To find the probability of selling more than 4 properties in one week, use the binomial probability formula for each number of sales above 4 and sum them, or subtract the sum of probabilities of 4 or fewer sales from 1.
Explanation:To compute the probability of selling more than 4 properties in one week, given that the chance of selling any one property is 50% and is independent of selling another property, we can use the binomial probability formula. In our scenario, we have a total of 14 properties, and the random variable X represents the number of properties sold in a week.
The binomial probability formula is:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
n is the number of trials (in this case, 14 properties)k represents the number of successes (properties sold)p is the probability of success on any given trial (50% or 0.5)However, instead of calculating the probability of selling exactly k properties, we are interested in selling more than 4 properties. This means we need to calculate the probabilities for selling 5, 6, ..., up to 14 properties and sum them up. Alternately, we can calculate 1 minus the probability of selling 4 or fewer properties to save time, since the probabilities must sum to 1.
The dimensions of the bed of a dumptruck are 12.05 feet long, 3.86 feet wide and 5.11 feet 5.11 feet high, what is the volume of the dumptruck
Show your work below.
The volume of dump truck is 237.7 feet cubed.
Step-by-step explanation:
Given dimensions are;
Length of dump truck = 12.05 feet
Width of dump truck = 3.86 feet
Height of dump truck = 5.11 feet
Volume = Length * Width * Height
Volume = 12.05 * 3.86 * 5.11
Volume = 237.681 ft³
Rounding off to nearest tenth
Volume = 237.7 ft³
The volume of dump truck is 237.7 feet cubed.
Keywords: volume, multiplication
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Determine whether the given value is a discrete or continuous variable. People are asked to state how many times in the last month they visited their family doctor.
Discrete
Continuous
Answer:
The given value is discrete variable.
Step-by-step explanation:
Discrete Variable:
Discrete Variable are those variables that can only take on a finite number of values are called "discrete variables." All qualitative variables are discrete. Some quantitative variables are discrete, such as performance rated as 1,2,3,4, or 5, or temperature rounded to the nearest degree.
Here They have visited the doctor many times so it will be a whole number for sure.
Categorical or Quantitative (Numerical)?Airbnb is a large online marketplace for peopleto list, discover, and book unique accommodations around the world. This online service hasgrown into a multi-billion dollar industry that is even popular right here in Ames, IA. Classifyeach variable below as categorical or quantitative.(a) Month of the year with the most Airbnb reservations in Ames, IA.(b) Airbnb’s total annual profit. (c) Type of rental on Airbnb ( Type 1= whole house, Type 2 = private room, Type 3 = shared room, etc.). (d) Unique 10-digit reservation number for each Airbnb stay. (e) Number of house rentals available in a given county of Iowa.
Answer:
a. Categorical
b. Quantitative
c. Categorical
d. Categorical
e. Quantitative
Step-by-step explanation:
a.
Month of year with most reservations is a qualitative or categorical variable because it can't be represented numerically in a meaningful way. For example, with most reservations month of a year can be June or July.
b.
Airbnb's total annual profit is a quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully interpreted.
c.
Type of rental on Airbnb is a qualitative or categorical variable because it can't be represented numerically in a meaningful way. Also, it can be divided into categories whole house, private room and shared room etc.
d.
Unique 10-digit reservation number is a qualitative or categorical variable as these exists in numerical form but these numbers are used only as identifiers. The mathematical operation on these numbers can't be meaningfully be interpreted.
e.
Number of house rentals is quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully interpreted.
The qualitative, categorical, and quantitative statements of the above cases are:
a. Categorical
b. Quantitative
c. Categorical
d. Categorical
e. Quantitative
What is quantitative?Quantitative is the term used mainly to describe the quantity of a particular case, but not describe it as an attribute.
What is categorical?Categorical means describing anything in a particular way or series.
a.
The month of the year with most reserves is a qualitative or categorical variable because it can not be equal numerically in a meaningful way.
For example, with most reserves the month of the year can be June or July.
b.
Airbnb's total annual profit is a quantitative variable because it can be shown in mathematical operations that can be meaningfully interpreted.
c.
The type of rental on Airbnb is a categorical variable because it can't be represented numerically in a meaningful way. Also, it can be divided into categories whole house, private room, shared room etc.
d.
A unique 10-digit reservation number is a qualitative or categorical variable as these exist in numerical form, but these numbers are used only as identifiers. The mathematical operation on these numbers can't be meaningfully be interpreted.
e.
The number of house rentals is a quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully interpreted.
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A scientist is working with 1.3m of gold wire. How long is the wire in millimeters
Answer:
1300 millimeters
Step-by-step explanation:
Answer:
1300 mm
Step-by-step explanation:
In humans, tongue rolling is a dominant trait, those with the recessive condition cannot roll their tongues. Dave can roll his tongue, but his father could not. He is married to Nancy, who can roll her tongue, but her mother could not. What is the probability that their first born child will be able to roll his tongue
Answer:
0.75 or 75%
Step-by-step explanation:
Let R be the dominant allele for rolling the tongue and r be the recessive allele for not rolling the tongue. If both Dave and Nancy can roll their tongues and had a parent that could not, they both have a heterozygous genotype (Rr).
The sample space for the genotype of their first born child is:
S={RR, Rr, rR, rr}
Only the homozygous recessive genotype rr makes it so that their child is not able to roll his tongue. Therefore, the probability that their first born child will be able to roll his tongue is:
[tex]P=\frac{3}{4}=0.75=75\%[/tex]
Find the area of the parallelogram that has adjacent sides Bold u equals Bold i minus 2 Bold j plus 2 Bold kand Bold v equals 3 Bold j minus Bold k.
Answer:
The area of the parallelogram is [tex]A=\sqrt{26}[/tex].
Step-by-step explanation:
Let's rewrite these two vectors:
[tex]u=i-2j+2k[/tex]
[tex]v=0i+3j-k[/tex]
Let's recall that the area of the parallelogram is the magnitude of the cross product between these vectors.
We can use the Determinant method to find it.
[tex]u \times v=\left[\begin{array}{ccc}i&j&k\\1&-2&2\\0&3&-1\end{array}\right] = i((-2)*(-1)-2*3)-j(1*(-1)-2*0)+k(1*3-(-2)*0)=i(2-6)-j(-1)+k(3)=-4i+j+3k[/tex]
Now, the magnitude is the square root of each component squared. It will be:
[tex]|u \times v|=\sqrt{(-4)^{2}+(1)^{2}+(3)^{2}}=\sqrt{16+1+9}=\sqrt{26}[/tex]
Therefore the [tex]A=\sqrt{26}[/tex].
I hope it helps you!
The area of the parallelogram formed by vectors u = i - 2j + 2k and v = 3j - k is 3[tex]\sqrt{10}[/tex] square units, calculated using the cross product formula.
To find the area of the parallelogram with adjacent sides u and v, where:
u = i - 2j + 2k
v = 3j - k
We can use the cross product of u and v to calculate the area. The magnitude of the cross product represents the area of the parallelogram formed by these vectors.
The cross product of two vectors u and v is given by:
u x v = |u| * |v| * sin(θ) * n
Where:
|u| and |v| are the magnitudes of u and v, respectively.
θ is the angle between u and v.
n is the unit vector perpendicular to the plane formed by u and v.
First, let's calculate the magnitudes of u and v:
|u| = [tex]\sqrt{(1^2 + (-2)^2 + 2^2)}[/tex] = [tex]\sqrt{(1 + 4 + 4)}[/tex] = [tex]\sqrt{9}[/tex]= 3
|v| = [tex]\sqrt{(0^2 + 3^2 + (-1)^2)}[/tex] = [tex]\sqrt{(0 + 9 + 1)}[/tex] = [tex]\sqrt{10}[/tex]
Now, let's find the angle θ between u and v. We can use the dot product formula:
u · v = |u| * |v| * cos(θ)
Since u · v = 0 (they are orthogonal), we have:
0 = 3 * [tex]\sqrt{10}[/tex] * cos(θ)
cos(θ) = 0
This means θ is 90 degrees (π/2 radians).
Now, we can calculate the area using the cross product formula:
Area = |u x v| = |u| * |v| * sin(θ)
Area = 3 * [tex]\sqrt{10}[/tex] * 1 (sin(π/2) = 1)
Area = 3[tex]\sqrt{10}[/tex] square units
So, the area of the parallelogram formed by the vectors u and v is 3[tex]\sqrt{10}[/tex]square units.
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Complete question below :
Given two vectors u = 2i - 3j + k and v = i + 4j - 2k, calculate the area of the parallelogram formed by these vectors.
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
The probability that she will finish her trip in 80 minutes or less is _____.
Answer:
0.8 or 80%
Step-by-step explanation:
Since the time is uniformly distributed, every possible travel time has the same likelihood of occurring.
Lower boundary (L) = 40 minutes
Upper boundary (U) = 90 minutes
The probability that a student finishes her trip in 80 minutes or less is:
[tex]P(t\leq 80) = \frac{80-L}{U-L}=\frac{80-40}{90-40}\\P(t\leq 80) = 0.8=80\%[/tex]
The probability is 0.8 or 80%.
Answer:
80%
Step-by-step explanation:
Jack and Rodger both produce Sandwiches and Pies, and they both have 300 minutes of time available. It takes Jack 1 minutes to make a Sandwich, and 7 minutes to make a Pie. It takes Rodger 7 minutes to make a Sandwich and 1 minutes to make a Pie. What is the largest number of Sandwiches that Jack would be willing to trade away to get 4 Pies from Rodger
Answer:
28 sandwiches
Step-by-step explanation:
If Jack takes 7 minutes to make a pie, the time that would take Jack to produce 4 pies is:
[tex]t=4*7=28\ minutes[/tex]
Jack would be willing to trade away the amount of sandwiches he is able to produce in 28 minutes to get 4 pies from Rodger. In 28 minutes, the number of sandwiches Jack can produce is:
[tex]S=1*28=28\ sandwiches[/tex]
Jack would be willing to trade away 28 sandwiches for 4 pies.
Suppose that P(A|B)=0.2, P(A|B')=0.3, and P(B)=0.7. What is the P(A)? Round your answer to two decimal places (e.g. 98.76).
Answer: [tex]P(A) = 0.23[/tex]
Step-by-step explanation:
Given :
[tex]P(A/B) = 0.2[/tex]
[tex]P(A/B^{1})=0.3[/tex]
[tex]P(B)= 0.7[/tex]
[tex]P(A) = ?[/tex]
From rules of probability :
[tex]P(A) = P(AnB) + P(A n B^{1})[/tex] ........................... equation *
[tex]P(A n B)[/tex] can be written as [tex]P(A/B)[/tex] x [tex]P(B)[/tex]
Also , [tex]P(A/B^{1})[/tex] can be written as [tex]P(A/B^{1})[/tex] x [tex]P(B^{1})[/tex]
substituting into equation * , we have
[tex]P(A) = P(A/B)[/tex][tex]P(B) + P(A/B^{1})P(B^{1})[/tex]
since P(B) = 0.7, then [tex]P(B^{1}) = 1 - P(B) = 0.3[/tex]
so , substituting each values , we have
[tex]P(A) = (0.2)(0.7) + (0.3)(0.3)[/tex]
[tex]P(A) = 0.14 + 0.09[/tex]
[tex]P(A) = 0.23[/tex]
Heights of women are normally distributed with mean 63.7 inches and standard deviation 2.47 inches. Find the height that is the 10th percentile. Find the height that is the 80th percentile.
Answer: for 10th percentile, X = 60.53inches
for 80th percentile, X = 61.53 inches
Step-by-step explanation:
the relationship between the mean, standard deviation and the standard normal distribution is given as
X = μ + σZ
where μ is the mean and σ is the standard deviation of the variable X, and Z is the value from the standard normal distribution for the desired percentile.
Hence to determine the 10th and 80th percentile, we lookup the standard normal distribution table attached below,
from the table,
at 10th percentile Z = -1.282
at 80th percentile we interpolate between 75th percentile and 90th
(80 - 75)/(90 - 75) = (Z - 0.675)/(1.282 - 0.675)
5/15 = Z - 0.675/0.607
0.333*(0.607) = Z - 0.675
Z = 0.8771
hence the Z value for the 80th percentile is 0.8771
hence
X value for 10th percentile and 80th is calculated as
X = μ + σZ
since mean = 63.7 and standard deviation = 2.47
For 10th percentile
X = 63.7 + 2.47*(-1.282)
X = 60.53
for 80th percentile
X = 63.7 + 2.47*(0.8771)
X = 61.53
Final answer:
Height at 10th percentile: 61.53 inches
Height at 80th percentile: 65.78 inches
Explanation:
Given that the heights of women are normally distributed with a mean of 63.7 inches and a standard deviation of 2.47 inches, we can find the 10th and 80th percentiles using the Z-score formula in the context of a normal distribution.
The Z-score formula is given by: Z = (X - μ) / σ, where X is the value whose Z-score we're calculating, μ is the mean, and σ is the standard deviation.
To find the 10th and 80th percentiles, we first use Z-scores corresponding to these percentiles from a standard normal distribution table:
For the 10th percentile, Z ≈ -1.28For the 80th percentile, Z ≈ 0.84We then apply the formula for each Z-score to find the heights corresponding to these percentiles:
Height at 10th percentile: X = Zσ + μ = (-1.28)(2.47) + 63.7 ≈ 61.53 inchesHeight at 80th percentile: X = Zσ + μ = (0.84)(2.47) + 63.7 ≈ 65.78 inchesA certain standardized test's math scores have a bell-shaped distribution with a mean of 530 and a standard deviation of 119. Complete parts (a) through (c) (a) What percentage of standardized test scores is between 411 and 649? 68% (Round to one decimal place as needed.) (b) What percentage of standardized test scores is less than 411 or greater than 649? 1 32% (Round to one decimal place as needed.) (c) What percentage of standardized test scores is greater than 768? % (Round to one decimal place as needed.)
You can convert the given normal distribution to standard normal distribution and then use the z tables to find the needed probabilities.
Rounding to one places of decimal, we get the answers as:
[tex]P( 411 < X < 649 ) \approx 0.6826 = 68.2\%[/tex][tex]P(X < 411) + P(X > 649) \approx 0.3174 \approx 31.7\%[/tex][tex]P(X > 768) \approx 0.0228 \approx 2.3\%[/tex]How to get the z scores?If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
Using the z scores will help to find the probabilities from the z tables(available online).
Let for the given test, the test scores be tracked by a random variable X, then by the given data, we have:
[tex]X \sim N(530, 119)[/tex]
The needed probabilities are
[tex]P( 411 < X < 649 ) = P(X < 649) - p(X < 411)\\[/tex][tex]P(X < 411) + P(X > 649) = 1 - P(411 \leq X \leq 649) = 1 - P(411 < X < 649)[/tex][tex]P(X > 768) = 1 - P(X \leq 768)[/tex]Converting the distribution to standard normal variate, the probabilities become
[tex]Z = \dfrac{X - 530}{119}\\\\Z \sim N(0, 1)[/tex]
The probabilities convert to
a) [tex]P( 411 < X < 649 ) = P(X < 649) - p(X < 411)\\[/tex]
[tex]P(\dfrac{411 - 530}{119} < Z < \dfrac{ 649 - 530}{119}) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write [tex]P(Z < a) = P(Z \leq a)[/tex] )
Also, know that if we look for Z = a in z tables, the p value we get is [tex]P(Z \leq a) = p \: value[/tex]
The p value at Z = 1 is 0.8413 and at Z = -1 is 0.1587,
Thus, [tex]P(411 < X < 649) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826[/tex]
b) [tex]P(X < 411) + P(X > 649) = 1 - P(411 \leq X \leq 649) = 1 - P(411 < X < 649)\\\\P(X < 411) + P(X > 649) = 1 - 0.6826 = 0.3174[/tex]
c) [tex]P(X > 768) = 1 - P(X \leq 768)[/tex]
[tex]P(X > 768) = 1 - P(Z < \dfrac{768 - 530}{119}) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228[/tex]
Rounding to one places of decimal, we get the answers as:
[tex]P( 411 < X < 649 ) \approx 0.6826 = 68.2\%[/tex][tex]P(X < 411) + P(X > 649) \approx 0.3174 \approx 31.7\%[/tex][tex]P(X > 768) \approx 0.0228 \approx 2.3\%[/tex]Learn more about standard normal distribution here:
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Question 5: A recent CNN News survey reported that 76% of adults think the U.S. pennies should still be made. Suppose we select a sample of 20 people.
How many of the 20 would you expect to indicate that the Treasury should continue making pennies?
What is the standard deviation?
What is the likelihood that exactly eight people would indicate the Treasury should continue making pennies?
What is the likelihood that 10 to 15 adults would indicate the Treasury should continue making pennies?
Answer:
a) [tex] E(X) =np = 20*0.76=15.2[/tex]
b) [tex] Sd(X) = \sqrt{3.648}=1.910[/tex]
c) [tex]P(X=8)=(20C8)(0.76)^8 (1-0.76)^{20-8}=0.000512[/tex]
That correspond to approximately 0.0512%, so then we can conclude that is very unlikely since is <1%
d) [tex] P(10 \leq X \leq 15)=0.541[/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=20, p=0.76)[/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
For this case the expected value for the binomial distribution is given by:
[tex] E(X) =np = 20*0.76=15.2[/tex]
Part b
The variance for the binomial distribution is given by:
[tex] Var(X) = np(1-p) = 20*0.76*(1-0.76) =3.648[/tex]
And the deviation would be ust the square root of the variance and we got:
[tex] Sd(X) = \sqrt{3.648}=1.910[/tex]
Part c
For this case we want this probability:
[tex]P(X=8)=(20C8)(0.76)^8 (1-0.76)^{20-8}=0.000512[/tex]
That correspond to approximately 0.0512%, so then we can conclude that is very unlikely since is <1%
Part d
For this case we want this probability:
[tex] P(10 \leq X \leq 15)=P(X=10)+....+P(X=15)[/tex]
If we find the individual probabilities we got:
[tex]P(X=10)=(20C10)(0.76)^{10} (1-0.76)^{20-10}=0.0075[/tex]
[tex]P(X=11)=(20C11)(0.76)^{11} (1-0.76)^{20-11}=0.0217[/tex]
[tex]P(X=12)=(20C12)(0.76)^{12} (1-0.76)^{20-12}=0.0515[/tex]
[tex]P(X=13)=(20C13)(0.76)^{13} (1-0.76)^{20-13}=0.100[/tex]
[tex]P(X=14)=(20C14)(0.76)^{14} (1-0.76)^{20-14}=0.159[/tex]
[tex]P(X=15)=(20C15)(0.76)^{15} (1-0.76)^{20-15}=0.201[/tex]
And if we add the values we got:
[tex] P(10 \leq X \leq 15)=0.541[/tex]
The response provides the expected number of people in the sample supporting the production of pennies, calculates the standard deviation, evaluates the probability of exactly eight respondents, and determines the likelihood of 10 to 15 adults favoring the production of pennies.
Expectation: Out of 20 people, you would expect 76% to indicate that the Treasury should continue making pennies. So, 20 x 0.76 = 15.2 people.
Standard Deviation: To find the standard deviation, use the formula: sqrt(n x p x (1 - p)), where n = 20 and p = 0.76. So, sqrt(20 x 0.76 x 0.24) = 1.95.
Probability: To find the probability of exactly 8 people indicating they should continue making pennies, use the binomial probability formula: C(20, 8) x (0.76⁸) x (0.24¹²) ≈ 0.029.
Likelihood (10 to 15 adults): To find the likelihood of 10 to 15 adults wanting pennies made, sum the probabilities of 10, 11, 12, 13, 14, and 15 people: P(10) + P(11) + P(12) + P(13) + P(14) + P(15).
device uses five silicon chips. Suppose the five chips are chosen at random from a batch of a hundred chips out of which five are defective. What is the probability that the de\"ice contains no defecth'e chip when it is made up from one batch?
The probability that a device using five silicon chips selected randomly from a batch of 100 chips, which includes five defective ones, contains no defective chip is calculated by the ratio of selecting five good chips to selecting any five chips from the batch.
Explanation:The question is asking to find the probability that a device, which uses five silicon chips selected from a batch of a hundred chips with five being defective, will have no defective chip. To solve this, we can calculate the probability step by step using the concept of combinations.
Firstly, we determine the number of ways to select five non-defective chips out of 95 good ones, which is C(95,5). Then, we calculate the total number of ways to select any five chips out of the whole batch, which is C(100,5). The probability that the device contains no defective chip is the ratio of these two numbers:
P(device has no defective chip) = C(95,5) / C(100,5)
Where C(n,k) represents the number of combinations of n items taken k at a time.
To calculate this, use factorials where C(n,k) = n! / [(n-k)!k!].
So, the probability that the device contains no defective chip, is:
P(device has no defective chip) = (95! / (90!*5!)) / (100! / (95!*5!))
Simplifying the factorials, we have:
P(device has no defective chip) = (95*94*93*92*91) / (100*99*98*97*96)
Finally, calculate this to get the decimal value, which would give the probability that the device contains no defective chips when made up from one batch.
What are some solutions to nonresponse? Select all that apply. A. reduce undercoverage B. use stratified sampling C. use convenience sampling D. change wording of questions E. offer rewards and incentives F. reduce interview error G. attempt callbacks H. use cluster sampling
A non responses is a failure to reply something and is a condition that is not responding.
There exists various factors that can create this effect, for example: type of survey, bad questions, un-probabilistic sample, etc.
By the offering of rewards and the incentives It is true as people get a reward or the incentive they would be more willing to rely. A reduce interview error is False as the interview error is not directly linked to the non response bias .Hence the options E and F are correct.
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To tackle nonresponse in surveys, strategies such as reducing undercoverage, using stratified sampling, changing the wording of questions, offering incentives, reducing interview error, and attempting callbacks can be effective. These methods help enhance response rates and the reliability of survey data.
Tackling nonresponse in surveys is crucial for ensuring accurate and reliable data. Here are some effective solutions:
→ Reduce Undercoverage: By ensuring the survey reaches all relevant subpopulations, you can minimize the chances of missing out on certain groups.
→ Use Stratified Sampling: This method can enhance response rates by making sure each subgroup is adequately represented.
→ Change Wording of Questions: Making questions clearer and more straightforward can increase the likelihood of responses.
→ Offer Rewards and Incentives: Providing incentives can motivate participants to complete the survey.
→ Reduce Interview Error: Training interviewers to minimize bias and errors can improve response quality.
→ Attempt Callbacks: Following up with nonrespondents can help in obtaining more responses.
These methods are essential to improve response rates and, consequently, the accuracy of survey results.
Suppose there are two neighbors, Jared and Paul. These two neighbors have the same size lawn and the same amount of hedges. Each week the two neighbors mow their own lawns and trim their own hedges. It takes Jared 30 minutes to mow his lawn and 60 minutes to trim his hedges for a total time of 90 minutes of yard work. It takes Paul 120 minutes to mow his lawn and 90 minutes to trim his hedges, for a total of 210 minutes of yard work. Could the two neighbors gain by specializing and trading lawn services?, If so where should they specialize and how much time could the save each week?.
Answer:
Paul should trim both hedges and Jared should mow both lawns.
Each neighbor would save 30 minutes per week
Step-by-step explanation:
The time each neighbor takes to finish each task is presented below:
[tex]\begin{array}{ccc}&Paul&Jared\\Mow&120&30\\Trim&90&60\end{array}[/tex]
Jared is better at mowing the lawn than trimming hedges, while Paul is better at trimming hedges than mowing the lawn. Therefore, the two neighbors could gain by specializing and trading lawn services if Paul were to trim both hedges and Jared were to mow the lawns.
The time saved by each is:
[tex]P = 210 -90-90=30\ min\\J = 90-30-30=30\ min[/tex]
Solve the initival value problem: y′=7 cos(5x)/(8−3y)y′=7 cos(5x)/(8−3y), y(0)=3y(0)=3. y=y= When solving an ODE, the solution is only valid in some interval. Furthermore, if an initial condition is given, the solution will only be valid in the largest interval in the domain of the solution that is around the xx-value given in the initial condition. In this case, since y(0)=3y(0)=3, then the solution is only valid in the largest interval in the domain of yy around x=0x=0.
Answer:
The solution to the differential equation
y' = (7cos5x)/(8 - 3y); y(0) = 3
is
16y - 3y² = 70sin5x + 21
Step-by-step explanation:
y' = (7cos5x)/(8 - 3y)
This can be written as
dy/dx = (7cos5x)/(8 - 3y)
Separate the variables
(8 - 3y)dy = (7cos5x)dx
Integrate both sides
8y - (3/2)y² = 35sin5x + C
Applying the initial condition y(0) = 3
8(3) - (3/2)(3)² = 35sin(5(0)) + C
24 - (27/2) = 0 + C
C = 21/2
Therefore,
8y - (3/2)y² = 35sin5x + 21/2
Or
16y - 3y² = 70sin5x + 21
Suppose the exchange rate of US dollar to Japanese yen exchange rate is $1 for every 107.35 yen, and the Japanese yen to Bitcoin exchange rate is 1,086,300 yen for every 1 Bitcoin. If someone traded $83,000 US dollars for Japanese yen, then traded the yen for Bitcoin, how many Bitcoin would that person end up with? Round your answer to the nearest whole Bitcoin.
Answer:
The person would end up with 8 Bitcoins.
Step-by-step explanation:
This question can be solved by consecutive rules of three.
If someone traded $83,000 US dollars for Japanese yen, then traded the yen for Bitcoin, how many Bitcoin would that person end up with?
Each US dollar is worth 107.35 yen. So how many yens are $83,000 US dollars worth?
$1 - 107.35 yen
$83,000 - x yen
[tex]x = 83000*107.35[/tex]
[tex]x = 8,910,050[/tex]
The person has 8,910,050 yens. Each bitcoin is worth 1,086,300 yens. How many bitcoins are worth 8,910,050 yens?
1 bitcoin - 1,086,300 yens
x bitcoins - 8,910,050 yens
[tex]1086300x = 8910050[/tex]
[tex]x = \frac{8910050}{1086300}[/tex]
[tex]x = 8.2[/tex]
Rouded to the nearest whole Bitcoin, is 8.
So the person would end up with 8 Bitcoins.
By first converting the US dollars to yen and then trading the yen for Bitcoin, using the provided exchange rates, we determine that the person would end up with roughly 8 Bitcoin.
Explanation:To answer this exchange rate problem, we must first convert the US dollars to yen, then convert the yen to Bitcoin.
First, we multiply the amount of US dollars, $83,000 by the US dollar to yen exchange rate, which is 107.35 yen for every 1 US dollar. This gives us:
$83,000 * 107.35 yen/US dollar = 8,910,050 yen
Next, we trade the yen for Bitcoin by dividing by the yen to Bitcoin exchange rate. Our yen to Bitcoin rate is 1,086,300 yen for 1 Bitcoin:
8,910,050 yen ÷ 1,086,300 yen/Bitcoin ≈ 8.2 Bitcoin.
Rounding this to the nearest whole number, we find that the person ends up with approximately 8 Bitcoin.
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The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Round your answers to four decimal places (e.g. 98.7654).
Full Question
The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour.
a. What is the probability that 6 messages are received in 1 hour?
b. What is the probability that 10 messages are received in 1.5 hours?
c. What is the probability that fewer than 2 messages are received in 0.5 hour?
Answer and Explanation
Given
λ = 6 per hour
Poisson Probability P(X = k) = (λ^k e^-λ)/k!
a. K = 6
P(X = 6) = (6^6 e^-6)/6!
P(X = 6) = 0.160623141047980
P(X = 6) = 0.1606--------- Approximated
b.
If 6 messages are received on average per hour then the number of messages received on average per 1.5 hours is
λ = 6 *1.5
λ = 9
For k = 10
P(X = 10) = (9^10 e^-9)/10!
P(X = 10) = 0.118580076008570
P(X=10) = 0.1186 ---------- Approximated
c.
If 6 messages are received on average per hour then the number of messages received on average per 0.5 hours is
λ = 6 *0.5
λ = 3
For messages fewer than 2 means than k = 0 or k = 1
For k = 0
P(X = 0) = (3^0 e^-3)/0!
P(X = 0) = 0.049787068367863
P(X = 0) = 0.0498 ------_--- Approximated
For X = 1
P(X = 1) = (3^1 e^-3)/1!
P(X = 1) = 0.149361205103591
P(X = 1) = 0.1494 ---------- Approximated
P(X <2) = P(X=0) + P(X=1)
P(X<2) = 0.0498 + 0.1494
P(X<2) = 0.1996
Each T-shirt that just tease produces cross $1.50 to me they sell their T-shirts for $15 at events what is the markup on the T-shirts
Answer: The markup on the T-shirts is $ 13.50.
Step-by-step explanation:
Markup is the difference between the selling price of a product and cost price.Given : The cost price of each t-shirt = $1.50
The selling price of each t-shirt = $15
Then ,the markup on the T-shirts = (Selling price of each t-shirt) -( Cost price of each t-shirt)
i.e. The markup on the T-shirts = $15- $1.50= $ 13.50
Hence, the markup on the T-shirts is $ 13.50.
Answer:
90%
Step-by-step explanation:
Convert the data to centimeters (1 inchequals=2.54 cm), and recompute the linear correlation coefficient. What effect did the conversion have on the linear correlation coefficient?
Answer:
it is not affected by a change of units
Step-by-step explanation:
Since the correlation coefficient has no dimensions, it is not affected by a change of units. Then it will remain the same after the conversion
In fact, the linear correlation coefficient ρ ,where
ρ = Cov (X,Y) / (σx*σy)
then the units [ ] of ρ are
[ρ] = [ Cov (X,Y) ] / [σx]*[σy] = σ²/σ² = 1 → dimensionless
is more useful than using covariance [ Cov (X,Y) ] , since dividing by the standard deviations eliminates the units and standardise the variable
Using the extended Euclidean algorithm, find the multiplicative inverse of a. 1234 mod 4321 b. 24140 mod 40902
(a) The inverse of 1234 (mod 4321) is x such that 1234*x ≡ 1 (mod 4321). Apply Euclid's algorithm:
4321 = 1234 * 3 + 619
1234 = 619 * 1 + 615
619 = 615 * 1 + 4
615 = 4 * 153 + 3
4 = 3 * 1 + 1
Now write 1 as a linear combination of 4321 and 1234:
1 = 4 - 3
1 = 4 - (615 - 4 * 153) = 4 * 154 - 615
1 = 619 * 154 - 155 * (1234 - 619) = 619 * 309 - 155 * 1234
1 = (4321 - 1234 * 3) * 309 - 155 * 1234 = 4321 * 309 - 1082 * 1234
Reducing this leaves us with
1 ≡ -1082 * 1234 (mod 4321)
and so the inverse is
-1082 ≡ 3239 (mod 4321)
(b) Both 24140 and 40902 are even, so there GCD can't possibly be 1 and there is no inverse.
The multiplicative inverse of a number is simply its reciprocal
The multiplicative inverse of 1234 mod 4321 is [tex]\mathbf{ -1082 \equiv 3239\ (mod\ 4321)}[/tex].24140 mod 40902 as no multiplicative inverse.To determine the multiplicative inverse of a mod b, one of a and b must not be an even number
(a) Multiplicative inverse of 1234 mod 4321
This can be written as:
[tex]\mathbf{1234 \times x \equiv 1\ (mod\ 4321)}[/tex]
When the extended Euclidean's algorithm is applied, we start by writing the expression in the following format:
[tex]\mathbf{Dividend = Quotient \times Divisor + Remainder}[/tex]
So, we have:
[tex]\mathbf{4321 = 1234 \times 3 + 619}[/tex]
Express 1234 using the above format
[tex]\mathbf{1234 = 619 \times 1 + 615}[/tex]
Repeat the process for all quotient
[tex]\mathbf{619 = 615 \times 1 + 4}[/tex]
[tex]\mathbf{615 = 4 \times 153 + 3}[/tex]
[tex]\mathbf{4= 3 \times 1 + 1}[/tex]
Next, we reverse the process as follows:
Make 1 the subject in [tex]\mathbf{4= 3 \times 1 + 1}[/tex]
[tex]\mathbf{1 = 4 - 3}[/tex]
Substitute an equivalent expression for 3
[tex]\mathbf{1 = 4 - (615 - 4 \times 153)}[/tex]
[tex]\mathbf{1 = 4 - 615 + 4 \times 153}[/tex]
Collect like terms
[tex]\mathbf{1 = 4 + 4 \times 153 - 615 }[/tex]
[tex]\mathbf{1 = 4 \times 154 - 615 }[/tex]
Substitute an equivalent expression for 615
[tex]\mathbf{1 = 619 \times 154 - 155 \times (1234 - 619) }[/tex]
[tex]\mathbf{1 = 619 \times 309 - 155 \times 1234 }[/tex]
Substitute an equivalent expression for 619
[tex]\mathbf{1 =(4321 - 1234 \times 3) \times 309 - 155 \times 1234}[/tex]
[tex]\mathbf{1 = 4321 \times 309 - 1082 \times 1234}[/tex]
Recall that:
[tex]\mathbf{1234 \times x \equiv 1\ (mod\ 4321)}[/tex]
So, we have:
[tex]\mathbf{1 \equiv -1082 \times 1234\ mod(4321)}[/tex]
Add 4321 and -1082
[tex]\mathbf{4321 -1082 = 3239}[/tex]
Hence, the required inverse is:
[tex]\mathbf{ -1082 \equiv 3239\ (mod\ 4321)}[/tex]
(b) Multiplicative inverse of 24140 mod 40902
Recall that:
To determine the multiplicative inverse of a mod b, one of a and b must not be an even number
Because 24140 and 40902 are both even numbers, then:
24140 mod 40902 has no multiplicative inverse
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Standby time is amount of time a phone can remain powered on while not being used. A cell phone company claims that the standby time of certain phone model is 16 days on average. A consumer report firm gathered a sample of 19 batteries and conducted tests on this claim. The sample mean was 15 days and 10 hours and the sample standard deviation was 30 hours. Assume that the standby time is distributed as normal. In testing if the average standby time is shorter than 16 days, compute the value of the test statistic (round off to second decimal place).
Answer:
[tex]t_{stat} = -2.03[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 16 days = 384 hours
Sample mean, [tex]\bar{x}[/tex] = 15 days 10 hours = 370 hours
Sample size, n = 19
Sample standard deviation, s = 30 hours
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 384\text{ hours}\\H_A: \mu < 384\text{ hours}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{370 - 384}{\frac{30}{\sqrt{19}} } = -2.03415 \approx -2.03[/tex]
The value of t-statistic is -2.03
The test statistic for the claim that the average standby time of the phone model is 16 days, with a sample mean of 15 days and 10 hours and standard deviation of 30 hours, is -2.50.
Explanation:The question requires computation of a test statistic for the claim that the average standby time of a certain phone model is 16 days, using a sample mean of 15 days and 10 hours. The sample standard deviation is given as 30 hours. The number of phones (or sample size) is 19.
First, convert the sample mean to the same unit as the standard deviation. In this case, convert 15 days and 10 hours to hours: (15 * 24) + 10 = 370 hours. The null hypothesis mean is also converted to hours (16 * 24 = 384 hours).
The formula for the test statistic in a one-sample z-test is z = (Xbar - μ) / (σ/√n), where Xbar is the sample mean, μ is the hypothesized population mean, σ is the sample standard deviation, and n is the sample size.
Substitute values into the formula to get: z = (370 - 384) / (30/√19) = -2.50 (rounded to the second decimal place). So the test statistic is -2.50.
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Kayla set up an outdoor digital thermometer to record the temperature overnight as part of her science fair project. She began recording the temperature, in degrees Fahrenheit, at 10:00 p.m. Kayla modeled the overnight temperature with function t, where h represents the number of hours since 10:00 p.m. t(h) = 0.5h2 − 5h + 27.5 What is the lowest temperature and at what time did it occur? A. 5°F at 3:00 a.m. B. 15°F at 5:00 a.m. C. 15°F at 3:00 a.m. D. 5°F at 5:00 a.m.
Answer:
C. 15°F at 3:00 a.m
Step-by-step explanation:
We will start seeing the function they give us, as we can see it is of the form ax ^ 2 + bx + c, this means that it is a parabola.
First we will look the term a of the function
t(h) = 0.5h2 − 5h + 27.5
in this case a = 0.5 , is a positive number so we have a minimum, this point shows us when the temperature reaches its minimum at night.
To obtain it we will have to apply this parabola formula
x = -b / 2a
in this case h = -( -5) / 2(0.5)
h = 5
This 5 represents the hours that have passed since 10:00 p.m.
We add 5 to 10:00 p.m. and get the time that is 3:00 a.m.
Finally we replace the function t with this value, and obtain the value of the minimum temperature
t(h) = 0.5h2 − 5h + 27.5
t(5) = 0.5(5)^2 - 5(5) + 27.5
t = 12.5 - 25 + 27.5
t = 15
C. 15°F at 3:00 a.m
Answer:
C
Step-by-step explanation: because i take the test
In engineering and product design, it is important to consider the weights of people so that airplanes or elevators aren't overloaded. Based on data from the National Health Survey, we can assume the weight of adult males in the US has a mean weight of 197 pounds and standard deviation of 32 pounds. We randomly select 64 adult males. What is the probability that the average weight of these 64 adult males is over 205 pounds?
Answer:
There is a 2.28% probability that the average weight of these 64 adult males is over 205 pounds.
Step-by-step explanation:
To solve this question, we have to understand 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 = 197, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]
What is the probability that the average weight of these 64 adult males is over 205 pounds?
This is 1 subtracted by the pvalue of Z when X = 205.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{205 - 197}{4}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
There is a 2.28% probability that the average weight of these 64 adult males is over 205 pounds.
The probability that the average weight of 64 randomly selected adult males is over 205 pounds is approximately 2.28%.
Explanation:This problem involves the concept of normal distribution and probability in statistics. Given the mean (μ) is 197 pounds and the standard deviation (σ) is 32 pounds, we want to find the probability that the average weight of 64 randomly selected adult males (n=64) is over 205 pounds.
Firstly, we need to calculate the standard error (SE), which is σ/√n, thus SE=32/√64= 4 pounds. Next, we calculate the Z-score, which is (X-μ)/SE, thus Z=(205-197)/4=2.
A Z-score of 2 refers to a value that is 2 standard deviations away from the mean. Looking this up on a Z-table or using statistical software, we can see that the area to the left of Z=2 is approximately 0.9772, meaning there is a 97.72% chance that a randomly selected adult male's weight is below 205 pounds. Hence the probability of the weight being over 205 pounds is 1-0.9772=0.0228 or 2.28%.
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Research seems to indicate that the optimum group size for problem solving is _____ members. Select one: a. 2 b. 15 c. 5 d. 25
Answer:
Correct answer is (c). 5
Step-by-step explanation:
It is important to note that solving problem requires techniques and intelligent people most especially when problem are complex or hard in nature. It is therefore important to ensure the numbers of problem solving experts should not be undersized than required to avoid over burden them and should not be too large to avoid conflict in their individual resolutions. Hence, most scientific reports state that problem solving experts should be within 3 to 5 members and as for this question, the optimum is 5 members.
A food truck operator has traditionally sold 75 bowls of noodle soup each day. He moves to a new location and after a week sees that he has averaged 85 bowls of noodle soup sales each day. He runs a one-sided hypothesis test to determine if his daily sales at the new location have increased. The p-value of the test is 0.031. How should he interpret the p-value?
a. There is a 3.1% chance that the true mean of soup sales at the new location is 85 bowls a day.
b. There is a 96.9% chance that the true mean of soup sales at the new location is greater than 75 bowls a day.
c. There is a 96.9% chance that the sample mean of soup sales at the new location is 85 bowls a day.
d. There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day.
e. There is a 96.9% chance that the true mean of soup sales at the new location is within 3.1 bowls of 85 bowls a day.
Option d correctly interprets the p-value, signifying there is a 3.1% chance of observing an average sales of 85 or more daily bowls given the true mean is 75 or less. It indicates significant evidence against the null hypothesis, suggesting increased sales at the new location.
When interpreting the p-value of the hypothesis test conducted by the food truck operator, option d is the correct interpretation: There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day. The p-value in a one-sided hypothesis test indicates the probability of observing a result as extreme as, or more extreme than, the sample result, under the assumption that the null hypothesis is true. The null hypothesis in this case is that the true mean daily sales have not changed and remain at 75 bowls per day or less. Hence, with a p-value of 0.031, there is significant evidence against the null hypothesis, and the operator has reason to believe that the average sales have indeed increased at the new location.
The p-value of 0.031 means there's a 3.1% chance of obtaining a sample mean of 85 bowls or higher if the true mean remains 75 bowls per day. Hence option d is the correct option. This suggests sufficient evidence to reject the null hypothesis and conclude that soup sales at the new location have likely increased.
The food truck operator has conducted a one-sided hypothesis test to determine if his daily sales at the new location have increased from the traditional 75 bowls of noodle soup.
A p-value is the probability of obtaining a sample mean as extreme as 85 bowls of soup per day or higher, assuming the true mean is still 75 bowls per day.The p-value of 0.031 means there is a 3.1% chance of obtaining such a sample mean if the null hypothesis is true. Therefore, we interpret the p-value as follows:d. There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day.
Since the p-value of 0.031 is less than the typical significance level of 0.05, there is sufficient evidence to reject the null hypothesis and conclude that the daily sales at the new location have likely increased.
Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car in front of them go faster. Suppose that you are driving a considerable distance on a two-lane highway and are passed by 12 vehicles.
(a) Let r be the number of vehicles that tailgate before passing. Make a histogram showing the probability distribution of r for r = 0 through r = 12.
(b) Compute the expected number of vehicles out of 12 that will tailgate. (Round your answer to two decimal places.)
vehicles
(c) Compute the standard deviation of this distribution. (Round your answer to two decimal places.)
vehicles
Answer:
(a) The histogram is shown below.
(b) E (X) = 4.2
(c) SD (X) = 2.73
Step-by-step explanation:
Let X = r = a driver will tailgate the car in front of him before passing.
The probability that a driver will tailgate the car in front of him before passing is, P (X) = p = 0.35.
The sample selected is of size n = 12.
The random variable X follows a Binomial distribution with parameters n = 12 and p = 0.35.
The probability function of a binomial random variable is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]
(a)
For X = 0 the probability is:
[tex]P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006[/tex]
For X = 1 the probability is:
[tex]P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037[/tex]
For X = 2 the probability is:
[tex]P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109[/tex]
Similarly the remaining probabilities will be computed.
The probability distribution table is shown below.
The histogram is also shown below.
(b)
The expected value of a Binomial distribution is:
[tex]E(X)=np[/tex]
The expected number of vehicles out of 12 that will tailgate is:
[tex]E(X)=np=12\times0.35=4.2[/tex]
Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.
(c)
The standard deviation of a Binomial distribution is:
[tex]SD(X)=np(1-p)[/tex]
The standard deviation of vehicles out of 12 that will tailgate is:
[tex]SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\[/tex]
Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.
To determine the probability distribution, create a histogram showing the possible values of r and their probabilities. The expected number of vehicles that will tailgate can be calculated by multiplying each value of r by its probability and summing up the results. The standard deviation can be found by calculating the variance and taking the square root of it.
Explanation:In order to determine the probability distribution of the number of vehicles that tailgate before passing, we need to consider the given information. We know that about 35% of all drivers tailgate. Since we are passed by 12 vehicles, the number of vehicles that tailgate can range from 0 to 12. To create a histogram showing the probability distribution, we need to calculate the probability of each possible value of r and represent them in a bar graph.
(b) To compute the expected number of vehicles that will tailgate, we need to multiply each possible value of r by its corresponding probability and sum up the results. This will give us the average number of vehicles that tailgate out of the 12 vehicles that passed us.
(c) The standard deviation of this distribution can be calculated by determining the variance and taking the square root of it. Variance is calculated by summing up the squared differences between each value of r and the expected value, multiplying each squared difference by its corresponding probability, and summing up the results.
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Find the balance of $7,000 deposited at 4% compounded semi-annually for 2 years
Answer:
The balance will be $7,577.03.
Step-by-step explanation:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.
In this problem, we have that:
[tex]P = 7000, r = 0.04[/tex]
Semianually means twice a year, so [tex]n = 2[/tex]
We want to find A when [tex]t = 2[/tex].
So
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 7000(1 + \frac{0.04}{2})^{2*2}[/tex]
[tex]A = 7577.03[/tex]
The balance will be $7,577.03.
A Lake Tahoe Community College instructor is interested in the mean number of days Lake Tahoe Community College math students are absent from class during a quarter. The instructor takes her sample by gathering data on five randomly selected students from each Lake Tahoe Community College math class. Which type of sampling did she use?
Answer:
She used the simple random sampling technique because there was no condition attached to the samples she took.
Step-by-step explanation:
we have basically four types of sampling
1.Simple random sampling.
2.Systematic sampling.
3.Stratified sampling.
4.Cluster sampling.
simple Random sampling: is a sampling technique where every item in the population has an even chance and likelihood of being selected in the sample.
Select all the values that cannot be probabilities A.) 1 B.) square root of 2 C.) 0 D.) 0.04 E.) -0.54 F.) 3/5 G.) 5/3 H.) 1.29
Answer:
B.) square root of 2
E.) -0.54
G.) 5/3
H.) 1.29
Step-by-step explanation:
A probability of an event is how likely the event is to occur. It is always positive values, between 0% and 100%, or as decimals, between 0 and 1.
A.) 1
Can be a probability
B.) square root of 2
The square root of 2 is 1.41. 1.41 is higher than 1, so square root of 2 cannot be a probability
C.) 0
Can be a probability
D.) 0.04
0.04 = 4%
Can be a probability
E.) -0.54
Negative values cannot be probabilities
F.) 3/5
3/5 = 0.6 = 60%
Can be a probability
G.) 5/3
5/3 = 1.67
Higher than 1, so cannot be a probability
H.) 1.29
Higher than 1, cannot be a probability