 1/20, solve for 'x' using trial and error methods.
Explanation:To answer the question, we need to use the combination formula. This formula in the field of statistics is used to find the number of possible combinations that can be obtained by taking 'r' elements from a set of 'n' elements.
The formula is: C(n, r) = n! / [(n - r)! * r!]
Given 25 cherries, 2 of which are rotten, Susan wants to choose some cherries such that the probability of getting exactly 2 rotten cherries is 1/20. Let's assume she needs to pick 'x' cherries.
Now we can write the probability equation: Probability = [C(2, 2) * C(23, x - 2)] / C(25, x) = 1/20
Unfortunately, we can't explicitly solve this equation because it would require checking different values of 'x'. However, it can be solved manually or through trial and error using software or a calculator. After checking different values, you can find the 'x' that satisfies the equation.
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Tristan and Iseult play a game where they roll a pair of dice alternatingly until Tristan wins by rolling a sum 9 or Iseult wins by rolling a sum of 6.
If Tristan rolled the dice first, what is the probability that Tristan wins?
Answer:
If Tristan rolled the dice first the probability that Tristan wins is 0.474.
Step-by-step explanation:
The probability of an event E is computed using the formula:
[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}[/tex]
Given:
Tristan and Iseult play a game where they roll a pair of dice alternatively until Tristan wins by rolling a sum 9 or Iseult wins by rolling a sum of 6.
The sample space of rolling a pair of dice consists of a total of 36 outcomes.
The favorable outcomes for Tristan winning is:
S (Tristan) = {(3, 6), (4, 5), (5, 4) and (6, 3)} = 4 outcomes
The favorable outcomes for Iseult winning is:
S (Iseult) = {(1, 5), (2, 4), (3, 3), (4, 2) and (5, 1)} = 5 outcomes
Compute the probability that Tristan wins as follows:
[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}\\ P(Tristan\ wins)=\frac{4}{36}\\P(T)=\frac{1}{9} \\\approx0.1111[/tex]
Compute the probability that Iseult wins as follows:
[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}\\ P(Iseult\ wins)=\frac{4}{36}\\P(I)=\frac{1}{9} \\\approx0.1111[/tex]
If Tristan plays first, then the probability that Tristan wins is:
= P(T) + P(T')P(I')P(T) + P(T')P(I')P(T')P(I')P(T)+...
=P(T) + [(1-P(T))(1-P(I))P(T)]+[(1-P(T))(1-P(I))(1-P(T))(1-P(I))P(T)]+...
[tex]=0.1111+(0.8889\times0.8611\times0.1111)+(0.8889\times0.8611\times0.8889\times0.8611\times0.1111))+...\\=0.1111[1+(0.8889\times0.8611)+(0.8889\times0.8611)^{2}+...]\\[/tex]This is an infinite geometric series.
The first term is, a = 0.1111 and the common ratio is, r = (0.8889×0.8611).
The sum of infinite geometric series is:
[tex]S_{\infty}=\frac{a}{1-r}\\ =\frac{0.1111}{1-(0.8889\times0.8611}\\ =0.47364\\\approx0.474[/tex]
Thus, the probability that Tristan wins if he rolled the die first is 0.474.
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. (Enter your answers to three decimal places.) (a) P(all of the next three vehicles inspected pass) (b) P(at least one of the next three inspected fails) (c) P(exactly one of the next three inspected passes)
Based on the given information the required probabilities are as follows:
(a) Probability that all of the next three vehicles pass: 0.343
(b) Probability that at least one of the next three vehicles fails: 0.657
Given that,
70% of all vehicles examined at a certain emissions inspection station pass the inspection.
Successive vehicles pass or fail independently of one another.
(a) To find the probability that all three vehicles pass,
Multiply the individual probabilities.
Given that 70% of vehicles pass,
The probability that a single vehicle passes is 0.7.
So, the probability that all three vehicles pass is 0.7³ = 0.343.
(b) To find the probability that at least one of the next three vehicles fails, We can find the complement probability and subtract it from 1.
The complement probability is the probability that all three vehicles pass, which we calculated in the previous part.
So, the probability that at least one vehicle fails is 1 - 0.343 = 0.657.
(c) To find the probability that exactly one of the next three vehicles passes,
Use the binomial probability formula:
[tex]P(X=k) = ^nC_k p^k (1-p)^{(n-k)}[/tex],
Where n is the number of trials,
k is the number of successes,
p is the probability of success, and
C(n,k) is the combination of n and k.
In this case,
n = 3,
k = 1,
p = 0.7
Plugging these values into the formula,
We get [tex]P(X=1) = ^3C_1\times 0.7 \times (1-0.7)^2[/tex]
P(X=1) = 3x0.7x0.3²
P(X=1) = 0.189
So, the probabilities are:
(a) P(all of the next three vehicles inspected pass) = 0.343
(b) P(at least one of the next three inspected fails) = 0.657
(c) P(exactly one of the next three inspected passes) = 0.189
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To calculate the probabilities, we use the binomial probability formula. The probability of all three vehicles passing is 0.343, the probability of at least one vehicle failing is 0.657, and the probability of exactly one vehicle passing is 0.189.
Explanation:To calculate the probabilities, we can use the binomial probability formula. Let's solve each part step by step:
(a) P(all of the next three vehicles inspected pass):
The probability of each vehicle passing is 70%, so the probability of all three passing is 0.7 x 0.7 x 0.7 = 0.343.
(b) P(at least one of the next three inspected fails):
The probability of a vehicle failing is 30%, so the probability of all three passing is 1 - (0.7 x 0.7 x 0.7) = 0.657.
(c) P(exactly one of the next three inspected passes):
There are three possible scenarios where exactly one vehicle passes: (Pass, Fail, Fail), (Fail, Pass, Fail), (Fail, Fail, Pass). The probability of each scenario is 0.7 x 0.3 x 0.3 = 0.063. Since there are three scenarios, the total probability is 0.063 x 3 = 0.189.
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