Answer:
Have a broad mix of organizational levels in the JAD session
Explanation:
It is not possible for Hamid to include every employee in the JAD session, what Hamid needed to do is to select participants from the different departments and other key important people to ensure every one is well represented at the JAD session. Selecting lower-level and mid-level managers from the affected departments as well as the some senior staff and the project sponsor for the JAD session will ensure everyone's interest is well represented at the session.
The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The students must qualify by meeting certain standards in terms of GPA and standardized test scores. The goal of this new policy is to increase the geographic diversity of students from states beyond Arkansas and its border states. Historically, 90% of all new students came from Arkansas or a bordering state. Ginger, a student at the U of A, sampled 180 new students the following year and found that 157 of the new students came from Arkansas or a bordering state. Does Ginger’s study provide enough evidence to indicate that this new policy is effective with a level of significance 10%? What would be the correct decision?
a. Reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansasb. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
c. Reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
d. Fail to reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansas
Answer:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
Step-by-step explanation:
Data given and notation n
n=180 represent the random sample taken
X=157 represent the students who came from Arkansas or a bordering state
[tex]\hat p=\frac{157}{180}=0.872[/tex] estimated proportion of students who came from Arkansas or a bordering state
[tex]p_o=0.9[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher or not than 0.9.:
Null hypothesis:[tex]p\geq 0.9[/tex]
Alternative hypothesis:[tex]p < 0.9[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
There are two machines available for cutting corks intended for use in bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. Which machine is more likely to produce an acceptable cork? What should the acceptable range for cork diameters be (from 3 − d cm to 3 + d cm) to be 90% certain for the first machine to produce an acceptable cork?
Answer:
a) The second machine is more likely to produce an acceptable cork.
b) Acceptable range for cork diameters produced by the first machine with a 90% confidence = (2.8355, 3.1645)
Step-by-step explanation:
This is a normal distribution problem
For the first machine,
Mean = μ = 3 cm
Standard deviation = σ = 0.1 cm
And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.
P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)
We first standardize this measurements.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 2.9 cm
z = (x - μ)/σ = (2.9 - 3.0)/0.1 = - 1.00
For 3.1 cm
z = (x - μ)/σ = (3.1 - 3.0)/0.1 = 1.00
P(x ≤ 3.1) = P(z ≤ 1.00) = 0.841
P(x ≤ 2.9) = P(z ≤ -1.00) = 0.159
P(2.9 ≤ x ≤ 3.1) = P(-1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) - P(z ≤ -1.00) = 0.841 - 0.159 = 0.682 = 68.2%
This means that 68.2% of the diameter of corks produced by the first machine lies between 2.9 cm and 3.1 cm.
For the second machine,
Mean = μ = 3.04 cm
Standard deviation = σ = 0.02 cm
And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.
P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)
We standardize this measurements.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 2.9 cm
z = (x - μ)/σ = (2.9 - 3.04)/0.02 = - 7.00
For 3.1 cm
z = (x - μ)/σ = (3.1 - 3.0)/0.02 = 3.00
P(x ≤ 3.1) = P(z ≤ 3.00) = 0.999
P(x ≤ 2.9) = P(z ≤ -7.00) = 0.0
P(2.9 ≤ x ≤ 3.1) = P(-7.00 ≤ z ≤ 3.00) = P(z ≤ 3.00) - P(z ≤ -7.00) = 0.999 - 0.0 = 0.999 = 99.9%
This means that 99.9% of the diameter of corks produced by the second machine lies between 2.9 cm and 3.1 cm.
Hence, we can conclude that the second machine is more likely to produce an acceptable cork.
b) Margin of error = (z-multiplier) × (standard deviation of the population)
For 90% confidence interval, z-multiplier = 1.645 (from literature and the z-tables)
Standard deviation for first machine = 0.1
Margin of error, d = 1.645 × 0.1 = 0.1645.
The acceptable range = (mean ± margin of error)
Mean = 3
Margin of error = 0.1645
Lower limit of the acceptable range = 3 - d = 3 - 0.1645 = 2.8355
Upper limit of the acceptable range = 3 + d = 3 + 0.1645 = 3.1645
Acceptable range = (2.8355, 3.1645)
Final answer:
To determine which machine is more likely to produce acceptable corks, we examine their distribution characteristics. Machine 2 may be more reliable due to its tighter control despite a slightly higher mean. To find a 90% certain acceptable range for Machine 1, we calculate using its standard deviation and the z-score for the 90th percentile.
Explanation:
The question involves comparing two machines based on their ability to produce corks within a specified acceptable diameter range using normal distribution properties, and calculating the range for diameters to ensure a 90% certainty of producing acceptable corks for the first machine.
Comparing the Two Machines
For the first machine with a mean diameter of 3 cm and a standard deviation of 0.1 cm, and the second machine with a mean diameter of 3.04 cm and a standard deviation of 0.02 cm, the question is which machine is more likely to produce corks within the acceptable range of 2.9 cm to 3.1 cm.
Machine 1 produces corks closer to the center of the acceptable range but with a wider spread (higher standard deviation), while Machine 2 produces corks that are skewed slightly larger but with a much tighter spread around their mean (lower standard deviation). To determine which machine is more likely to produce acceptable corks, we would need to calculate the z-scores for the acceptance limits for both machines and compare the probabilities. However, intuitively, Machine 2 might be seen as more reliable due to its tighter control (lower standard deviation), assuming its mean is not too far out of the acceptable range.
Finding the Acceptable Range for 90% Certainty
To ensure 90% certainty that a cork produced by Machine 1 falls within an acceptable diameter range, we need to determine d in the range of 3 − d cm to 3 + d cm. This involves finding the z-score that corresponds to the 5th and 95th percentiles due to the symmetric nature of normal distribution, then solving for d using the properties of normal distribution and the given standard deviation of 0.1 cm.
The z-score corresponding to the 5th and 95th percentiles (for a 90% certainty) typically falls around ±1.645. Using the formula for z-score, which is (X − μ) / σ, and solving for d, we can find the acceptable range of diameters for the first machine to produce an acceptable cork with 90% certainty.
g On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any other day. Calculate the probability that, starting with tomorrow, the second day ofrain will occur within 5 days
Answer:
0.249579
Step-by-step explanation:
P(rain) = 0.3
P(no rain) = 1 - 0.3 = 0.7
The event of rain falling a second time within the next 5 days is possible in these ways
1. Rain on days 1 and 2
2. Rain on days 1 and 3; none on day 2
3. Rain on days 1 and 4; none on days 2 and 3
4. Rain on days 1 and 5; none on days 2, 3 and 4
5. Rain on days 1 and 6; none on day 2, 3, 4 and 5
[tex]P(\text{second rain within 5 days}) = 0.3^2+0.3^2\times0.7+0.3^2\times0.7^2+0.3^2\times0.7^3+0.3^2\times0.7^4 = 0.3^2(1+0.7+0.7^2+0.7^3+0.7^4)= 0.09\times2.7731=0.249579[/tex]
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)
Answer:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
Given that,
A company has found that in carton of transistors: 92% contain no defective transistors,
3% contain one defective transistor, 3% contain two defective transistors,
and 2% contain three defective transistors.
We have to find,
Calculate the mean and variance for the defective transistors. Mean = Variance.
According to the question,
Let, X the random variable who represent the number of defective transistors.
The following probability distribution for X,
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
To calculate the expected value with the following formula:
[tex]E(X) = \sum^{n}_{i=1} X_i. P.(X_i)[/tex]
On substitute all the values in the formula,
[tex]E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15[/tex]
To find the variance first the second moment given by:
[tex]E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)[/tex]
On substitute all the values in the formula;
[tex]E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330[/tex]
Therefore, The variance is calculated with this formula:
[tex]Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance,
[tex]Standard \ deviation = \sqrt{0.0375} = 0.5545[/tex]
Hence, The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
To know more about Mean click the link given below.
https://brainly.com/question/15167067
A blood bank asserts that a person with type O blood and a negative Rh factor (Rh?) can donate blood to any person with any blood type. Their data show that 43% of people have type O blood and 19% of people have Rh? factor; 45% of people have type O or Rh? factor.1.) Find the probability that a person has both type O blood and the Rh? factor.2.) Find the probability that a person does NOT have both type O blood and the Rh? factor.
Answer:P(O)UP(Rh)=17% while n(O)U(Rh) complement is 87%
Step-by-step explanation:
Since 43% represent those blood group O and 19% those with blood Rh,summing up gives 62%
Subtract 45% from 62%=17%
The probability that the person doesn't have either blood group is1 -17%=83%
Answer:
The probability that a person has a positive Rh factor given that he/she has type O blood is 82 percent.
There is a greater probability for a person to have a Positive Rh factor given type A blood than a person to have a positive Rh factor given type O blood.
3. It is known that 80% of all college professors have doctoral degrees. If 10 professors are randomly selected, find the probability that a. five have a doctoral degree b. fewer than 4 have doctoral degrees. c. At least 6 have doctoral degrees. d. Between 5 and 7 (inclusive) have doctoral degrees. e. What is the expected number of college professors with doctoral degrees
Answer:
a) [tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
b) [tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
c) [tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
d) [tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
e) [tex] E(X) = n*p = 10*0.8 =8[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=10, p=0.8)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
We want to find this probability:
[tex] P(X=5)[/tex]
And using the pmf we got:
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
Part b
[tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
Part c
For this case we can use the complement rule like this:
[tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
Part d
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
Part e
The expected number is given by:
[tex] E(X) = n*p = 10*0.8 =8[/tex]
This answer provides step-by-step explanations for finding probabilities of professors with doctoral degrees and the expected number of professors with doctoral degrees.
Expected number of college professors with a doctoral degree:
Find the probability that five have a doctoral degree: Using the binomial probability formula, [tex]P(X = 5) = 10C5 * 0.8^5 * 0.2^5[/tex]
Find the probability that fewer than 4 have a doctoral degree: Calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Find the probability that at least 6 have a doctoral degree: Calculate P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)).
Find the probability that between 5 and 7 inclusive have a doctoral degree: Calculate P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7).
Expected number of professors with a doctoral degree: Multiply the total number of professors by the probability (0.8).
Bob can row 13mph in still water. The total time to travel downstream and return upstream to the starting point is 2.6 hours. If the total distance downstream and back is 32 miles. Determine the speed of the river (current speed)
Answer:
"s" is the river speed 16 miles is the one-way distance
Time = distance / velocity
2.6 = [16 / (13 + s) ] + [16 / (13 -s) ]
That solves to s = 3 miles per hour
Step-by-step explanation:
Answer: the speed of the current is 3 mph
Step-by-step explanation:
Let x represent the speed of the river current.
Bob can row 13mph in still water. Assuming that while rowing upstream, he rowed against the current, this means that his total speed upstream is (13 - x) mph
Assuming that while rowing downstream, he rowed with the current, this means that his total speed downstream is (13 + x) mph.x
If the total distance downstream and back is 32 miles. Assuming the distance upstream and downstream is the same, then the distance upstream is 32/2 = 16 miles. Distance downstream is also 16 miles.
Time = Distance/speed
Time taken to row upstream is
16/(13 - x)
Time taken to row downstream is
16/(13 + x)
If total time spent is 2.6 hours, it means that
16/(13 - x) + 16/(13 + x) = 2.6
Cross multiplying, it becomes
16(13 + x) + 16(13 - x) = 2.6(13 + x)(13 - x)
= 208 + 16x + 208 - 16x = 2.6(169 - 13x + 13x - x²)
416 = 2.6(169 - x²)
416/2.6 = 169 - x²
160 = 169 - x²
x² = 169 - 160 = 9
x =√9
x = 3
The number of hours needed to paint a house, h, varies inversely with the number of painters, n. Four painters need 6 hours to paint a house. How many hours would it take 3 painters to paint the house?
Answer:
It would take 3 painters, a total number of 8 hours, to paint the house
Step-by-step explanation:
Let the number of hours be denoted by N, and the number of painters be denoted by P. The inverse relationship can be expressed as:
N α [tex]\frac{1}{P}[/tex]
Removing the proportionality symbol (α), introduces a constant of proportionality, which can be assumed to be C, in this example.
So, N = [tex]\frac{C}{P}[/tex]
When N = 6, P = 4
6 = [tex]\frac{C}{4}[/tex]
C = 24
So that, N = [tex]\frac{24}{P}[/tex]. This is the formula to use for determination of the other variable when one of them is known.
Now, given 3 painters, the number of hours needed can be obtained as follows:
N = [tex]\frac{24}{3}[/tex] = 8 hours
A soda filling machine is supposed to fill cans of soda with 12 fluid ounces. Suppose that the fills are actually normally distributed with a mean of 12.1 oz and a standard deviation of 0.2 oz. a) What is the probability of one can less than 12 oz
Answer:
30.85% probability of one can less than 12 oz
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 12.1, \sigma = 0.2[/tex]
a) What is the probability of one can less than 12 oz
This is the pvalue of Z when X = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 12.1}{0.2}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085
30.85% probability of one can less than 12 oz
g Suppose that when a certain lake is stocked with fish, the birth and death rates ˇ and ı are both inversely propor- (a) Show that ????1 p????2 P.t/D 2ktC P0 ; where k is a constant. (b) If P0 D 100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?
Answer:
k=1
P(12)=256
Step-by-step explanation:
The model for population involving birth and death is given as:
[tex]\frac{dP}{dt}=(b-d)P[/tex]
where b=birth rate and d=death rate.
If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]
[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.
Substituting b and d into our model
[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant
Simplifying using indices
[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]
Next, we Separate Variables and Integrate both sides
[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]
[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]
[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration
[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]
When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.
(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]
Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]
Therefore the population after a year is 256.
The population of fish after 1 year would be 285.
Let P be the population of fish at any time t months.
We are given:
Initial population: [tex]P_0[/tex] = 100 fishPopulation after 6 months: P(6) = 169 fishWe need to determine the population after 12 months.
The logistic growth model can be expressed as:
dP/dt = rP(1 - P/K)
where r is the intrinsic growth rate and K is the carrying capacity of the environment. Here, since the birth and death rates are inversely proportional to √P, we can infer that the growth rate might be proportional to √P, leading us to use a simplified logistic growth equation.
Given P(6) = 169, let’s find the intrinsic growth rate , r. From population changes between 0 and 6 months:
Let’s assume the logistic growth function in the form:
P(t) = [tex]P_0 e^{rt}[/tex]
We solve for r:
169 = [tex]100 e^{6r}[/tex]
1.69 = [tex]e^{6r}[/tex]
Taking natural logarithms:
ln(1.69) = 6r
0.524 = 6r
r = 0.0873/month
Now, to find P(12):
P(12) = [tex]100 e^{0.0873*12}[/tex]
= [tex]100 e^{1.048}[/tex]
= 100 * 2.853
= 285.3 fish
So, the population after 1 year is approximately 285 fish.
Using the fixed-time period inventory model, and given an average daily demand of 200 units, 4 days between inventory reviews, 5 days for lead time, 120 units of inventory on hand, a "z" of 1.96, and a standard deviation of demand over the review and lead time of 3 units, which of the following is the order quantity?
A. About 1,086
B.About 1,686
C. About 1,806
D. About 2,206
E. About 2,686
Answer:
Correct option: B. About 1,686.
Step-by-step explanation:
The formula to compute the order quantity (Q) is:
[tex]Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}[/tex]
Here
[tex]q_{d}=average\ daily\ semand=200\\I = Inventory\ review\ time=4\\L=lead\ time=5\\\sigma_{I+L}=standard\ deviation\ over\ the\ review\ and\ lead\ time=3\\I_{n}=number\ of\ units\ of\ inventory\ on\ hand=120[/tex]
Compute the order quantity as follows:
[tex]Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}\\=(200\times(4+5))+(1.96\times 3)-120\\=1800+5.88-120\\=1685.88\\\approx1686[/tex]
Thus, the order quantity was about 1,686.
Which of the following represents a null hypothesis? Group of answer choices Class A high school basketball teams who employ a sports psychologist will have a higher proportion ofwins over the course of the season than comparable teams who do not employ a sports psychologist. There will be no difference in rate of skill improvement between college gymnasts who practice meditation and those who do not. Does incorporating relaxation exercises into the daily practice routine of college vocal majors enhance their performance confidence? None of the above
In factual tests, an invalid speculation recommends that there is no importance between factors in a bunch of noticed information. In the given decisions, the assertion shows no distinction in the pace of ability improvement between school gymnasts who practice contemplation and the people who don't address the invalid speculation.
Explanation:In factual theory testing, an invalid speculation is an assertion recommending that no measurable relationship and importance exists in a bunch of noticed information between factors. It expects that any noticed contrasts are because of possibility. So in the given choices, the invalid speculation is: 'There will be no distinction in the pace of expertise improvement between school gymnasts who practice contemplation and the people who don't.' This assertion proposes no effect of the variable (reflection) on the result (ability improvement), which is what an invalid theory addresses in a trial of importance.
Learn more about the Null Hypothesis here:https://brainly.com/question/32386318
#SPJ3
In the last homework, you analyzed the equation ˙x = cx − x 3 graphically, for c positive, zero and negative. Now perform a formal linear stability analysis, and compare your results with a graphical analysis
Answer:
Please find attached file for complete answer solution and explanation of same question.
Step-by-step explanation:
We have a biased coin (probability of heads is equal to 1/4). Consider the following 2 step process: In the first step we flip the coin until we get a heads. Let X denote the trial on which the first heads occurs. In the second step we flip the coin X more times. Let Y be the number of heads in the second step. (a) For each non-negative integer, k, what is the probability that X = k? (b) Conditioned on the event {X = k}. What is the probability Y = 0? (c) Use the Law of Total Probability to compute the unconditional probability that Y = 0.
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached picture for detailed explanation.
suppose you have 3 bags containing only apples and oranges. bag a has 2 apples and 4 oranges, bag b has 8 apples and 4 oranges, and bag c has 1 apple and 3 oranges. you pick 1 fruit (at random) from each bag. a) what is the probability that you picked exactly 2 apples? b) suppose you picked 2 apples but forgot which bag they came from. what is the probability that you picked an apple from bag a?
Answer:
Step-by-step explanation:
There are three bags
Bag A
2apples and 4 oranges
P(A¹)=2/6
P(A¹)=⅓
P(O¹)=4/6
P(O¹)=⅔
Bag B
8 apples and 4 oranges
P(A²)=8/12
P(A²)=⅔
P(O²)=4/12
P(O²)=⅓
Bag C
1 apple and 3 oranges
P(A³)=¼
P(O³)=¾
Note
P(A¹) means probability of Apple in bag A
P(A²) means probability of Apple in bag B
P(A³) means probability of Apple in bag C
P(O¹) means probability of oranges in bag A.
P(O²) means probability of oranges in bag B.
P(O³) means probability of oranges in bag C.
a. The probability of picking exactly two apples can be analyzed as
Picking apple in bag A and picking apple in bag B and picking orange in bag C or picking apple in bag A and picking orange in bag B and picking apple in bag C or picking orange in bag A and picking apple in bag B and picking apple in bag C.
Then,
P(exactly two apples)=P(A¹ n A² n O³) + P(A¹ n O² n A³) + P(O¹ n A² n A³)
Since they are mutually exclusive
Then,
P(exactly two apples) =
P(A¹) P(A²)P(O³) + P(A¹)P(O²)P(A³) + P(O¹) P(A²) P(A³)
P(exactly two apples) =
(⅓×⅔×¾)+(⅓×⅓×¼)+(⅔×⅔¼)
P(exactly two apples)=1/6 +1/36 +1/9
P(exactly two apples)= 11/36
b. Probability that an apple comes from Bag A out of the two apple will be Picking apple in bag A and picking apple in bag B and picking orange in bag C or picking apple in bag A and picking orange in bag B and picking apple in bag C.
P(an apple belongs to bag A)=P(A¹ n A² n O³) + P(A¹ n O² n A³)
Since they are mutually exclusive
Then,
P(an apple belongs to bag A) =
P(A¹) P(A²)P(O³) + P(A¹)P(O²)P(A³)
P(an apple belongs to bag A) =
(⅓×⅔×¾)+(⅓×⅓×¼)
P(an apple belongs to bag A)=1/6 +1/36
P(an apple belongs to bag A)= 7/36
"According to contractarian logic, we should be willing to make concessions to others if they agree to make comparable and reciprocal concessions, with the overall result being that everyone gets a desired benefit with an acceptable minimum of sacrifice on each side. Please identify, and briefly analyze, a situation or scenario that illustrates this principle. "
Answer:In a couple with a newborn baby at home, to take turns on feeding the baby at night.
Step-by-step explanation: Here both parents are willing to sacrifice a few minutes, if not hours of sleeptime with the promise to be allowed to rest the next time the baby needs to be fed. There is no certainty in how long it will take for the baby to go back to sleep or how long it will be for the baby to be awake again, but the chances are the same for both parents, so they both agree to take care of the child one at a time with the promise to be in turns, this is an example of contractarian logic.
Final answer:
Contractarian logic is exemplified in international trade agreements where nations mutually lower tariffs under the expectation of reciprocal actions, demonstrating the principle of reciprocity and mutual advantage.
Explanation:
According to contractarian logic, we should be willing to make concessions to others if they agree to make comparable and reciprocal concessions, leading to a situation where everyone benefits with a minimal level of sacrifice. A classic example illustrating this principle is international trade agreements. Nations often agree to lower tariffs and grant each other favorable trade terms under the condition that the other nation reciprocates. These agreements are founded on the expectation that both sides will adhere to the agreements, benefiting both by expanding their markets and reducing costs for consumers. Here, the benefit is mutual economic growth, and the sacrifice might involve foregoing the protection of certain domestic industries in the interest of broader gains. This scenario mirrors the foundational ideas of social contract theory where rational, self-interested agents come together to agree on a set of rules or actions that benefit all parties involved, thereby demonstrating the principle of reciprocity and mutual advantage that is central to contractarian logic.
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack _____________.
Answer:
Reliability
Step-by-step explanation:
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack reliability.
Reliability can be defined as the extent to which an experiment or test provides the same result on repeated trials.
The huge difference in Dylan's IQ test scores in a short period of time suggests that the Whaddayanno IQ test is not reliable.
Maya is planning a bridal shower for her best friend. At the party, she wants to serve 3 beverages, 3 appetizers, and 3 desserts, but she does not have time to cook. She can choose from 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts at the supermarket. How many different ways can Maya pick the food and drinks to serve at the bridal shower
Using the combinations formula C(n, r), the number of ways Maya can choose the food and drinks for the bridal shower is computed as: C(12,3) for drinks * C(12,3) for appetizers * C(10,3) for desserts.
Explanation:The subject of this question is combinations in Mathematics. Maya has choices of 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts. She wants to pick 3 of each, and the order of selection does not matter. We can use the combination formula to calculate the number of ways she can do this:
For the bottled drinks, the number of combinations can be calculated as C(12,3).For the frozen appetizers, the number of combinations can be calculated as C(12,3).For the prepared desserts, the number of combinations can be calculated as C(10,3).So the total number of different ways Maya can pick the food and drinks to serve at the bridal shower is C(12,3) * C(12,3) * C(10,3).
Learn more about Combinations here:https://brainly.com/question/24703398
#SPJ3
A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to
Answer:
20 m
Step-by-step explanation:
We are given that
Initial horizontal speed ,[tex]u_x=15 m/s[/tex]
Time, t=2 s
Initial vertical velocity, [tex]u_y=0[/tex]
We know that
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
Where [tex]g=-9.8m/s^2[/tex]
Using the formula
[tex]h=0(2)+\frac{1}{2}(-9.8)(2)^2[/tex]
[tex]h=-19.6 m\approx -20 m[/tex]
The negative sign indicates that the displacement of stone is in downward direction.
Hence, the height of the cliff is closest to 20 m
1. Hank is an intelligent student and usually makes good grades, provided that he can review the course material the night before the test. For tomorrow's test, Hank is faced with a small problem: His fraternity brothers are having an all-night party in which he would like to participate. Hank has three options: a1 - party all night: 2-divide the night equally between studying and partying: 3 - study all night. Tomorrow's exam can be easy (s1), moderate (S2) or tough (53), depending on the professor's unpredictable mood. Hank anticipates the following scores:
S1 S2 S3
a1 85 60 40
a2 92 85 81
a3 100 88 82
(a) Recommend a course of action for Hank based on each of the four criteria of decisions under uncertainty.
(b) Suppose that Hank is more interested in the letter grade he will get. The dividing scores for the passing letter grades A to Dare 90, 80, 70 and 60, respectively. What should the decision/s be?
Answer and Step-by-step explanation:
The answer is attached below
The expected pay-off for a₃ is maximum. Then the decision a₃ (study all night) is considered.
What are statistics?Statistics is the study of collection, analysis, interpretation, and presentation of data or to discipline to collect, summarise the data.
Hank is an intelligent student and usually makes good grades, provided that he can review the course material the night before the test.
For tomorrow's test, Hank is faced with a small problem: His fraternity brothers are having an all-night party in which he would like to participate.
Hank has three options:
a₁ - party all night
a₂ - divide the night equally between studying and partying
a₃ - study all night
Tomorrow's exam can be easy (S₁), moderate (S₂) or tough (S₃), depending on the professor's unpredictable mood. Hank anticipates the following scores:
S₁ S₂ S₃
a₁ 85 60 40
a₂ 92 85 81
a₃ 100 88 82
Decision under uncertainty
1. For maximum criterion when the exam is tough.
a₁ = 40, a₂ = 81, and a₃ = 82
Since 82 is the maximum out of the minimum. Then the optional action is a₃ (study all night).
2. For maximum criterion when the exam is easy.
a₁ = 85, a₂ = 92, and a₃ = 100
Since 82 is the maximum out of the maximum. Then the optional action is a₃ (study all night).
3. Regret criterion
First, find the regret matrix.
S₁ S₂ S₃ Max. regret
a₁ 15 28 42 42
a₂ 8 3 1 8
a₃ 0 0 0 0
From the maximum regret column, we find that the regret corresponding to the course of action is a₃ is minimum. Therefore, decision a₃ (study all night) will be considered.
4. Laplace criterion
The probability of occurrence is 1/3.
Therefore, the expected pay-off for each decision will be
E(a₁) = 61.67, E(a₂) = 86, and E(a₃) = 90
Therefore, the expected pay-off for a₃ is maximum.
Thus, decision a₃ (study all night) is considered.
More about the statistics link is given below.
https://brainly.com/question/10951564
A sample of 100 items has a population standard deviation LaTeX: \:\sigma\:\:σof 5.1 and a sample mean of 21.6. What is the value of the point estimate for the population mean?
Answer:
The point estimate of population mean is 21.6.
Step-by-step explanation:
A point estimate is a numerical value that is used to estimate the value of the parameter under study.
The point estimate is calculated using the sample values.
For example, sample ([tex]\bar x[/tex]) is the point estimate of population mean (μ), sample standard deviation (s) is the point estimate of population standard deviation (σ), sample proportion ([tex]\hat p[/tex]) is the point estimate of population proportion (p).
It is provided that the sample mean of 100 items is,
[tex]\bar x=21.6[/tex]
Then the point estimate of population mean is:
[tex]\mu=\bar x=21.6[/tex]
Thus, the point estimate of population mean is 21.6.
Find the zeros of each function
f(x)=(x-3)(x+5)
Answer:
The zeros of this function are 3 and -5.
Step-by-step explanation:
One note: The question says "Find the zeros of EACH function", but there seems to be only one function in the question. I hope I answered it well.
To complete the function provided, you need to get both terms inside both parenthesis to equal zero. So, if its (x-3)(x+5), the X's need to equal the opposite of the number that is being added or subtracted.
For the first parenthesis, (x-3), the number is -3, so the zero of that side is positive 3. (3-3 = 0)
For the second parenthesis, (x+5), the number is positive 5, so the zero of that side is -5. (-5+5 = 0)
A supervisor must split 60 hours of overtime between five people. One employee must be assigned twice the number of hours as each of the other four employees. How many hours of overtime will be assigned to each employee?
Solution:
Given that,
A supervisor must split 60 hours of overtime between five people
One employee must be assigned twice the number of hours as each of the other four employees
Let "x" be the number of hours overtime per person.
One person does 2x hours of overtime
Which means,
(number of hours overtime per person)(6 person) = 60 hours
6x = 60
x = 10
Thus, 4 people do 10 hours and one person does 20 hours
Each of the ODEs below is second order in y, with y1 as a solution. Reduce the ODE from being second order in y to being first order in ????, with ???? being the only response variable appearing in the ODE. Combine like terms. Show your work.
Answer:
Step-by-step explanation:
The detailed steps and appropriate workings is as shown in the attached file.
The question asks about reducing a second-order ODE to first-order given a known solution, which is achieved using the method of reduction of order to find a new function v(y) leading to a first-order ODE.
To reduce a second-order ordinary differential equation (ODE) to a first-order ODE, given that y1 is a solution, you can apply the method of reduction of order. This involves introducing a new function v(y) such that y can be expressed as a product of the known solution y1 and this new function v(y). Essentially, you substitute y = y1*v into the original second-order ODE and differentiate as necessary to obtain an equation in terms of v and its derivatives only, effectively transforming the equation into first-order.
The steps include (i) expressing y in terms of y1 and v, (ii) differentiating this expression to find the derivatives of y, and (iii) substituting all of this into the equation that y1 obeys. The resultant first-order equation will only involve the function v and its first derivative, v'. By solving this simplified equation, you can find v and thus the second solution to the original second-order ODE.
A movie theater offers 6 showings of a movie each day. A total of 1000 people come to see the movie on a particular day. The theater is interested in the number of people who attended each of the six showings. How many possibilities are there for the tallies for each showing for that day?
Answer:
2.03 × 10¹⁴ different possibilities.
Step-by-step explanation:
We want to know the different ways 1000 people can come out to watch a movie at 6 different times of the day.
This is a permutation and combination problem.
Dividing 1000 people amongst 6 movie showings = 1005C5 = 1005!/(1005-5)!5!
= (1005×1004×1003×1002×1001×1000!)/(1000!5!) = (1005×1004×1003×1002×1001)/5! = 2.03 × 10¹⁴ different possibilities.
A 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia). At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. What is the probability that at least one of them suffers from arachnophobia
Answer:
Probability that at least one of them suffers from arachnophobia is 0.5160.
Step-by-step explanation:
We are given that a 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia).
Also, At a summer camp there are 10 teenagers sleeping in each tent.
Firstly, the binomial probability is given by;
[tex]P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
where, n = number of trials(teenagers) taken = 10
r = number of successes = at least one
p = probability of success and success in our question is % of
the teenagers suffering from arachnophobia, i.e. 7%.
Let X = Number of teenagers suffering from arachnophobia
So, X ~ [tex]Binom(n= 10,p=0.07)[/tex]
So, probability that at least one of them suffers from arachnophobia
= P(X >= 1) = 1 - probability that none of them suffers from arachnophobia
= 1 - P(X = 0) = 1 - [tex]\binom{10}{0}0.07^{0}(1-0.07)^{10-0}[/tex]
= 1 - (1 * 1 * [tex]0.93^{10}[/tex] ) = 1 - 0.484 = 0.5160 .
Therefore, Probability that at least one of them suffers from arachnophobia is 0.5160 .
Which of the following is the cheapest route to visit each city using the "Brute Force Method"
starting from A and ending at A.
Answer:
ACDBA, $900Step-by-step explanation:
Since there are 4 cities, there are (4-1)! = 6 possible routes. Half of those are the reverse of the other half, so there are 6/2 = 3 different possible routes. All of those are listed among the answer choices, along with their cost. All you need to do is choose the answer with the lowest cost:
ACDBA, $900
__
At $960, the other two routes are higher cost.
You are certain to get a heart comma diamond comma club comma or spade when selecting cards from a shuffled deck. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive
Answer:
Probability = 1
Step-by-step explanation:
The number of each type of card described in the question from within a full deck of cards is as follows;
Hearts = 13
Clubs = 13
Diamonds = 13
Spades = 13
These add up to a total of 52 cards. Since a deck only has 52 cards, these make up all the cards in the deck.
Since the probability of taking out a card from these four suits is going to be as follows:
Probability = number of ways we can take out a corresponding suit card / total number of cards
Probability = 52 / 52
Probability = 1
Thus, we can see that the probability of taking out a card belonging to one of the four suits (heart,diamond,club,spade) is 1.
Solve 0 = (x – 4)2 – 1 by graphing the related function.
What are the solutions to the equation?
3 and 5 is the answer
AND THAT'S JUST ON PERIOD POOH!
Answer:
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached.
Step-by-step explanation:
The solution of the graph could be obtained by finding the x-intercept.
[tex]y=\left(x-\:4\right)^2-1[/tex]
Finding the x-intercept by substituting the value y = 0
so
[tex]y=\left(x-\:4\right)^2-1[/tex]
[tex]\:0\:=\:\left(x\:-\:4\right)^2\:-\:1[/tex] ∵ y = 0
[tex]\left(x-4\right)^2-1=0[/tex]
[tex]\left(x-4\right)^2-1+1=0+1[/tex]
[tex]\left(x-4\right)^2=1[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]\mathrm{Solve\:}\:x-4=\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=1[/tex]
[tex]x=5[/tex]
[tex]\mathrm{Solve\:}\:x-4=-\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=-1[/tex]
[tex]x=3[/tex]
So, when y = 0, then x values are 3, and 5.
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.
Answer:
3 and 5 is the answer
Step-by-step explanation:
A drawer contains 4 different pairs of gloves. Suppose we choose 3 gloves randomly, what is the probability that there is no matching pair?
Answer:
The probability that there is no matching pair is 4/7 = 0.5714286.
Step-by-step explanation:
For the first glove we have no restrictions. For the second glove, we have 6 gloves that works for us and 1 that doesnt work (the one that matches the first glove), hence we have 6 possibilities out of 7. Once we pick a good second glove, for the last glove, we only have 4 cases that doesnt match the other two pairs, out of 6 total. This means that the probability that there is no matching pair is 6/7*4/6 = 4/7.