Answer:
a) 3⁵5³.
b) 1
c) 23³
d) 41·43·53
e) 1
f) 1111
Step-by-step explanation:
The greatest common divisor of two integers is the product of their common powers of primes with greatest exponent.
For example, to find gcd of 2⁵3⁴5⁸ and 3⁶5²7⁹ we first identify the common powers of primes, these are powers of 3 and powers of 5. The greatest power of 3 that divides both integers is 3⁴ and the greatest power if 5 that divides both integers is 5², then the gcd is 3⁴5².
a) The greatest common prime powers of 3⁷5³7³ and 2²3⁵5⁹ are 3⁵ and 5³ so their gcd is 3⁵5³.
b) 11·13·17 and 2⁹3⁷5⁵7³ have no common prime powers so their gcd is 1
c) The only greatest common power of 23³ and 23⁷ is 23³, so 23³ is the gcd.
d) The numbers 41·43·53 and 41·43·53 are equal. They both divide themselves (and the greatest divisor of a positive integer is itself) then the gcd is 41·43·53
e) 3³5⁷ and 2²7² have no common prime divisors, so their gcd is 1.
f) 0 is divisible by any integer, in particular, 1111 divides 0 (1111·0=0). Then 1111 is the gcd
Greatest common divisors were calculated for each of the given pairs. Many pairs had no common factors and their GCD was 1, while others had a GCD equal to a shared prime factor or to one of the pair elements.
Explanation:In the subject of mathematics, specifically number theory, the greatest common divisor (GCD) is the largest number that divides two or more numbers without a remainder. Let's determine the greatest common divisors of the given pairs:
3⁷, 5³, 7³, 2ⁱⁱ, 3⁵, 5⁹: These numbers do not have any prime number as a common factor. Hence, their GCD is 1. 11, 13, 17, 2⁹, 3⁷, 5⁵, 7³: None of these numbers share any common factor. Therefore, their GCD is also 1. 23³ⁱ, 23ⁱ⁷: The least power of 23 is 3i. Hence, the GCD is 23³ⁱ. 41, 43, 53, 41, 43, 53: Here, the GCD is the common prime factor, 41. 3ⁱ³, 5 ⁱ⁷, 2ⁱ², 7²ⁱ: These numbers have no common factors, and so, their GCD is 1. 1111, 0: The GCD of 0 and any number is the number itself, thus in this case the GCD is 1111. Learn more about Greatest Common Divisors here:
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Management at a seaside resort is publishing a brochure and wants to include a statement about the proportion of clear days during their peak season. Out of a random sample of 150 days from over the last two peak seasons, 117 days were recorded as clear. They want to estimate the proportion of clear days to within a 5% margin of error with a 95% confidence interval. What's the sample size necessary to construct this interval?A. 384B. 264C. 383D. 385E. 263
Answer: B. 264
Step-by-step explanation:
Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:
[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex]
, where z = Critical z-value corresponds to the given confidence interval
E= margin of error
Let p be the population proportion of clear days.
As per given , we have
Prior sample size : n= 150
Number of clear days in that sample = 117
Prior estimate of the population proportion of clear days = [tex]p=\dfrac{117}{150}[/tex]
E= 0.05
The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)
Then, the required sample size will be :
[tex]n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2[/tex]
Simplify ,
[tex]n= (0.1716)(39.2)^2[/tex]
[tex]n= 263.687424\approx264[/tex]
Hence, the sample size necessary to construct this interval =264
Thus the correct option is B. 264
The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceiving a boy. As this book was being written, 51 babies were born to parents using the YSORT method, and 39 of them were boys. Use the sample data with a 0.01 significance level to test the claim that with this method, the probability of a baby being a boy is greater than 0.5. Does the method appear to work?
Answer:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]
[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.
Step-by-step explanation:
1) Data given and notation
n=51 represent the random sample taken
X=39 represent the number of boys
[tex]\hat p=\frac{39}{51}=0.765[/tex] estimated proportion of boys
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.
A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.
Answer:
The 90% confidence interval would be given by (57.006;62.994)
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X=60[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=10[/tex] represent the population standard deviation
n=30 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=10)[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=1.64[/tex]
Since we have all the values we can replace:
[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]
[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]
So on this case the 90% confidence interval would be given by (57.006;62.994)
Final answer:
To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.
Explanation:
To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:
CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)
Where:
\(\bar{x}\) is the sample mean (60 days)
\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)
\(\sigma\) is the population standard deviation (10 days)
\(n\) is the sample size (30 customers)
Plugging the values into the formula gives:
CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})
Calculating the margin of error:
ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00
Now compute the confidence interval:
CI = [60 - 3.00, 60 + 3.00]
CI = [57.00, 63.00]
So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.
Use the data from problem:
52.2 43.8 50.3 51.1 48.3 47.8 48.3 47.4 50.1 50.5 51.4 54.2 54.4 48.6 54.5 47.3 50.3 48.1 46.6 50.2 50.5 48.2 46.3 48.1 49.4 50.5 47.7 50.1 45.6 49.3 44.4 47.2 47.6 56.9 48.9 49.9 46.3 44.9 51.2 48.5 49.2 46.6 47.3 45.3 49.2 51.1 49.2 50.0 49.8 48.2 47.2 42.6 46.9 46.5 47.3 46.5 47.7 49.2 46.3 48.5 53.4 48.0 50.0 49.7 48.8 48.3 48.7 48.1 48.2 48.6 48.3 48.3 48.3 48.3 48.6 48.2 48.3 48.7 48.1 48.5
a. Calculate the sample mean, sample median, sample variance, and sample standard deviation.
b. Construct a Stem and Leaf Plot, Histogram, and Box Plot.
Answer:
a) Sample mean: 48.71
Sample median: 48.3
Sample variance: 5.41
Sample standard deviation: 2.37
Step-by-step explanation:
a) Sample mean:
[tex]\bar{X}=\frac{1}{N} \sum X_i=\frac{1}{80}*3896.9=48.71[/tex]
Sample median: M=48.3
Note: I order the data increasingly and take the value N / 2 = 40. In this way there are 39 values above and 39 values below the median.
Sample variance:
[tex]s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41[/tex]
Sample standard deviation
[tex]s=\sqrt{s^2}=\sqrt{5.41}=2.37[/tex]
b)
how much mg of a metal containing 4% silver must be combined with 11 mg of a metal containing 38% silver to form an alloy containing 26% silver
Answer:
6 mg of the metal needs to be added.
Step-by-step explanation:
Let the amount (in mg) of metal that needs to be added by y.
Therefore, the amount of silver in the above metal is 0.04y.
Prior to mixing, 11 mg of a metal contained 38% of silver (Given).
Therefore, the amount of silver before= [tex]\frac{38}{100}*11[/tex]= 4.18 mg
The total amount of silver after mixing, 4.18 + 0.04y mg
The total amount of metal after mixing, 11 + y mg
New percentage of silver = 26% .
Thus, [tex]\frac{4.18+0.04y}{11+y}*100= 26[/tex]
[tex](4.18 +0.04 y) * 100 = 26 *(11 +y)[/tex]
[tex]418+4y=286+26y[/tex]
[tex]132=22y[/tex]
y=6 mg
Therefore, the amount of metal that needs to be added is 6 mg.
This exercise uses the population growth model. It is observed that a certain bacteria culture has a relative growth rate of 15% per hour, but in the presence of an antibiotic the relative growth rate is reduced to 8% per hour. The initial number of bacteria in the culture is 28. Find the projected population after 24 hours for the following conditions. (Round your answers to the nearest whole number.) (a) No antibiotic is present, so the relative growth rate is 15%. (b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
Answer:
a) P(24) = 1025.
b) P(24) = 191.
Step-by-step explanation:
This population can be modeled by the following exponential model.
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.
The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].
Population after 24 hours.
(a) No antibiotic is present, so the relative growth rate is 15%.
So r = 0.15.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]
(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
So r = 0.08.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]
Shane receives an hourly wage of $30.40 an hour as an emergency room nurse. When called in at night, he receives time and half. How much does he make if he works 15 hours at night?
Mutliply his hourly rate by 1.5 to find his night pay:
30.40 x 1.5 = $45.60 per hour at night.
Multiply his rate by number of hours:
45.60 x 15 = $684
ANYTHING WILL HELP!!!! ASAP!!!!
Washington High School won the meet.
Johnson High School came in second with 159 points.
Difference between first and second was 3 points.
Step-by-step explanation:
At a particular swim meet, the details of the points which were awarded to the first three places finish are given.
First place finish got 10 points.Second place finish got 8 points.Third place finish got 7 points.To find the total number of points scored by each school is equal to sum of multiplying the respective points which were awarded to number of respective places finish.
i.e. Washington High School had 10 first places finish, so total points for first place finish [tex]= 10 \times10=100[/tex] points.
Similarly, for second places finish = [tex]6\times 8=48[/tex]
and for third places finish = [tex]=2\times 7=14[/tex]
Therefore, Washington High School had total points = 100 + 48 + 14 = 162 points
By using this method, we can make the matrices (Please refer the below attachment).
From the matrices,
Total number of points:
Washington High School = 162 pointsJohnson High School = 159 pointsRoosevelt High School = 108 pointsLewis High School = 96 pointsAmong these scores of four schools, 162 is the highest score. So, Washington High School won the meet.
Johnson High School came in second with 159 points.
Difference between first and second = 162 - 159 = 3 points.
Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective and why it is not injective. Hint: To show that a function f : N → N is surjective, you need to show that for all y ∈ N there is some x ∈ N such that f(x) = y. To show that a function is not injective, simply show that there are two points x1 6= x2 in the domain such that f(x1) = f(x2).
The function f(x) = x // 2 (integer division) is an example of a function from N to N that is surjective because every natural number is covered, but not injective because different numbers can result in the same output.
Explanation:To provide an example of a function f from the natural numbers N to N that is surjective but not injective, consider the function f(x) = x // 2, where '//' denotes integer division. For the function to be surjective, each element y in N must have at least one x such that f(x) = y. This is indeed the case here since for any y > 0, we can choose x = 2y or x = 2y + 1, and f(x) will equal y. To show that it is not injective, we can find two different numbers, x1 and x2, such that f(x1) = f(x2). For instance, if x1 = 4 and x2 = 5, both f(x1) and f(x2) equal 2, thus violating the definition of injectivity. Hence, f(x) = x // 2 is surjective because every y in N is an image of some x, but it is not injective because at least two different values in the domain map to the same value in the codomain.
What percentage of youth sport participants has experienced a sport related injury?
A.50%100%
B. 75%
C. 25%
D. 95%
Answer:
B.75%
Step-by-step explanation:
Answer:
A. 50%
Step-by-step explanation:
about 50% b/c over half the people who play sports all get injuries
Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believed to be a very large number of planets that can be found this way, actually finding one takes time and luck; and finding one planet does not help at all with finding planets of other stars in the same part of the sky. Audrey is good at it, and finds one planet at a time, on average once every three months.
a.) Find the expected value and standard deviation of the number of planets she will find in the next two years.
b.) When she finds her sixth new planet, she will be eligible for a prize. Find the expected value and standard deviation of the amount of time until she is eligible for that prize.
c.) Find the probability that she will become eligible for that prize within one year.
Answer:
Step-by-step explanation:
The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]
a)
The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month
[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}
(24)\\\\=8[/tex]
(Here. t = 24)
For Poisson process mean and variance are same,
[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]
(Poisson distribution mean and variance equal)
The standard deviation of the number of planets is,
[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828
[/tex]
b)
For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter [tex]\lambda[/tex]. The sum of [tex]K[/tex] of these independent exponential has distribution Gamma [tex](K, \lambda)[/tex].
From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}
[/tex]
Calculate the expected value.
[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}
\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]
(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])
C)
Calculate the probability that she will become eligible for the prize within one year.
Here, 1 year is equal to 12 months.
P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)
[tex]=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]
Hence, the required probability is 0.2149 or 21.49%
We can calculate the expected value and standard deviation of the number of planets Audrey will find in the next two years using probability and statistics. Using geometric distribution, we can also calculate the expected value and standard deviation of the amount of time until Audrey is eligible for a prize after finding her sixth new planet. Finally, we can calculate the probability of Audrey becoming eligible for the prize within one year using the binomial distribution formula.
Explanation:To find the expected value and standard deviation of the number of planets Audrey will find in the next two years, we can use the concepts of probability and statistics. Since she finds one planet every three months on average, the expected value can be calculated by multiplying the average number of planets found per month (1/3) by the number of months in two years (24). The standard deviation can be calculated using the formula sqrt(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.
To find the expected value and standard deviation of the amount of time until Audrey is eligible for the prize after finding her sixth new planet, we can use the concept of geometric distribution. The expected value can be calculated by taking the reciprocal of the probability of success (1/6) and the standard deviation can be calculated using the formula sqrt((1-p) / (p^2)).
To find the probability that Audrey will become eligible for the prize within one year, we can calculate the cumulative probability of finding six or more new planets in one year using the binomial distribution formula.
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The weights of certain machine components are normally distributed with a mean of 4.81 ounces and a standard deviation of 0.04 ounces. Find the two weights that separate the top 6% and the bottom 6%. These weights could serve as limits used to identify which components should be rejected. Round your answer to the nearest hundredth, if necessary.
Suppose SAT Writing scores are normally distributed with a mean of 496 and a standard deviation of 109. A university plans to award scholarships to students whose scores are in the top 7%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.
Answer:
First question:
Top 6%: 4.87 ounces
Bottom 6%: 4.75 ounces
Second question:
Top 7%: Score of 649.4.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For the first problem, we have that:
[tex]\mu = 4.81, \sigma = 0.04[/tex]
Top 6%
The value of X when Z has a pvalue of 0.94. This is [tex]Z = 1.555[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.555 = \frac{X - 4.81}{0.04}[/tex]
[tex]X - 4.81 = 1.555*0.04[/tex]
[tex]X = 4.8722[/tex]
Bottom 6%
The value of X when Z has a pvalue of 0.06. This is [tex]Z = 1.555[/tex]
For the second problem, we have that:
[tex]\mu = 496, \sigma = 109[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.555 = \frac{X - 4.81}{0.04}[/tex]
[tex]X - 4.81 = -1.555*0.04[/tex]
[tex]X = 4.7477[/tex]
Top 7%
The value of X when Z has a pvalue of 0.93. This is [tex]Z = 1.475[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.475 = \frac{X - 496}{104}[/tex]
[tex]X - 496 = 104*1.475[/tex]
[tex]X = 649.4[/tex]
What is 6 to the power of 8 in exponential form?
Answer: 1.7 x 10^6
Step-by-step explanation:
6^8 = 1,679,616
1,679,616 = 1.7 x 10^6
A geologist manages a large museum collection of minerals, whose mass (in grams) is known to be normally distributed. She knows that 60% of the minerals have mass less than 5000 g, and needs to select a random sample of n = 16 specimens for an experiment. With what probability will their average mass be less than 5000 g?
Answer:
[tex]P(\bar X < 5000)=P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Let X the random variable that represent the mass of minerals of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
2) Solution to the problem
For this case we know this condition given :
[tex]P(X<5000)=0.6[/tex]
We can use the Z score given by this formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
And using this formula we got:
[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]
And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"
So then we can do this:
[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex] (1)
By the other hand when we find the z score for the sample mean we have this:
[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And we want to find this probability:
[tex]P(\bar X < 5000)[/tex]
And if we use the z score formula we got:
[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex] (2)
And replacing condition (1) into equation (2) we got:
[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]
And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"
Suppose that for all t, a particle moving with constant speed is parameterized by r(t). Given that the length of the path from t = 5 to t = 7 is equal to 8, find the value of the speed, llv(t)ll.
Answer: value of the speed, llv(t)ll = 4.0
Step-by-step explanation:
Given;
time interval t1 =5, t2= 7
Length of path ( distance) L = 8
Speed = distance travelled/ time taken
Speed = dL/dt
Speed = 8/(t2-t1) = 8/(7-5)
Speed = 8/2
Speed = 4.0
Since it's moving at constant speed the speed = 4.0
Suppose the interval [4,6] is partitioned into n=4 subintervals. What is the subinterval length Δx? List the grid points x0, x1, x2, x3, x4. Which points are used for theleft, right, and midpoint Riemann sums?
If a generic interval [tex][a,b][/tex] is partitioned into [tex]n[/tex] subintervals, each one has a length:
[tex]\Delta x = \dfrac{b-a}{n}.[/tex]
In this case, [tex]a = 4[/tex], [tex]b=6[/tex] and [tex]n=4[/tex], so:
[tex]\Delta x = \dfrac{6-4}{4} = \dfrac{2}{4} = \dfrac{1}{2} = 0.5.[/tex]
The grid points are given by:
[tex]x_k = a + k\Delta x, \quad\textrm{with } k \in \{0,1,2,3,4\}.[/tex]
Since [tex]a = 4[/tex] and [tex]\Delta x = 0.5[/tex], we have:
[tex]x_0 = 4 + 0 \times 0.5 = 4 \\x_1 = 4 + 1 \times 0.5 = 4 + 0.5 = 4.5\\x_2 = 4 + 2 \times 0.5 = 4 + 1 = 5\\x_3 = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5\\x_4 = 4 + 4 \times 0.5 = 4 + 2 = 6[/tex]
The [tex]4[/tex] subintervals are of the form [tex]I_k = [x_{k-1}, x_k], \quad\textrm{with } k \in \{1, 2, 3,4\}[/tex]:
[tex]I_1 = [x_0, x_1] = [4,4.5]\\I_2 = [x_1, x_2] = [4.5,5]\\I_3 = [x_2, x_3] = [5,5.5]\\I_4 = [x_3, x_4] = [5.5, 6][/tex]
For the left Riemann sums we will use the left-handed points, namely:
[tex]\{x_0, x_1, x_2, x_3\} = \{4,4.5,5,5.5\}.[/tex]
For the right Riemann sums we will use the right-handed points, namely:
[tex]\{x_1, x_2, x_3, x_4\} = \{4.5,5,5.5,6\}.[/tex]
For the midpoint Riemann sums we will use the average between the two extrema of each subinterval, given by
[tex]I_k \to \tilde{x}_k = \dfrac{x_{k-1}-x_k}{2}, \quad \textrm{with } k \in\{1,2,3,4\}.[/tex]
This gives the midpoints:
[tex]I_1 = [x_0, x_1] = [4,4.5] \to \tilde{x}_1 = \dfrac{x_0 + x_1}{2} = \dfrac{4+4.5}{2} = \dfrac{8.5}{2} = 4.25\\\\I_2 = [x_1, x_2] = [4.5,5] \to \tilde{x}_2 = \dfrac{x_1 + x_2}{2} = \dfrac{4.5+5}{2} = \dfrac{9.5}{2} = 4.75\\\\I_3 = [x_2, x_3] = [5,5.5] \to \tilde{x}_3 = \dfrac{x_2 + x_3}{2} = \dfrac{5+5.5}{2} = \dfrac{10.5}{2} = 5.25\\\\I_4 = [x_3, x_4] = [5.5,6] \to \tilde{x}_4 = \dfrac{x_3 + x_4}{2} = \dfrac{5.5+6}{2} = \dfrac{11.5}{2} = 5.75[/tex]
The points used for the midpoint Riemann sums are therefore:
[tex]\{\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\tilde{x}_4\} =\{4.25,4.75,5.25,5.75\}.[/tex]
The interval and sub intervals are illustrations of Riemann sums
The length of the sub-intervals is 0.5The grid points are 4, 4.5, 5, 5.5 and 6The midpoints are 4.25, 4.75, 5.25 and 5.75The given parameters are:
[tex]\mathbf{Interval = [4,6]}[/tex]
[tex]\mathbf{n =4}[/tex] --- sub intervals
(a) The sub interval length
This is calculated as:
[tex]\mathbf{\Delta x = \frac{b - a}{n}}[/tex]
Where:
[tex]\mathbf{[a,b] =[4,6]}[/tex]
So, we have:
[tex]\mathbf{\Delta x = \frac{6 - 4}{4}}[/tex]
[tex]\mathbf{\Delta x = \frac{2}{4}}[/tex]
[tex]\mathbf{\Delta x = 0.5}[/tex]
Hence, the length of the sub-intervals is 0.5
(b) The grid points
This is calculated as:
[tex]\mathbf{x_k = a + k\Delta x}[/tex]
So, we have:
[tex]\mathbf{x_0 = 4 + 0 \times 0.5 = 4}[/tex]
[tex]\mathbf{x_1 = 4 + 1 \times 0.5 = 4.5}[/tex]
[tex]\mathbf{x_2 = 4 + 2 \times 0.5 = 5}[/tex]
[tex]\mathbf{x_3 = 4 + 3 \times 0.5 = 5.5}[/tex]
[tex]\mathbf{x_4 = 4 + 4 \times 0.5 = 6}[/tex]
So, the grid points are 4, 4.5, 5, 5.5 and 6
(c) The left, right and midpoint Riemann sums
The left points are 4, 4.5, 5 and 5.5The right points are 4.5, 5, 5.5 and 6The midpoint is the average of the left and right points.
So, we have:
[tex]\mathbf{x_0 = 0.5 \times (4 + 4.5) = 4.25}[/tex]
[tex]\mathbf{x_1 = 0.5 \times (4.5 + 5) = 4.75}[/tex]
[tex]\mathbf{x_2 = 0.5 \times (5 + 5.5) = 5.25}[/tex]
[tex]\mathbf{x_3 = 0.5 \times (5.5 + 6) = 5.75}[/tex]
So, the midpoints are 4.25, 4.75, 5.25 and 5.75
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In a test of hypothesis, the null hypothesis is that the population mean is equal to 74 and the alternative hypothesis is that the population mean is less than 74. A sample of 20 elements selected from this normal population produced a mean of 68.5 and a standard deviation of 6.4. The significance level is 1%. What is the value of the test statistic, t?a) 6.372
b) -4.076
c) -2.509
d) -3.843
Answer:
Step-by-step explanation:
The test statistic, t, in this hypothesis testing scenario, is calculated by using the formula t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n). Substituting given values into the formula, the answer is b) -4.076.
Explanation:In this hypothetical testing question, the test statistic, t, is calculated by dividing the difference between the sample mean and the null hypothesis mean by the standard error.
By dividing the standard deviation by the square root of the sample size, the standard error is determined. Here, the sample mean is 68.5, the null hypothesis mean is 74, the standard deviation is 6.4, and the sample size is 20.
The formula to calculate t is:
t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n)
Applying the given figures to the formula, the calculation is:
t = (68.5 - 74) / (6.4 / √20) = -4.076
So, the correct answer is b) -4.076.
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The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3. The probability that there are 8 occurrences in ten minutes is
A. .0241
B. .0771
C. .1126
D. .9107
Answer:
The probability that there are 8 occurrences in ten minutes is
option B. 0 .0771
Step-by-step explanation:
Given:
Random Variable = x
Mean number of occurrences in ten minutes is 5.3.
The probability of an occurrence is the same in any two time periods of an equal length
To Find:
The probability that there are 8 occurrences in ten minutes = ?
Solution:
Let X be the number of occurrences of the event X
[tex]X \sim {Pois} (\lambda)[/tex]
[tex]\lambda = E(X) = 5.3[/tex]
Possion of distribution is given by ,
[tex]P(X=x) = \frac{e^{- \lambda} \lambda^{x}}{x!}[/tex]
Substituting the values,
[tex]P(X=8) = \frac{e^{- 5.3} 5.3^{8}}{8!}[/tex]
[tex]P(X=8) = \frac{(0.004994) ( 622596.904)}{40320}[/tex]
[tex]P(X=8) = \frac{(3109.24894)}{40320}[/tex]
P(X=8) = 0.0771
To find the probability of 8 occurrences in ten minutes with a mean of 5.3, use the Poisson distribution formula to calculate the probability, resulting in approximately 0.0771.
Given: Mean number of occurrences in ten minutes = 5.3
Formula: Probability mass function of a Poisson distribution is given by: P(x) = (e^-λ * λ^x) / x!
Calculations: Plugging in the values, P(X=8) = (e^(-5.3) * 5.3^8) / 8! ≈ 0.0771
Average expenditure on different items such as food, clothes, fuel comes under ____. Select one: a. Descriptive statistics b. Inferential statistics c. Applied statistics d. Business statistics e. Industrial statistics
Answer:
Descriptive
Step-by-step explanation:
Statistics can be broadly classified into two main branches
i) Descriptive and ii) inferential
Descriptive statistics deal with values such as mean, standard deviation from the data.
Inferential statistics is used to predict unknown values from the observed values.
Applied statistics mainly analyses the data according to the needs of business or industries.
Hence the average expenditure on different items such as food, clothes, fuel comes under
Descriptive Statistics
Final answer:
Average expenditure on different items like food, clothes, and fuel falls under descriptive statistics, essential for summarizing trends in expenditure data. The correct option is a.
Explanation:
Descriptive statistics involve summarizing trends in data, such as calculating averages and standard deviations, making them suitable for analyzing average expenditures on items like food and clothes.
Inferential statistics are used to make inferences about populations based on sample data, aiming to find cause and effect relationships or correlations, which is essential when studying average expenditures across different categories.
Therefore, analyzing average expenditures on various items falls under the domain of descriptive statistics as it involves summarizing and interpreting trends in expenditure data.
Complete the two column proof Given: 22= 24,m_2 = 110°
Prove: m 23 = 70°
Statement
Proof
22 24, mZ2=1109
Given
m 2 2 = m 24
Definition of congruent angles
m 24 = 1100
m 23 and m 24 are a linear pair
Definition of a linear pair (shown in
diagram)
Step-by-step explanation:
I took 3 indicated as 5 and its adjacent angle to be 3
<2 = <4
As <2 =<3( corresponding angle)
And <3 = <4 ( Vertically opp.angle)
hence <2 = <4
<2 =>110
so , <4 = 110
So, <4 =>110
As <4 and <5 form linear pair
So <4 + <5 =>180
<5 = 180 -110 =>70
As i took <5 as replacing angle to <3
So According to Question fig
<3 =>70
Hence proved
A disadvantage of using an arithmetic mean to summarize a set of data is that __________. Select one: a. The arithmetic mean sometimes has two values b. It can be used for interval and ratio data c. It is always different from the median d. It can be biased by one or two extremely small or large values
Answer:
d. It can be biased by one or two extremely small or large values
Step-by-step explanation:
Specially in a small sample, a single measure can cause a large difference.
For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.
So the correct answer is:
d. It can be biased by one or two extremely small or large values
The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.
Explanation:The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.
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Complete parts (a) through (c) below.
a) Determine the critical value(s) for a right-tailed test of a population mean at the α = 0.10 level of significance with 15 degrees of freedom.
b) Determine the critical value(s) for a left-tailed test of a population mean at the α = 0.10 level of significance based on a sample size of n = 20.
c) Determine the critical value(s) for a two-tailed test of a population mean at the α = 0.05 level of significance based on a sample size of n = 18.
Answer:
a) [tex]t_{crit}=1.34[/tex]
b) [tex]t_{crit}=-1.33[/tex]
c) [tex]t_{crit}=\pm 2.11[/tex]
Step-by-step explanation:
Part a
[tex]\alpha=0.1[/tex] represent the significance level
df =15
Since is a right tailed test the critical value is given by:
[tex]t_{crit}=1.34[/tex]
And we can use the following excel code to find it: "=T.INV(0.9,15)"
Part b
[tex]\alpha=0.1[/tex] represent the significance level
n=20 represent the sample size
First we need to find the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since is a left tailed test the critical value is given by:
[tex]t_{crit}=-1.33[/tex]
And we can use the following excel code to find it: "=T.INV(0.1,19)"
Part c
[tex]\alpha=0.05[/tex] represent the significance level
n=18 represent the sample size
First we need to find the degrees of freedom given by:
[tex]df=n-1=18-1=17[/tex]
The value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]
Since is a two tailed tailed we have two critical values is given by:
[tex]t_{crit}=\pm 2.11[/tex]
And we can use the following excel code to find it: "=T.INV(0.025,17)"
We guess, based on historical data, that 30% of graduating high-school seniors in a large city will have completed a first-year calculus course. What's the minimum sample size needed to construct a 95% confidence interval for a proportion with a margin of error of 2.5%?
Answer:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
2) Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.025[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
Find equation of set of points pieces that its distance from the point 3, 4, -5 and -2, 1, 4 are equal.
Answer:
Step-by-step explanation:
Suppose we a point [tex]P(x,y,z)[/tex] such that its distance from either the point [tex]A(3,4,-5)[/tex] or [tex]B(-2,1,4)[/tex] is the same.
Using this information we can formula:
distance AP = distance BP
first, let's find the distance from AP, using the distance formula.
[tex]r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}[/tex]
[tex]AP = \sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2}[/tex]
similarly, we can find the distance BP
[tex]BP = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]
since both distances are exactly the same we can equate them
[tex]AP = BP[/tex]
[tex]\sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2} = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]
we can simplify it a bit squaring both sides, and getting rid of the subscripts.
[tex](3 - x)^2 + (4 - y)^2 + (-5 - z)^2 = (-2 - x)^2 + (1 - y)^2 + (4 - z)^2[/tex]
what we have done here is formulated an equation which consists of any point P that will have the same distance from (3,4,-5) and (-2,1,4).
To put it more concretely,
This is the equation of the the plane from that consists of all points (P) from which the distance from both (3,4,-5) and (-2,1,4) are equal.
There are 3 alternative routes by which you may drive to work: Alabaster Street, Brillantine Street, and Clancy Street. It is the beginning of rush hour, and by experience Alabaster street will be closed by a car crash on average in 25 minutes, Brillantine street in 50 minutes, and Clancy Street in 40 minutes. Accident times are completely unpredictable. If you leave for work in one hour (60 minutes), what is the probability that (at the moment you leave) a route to work will still be open?
The probability that at least one route to work will still be open is [tex]\( \frac{209}{216} \)[/tex] when leaving.
To find the probability that at least one route to work will still be open when you leave for work, we need to find the probability that none of the streets will be closed within the next hour.
Let [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] represent the events that Alabaster, Brillantine, and Clancy Streets are closed due to accidents within the next hour, respectively.
The probability of each street being closed within the next hour is:
[tex]\[ P(A) = \frac{25}{60} = \frac{5}{12} \][/tex]
[tex]\[ P(B) = \frac{50}{60} = \frac{5}{6} \][/tex]
[tex]\[ P(C) = \frac{40}{60} = \frac{2}{3} \][/tex]
Since the accidents on each street are independent events, the probability that all three streets remain open is the product of the probabilities that each street remains open:
[tex]\[ P(\text{})[/tex] = [tex]P(\neg A \cap \neg B \cap \neg C) = P(\neg A) \times P(\neg B) \times P(\neg C) ][/tex]
where [tex]\( \neg A \)[/tex], [tex]\( \neg B \)[/tex], and[tex]\( \neg C \)[/tex] represent the complementary events that the streets are not closed.
So, we have:
[tex]\[ P(\text{})[/tex] = [tex](1 - P(A)) \times (1 - P(B)) \times (1 - P(C)) ][/tex]
[tex]\[ = \left(1 - \frac{5}{12}\right) \times \left(1 - \frac{5}{6}\right) \times \left(1 - \frac{2}{3}\right) \][/tex]
[tex]\[ = \left(\frac{7}{12}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{3}\right) \][/tex]
[tex]\[ = \frac{7}{12} \times \frac{1}{6} \times \frac{1}{3} \]\[ = \frac{7}{216} \][/tex]
Therefore, the probability that at least one route to work will still be open when you leave for work is [tex]\( 1 - \frac{7}{216} = \frac{209}{216} \)[/tex].
A few years ago, a census bureau reported that 67.4% of American families owned their homes. Census data reveal that the ownership rate in one small city is much lower. The city council is debating a plan to offer tax breaks to first-time home buyers in order to encourage people to become homeowners. They decide to adopt the plan on a 2-year trial basis and use the data they collect to make a decision about continuing the tax breaks. Since this plan costs the city tax revenues, they will continue to use it only if there is strong evidence that the rate of home ownership is increasing.
Who would be harmed by a Type II error?
(A) The city, because it would lose tax revenue. Faster pace
(B) The citizens of the city, because they lose help they could have used to buy a home.
(C) The city, because it would lose homeowners.
(D) The citizens of the city, because they would have to pay higher taxes than before.
(E) There is no Type Il error in this context.
Answer:
(B) The citizens of the city, because they lose help they could have used to buy a home.
Step-by-step explanation:
Nul and alternative hypotheses are:
[tex]H_{0}:[/tex] the rate of home ownership is the same after tax cut[tex]H_{a}:[/tex] the rate of home ownership is increasing after tax cutType II error occurs when one fails to reject null hypothesis when the null hypothesis was wrong.
In this case Type II error happens when the conclusion is the rate of home ownership is not increasing after tax cut, where actually it is.
With this conclusion city council does not continue tax cut, and citizens of the city is harmed because they lose help they could have used to buy a home.
Final answer:
The citizens of the city would be harmed by a Type II error as they would miss out on the help to buy homes.
Explanation:
Who would be harmed by a Type II error?
(B) The citizens of the city, because they lose help they could have used to buy a home.
A Type II error in this context would harm the citizens of the city as they would miss out on the intended help in buying homes due to the failure to detect an increase in the rate of home ownership.
Provide an appropriate response. Find the variance of the binomial distribution for which n = 800 and p = 0. 87 a. 0.90 b. 0.48 c. 696 d. 32.54 e. 9.51
Final answer:
To find the variance of a binomial distribution with n = 800 and p = 0.87, use the formula Variance = n × p × (1-p). The variance in this case is approximately 90.64.
Explanation:
To find the variance of a binomial distribution, we use the formula:
Variance = n × p × (1-p)
Where n is the number of trials and p is the probability of success.
So, in this case, with n = 800 and p = 0.87, we can calculate the variance as:
Variance = 800 × 0.87 × (1-0.87)
Variance = 800 × 0.87 × 0.13
Variance = 90.64
Therefore, the variance of the binomial distribution is approximately 90.64.
Sheilas monthly periodic rate is 2.41%. What is her APR
Answer:
APR = 2.41% x 12 = 28.92%
Step-by-step explanation:
Her APR is 28.92%.
Sheila's APR is calculated by multiplying the monthly periodic rate of 2.41% by 12, yielding an APR of 28.92%.
Explanation:The question refers to the process of calculating an Annual Percentage Rate (APR) from a given monthly periodic rate. Sheila's monthly periodic rate is 2.41%. The APR can be calculated by multiplying this monthly rate by the number of months in a year, which is 12.
To find Sheila's APR, we perform the following calculation:
APR = Monthly Periodic Rate × Number of Periods in a Year = 2.41% × 12 = 28.92%.
Hence, Sheila's APR is 28.92%.
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Which function represents a vertical stretch of an exponential function? f (x) = 3 (one-half) Superscript x f (x) = one-half (3) Superscript x f (x) = (3) Superscript 2 x f (x) = 3 Superscript (one-half x)
Answer:
f(x) = 3[tex](\frac{1}{2})^{x}[/tex]
Step-by-step explanation:
hope it helps!
The function that represents a vertical stretch of an exponential function is f(x) = (3)^x.
Explanation:The function that represents a vertical stretch of an exponential function is f(x) = (3)x.
In this function, the base of the exponential term is 3, and the exponent, x, determines the position on the graph. When the value of x increases, the function values also increase at an exponential rate.
For example, when x = 1, f(1) = (3)1 = 3. When x = 2, f(2) = (3)2 = 9. The function values double with each increase of x.
A continuous random variable may assume :
a. any numerical value in an interval or collection of intervals.b. finite number of values in a collection of intervals.c. an infinite sequence of values.d. only the positive integer values in an interval.
Answer:
Option A) any numerical value in an interval or collection of intervals
Step-by-step explanation:
Continuous Random Variable:
A continuous random variable can take any value within an interval.Thus, it can take infinite values since there are infinite numbers in an interval.A continuous variable is a variable whose value is obtained by measuring.Examples: height of students in class , weight of students in class, time it takes to get to school, distance traveled between classes.Thus, the correct meaning of continuous random variable is explained by Option A)Option A) any numerical value in an interval or collection of intervals
A continuous random variable can take on any numerical value within a given range or collection of ranges, and it's a characteristic feature of it to take an infinite number of values in any interval. Some examples of this can be a person's height, time, temperature, and weight in physics.
Explanation:A continuous random variable is a type of random variable that can assume any numerical value in a given interval or collection of intervals, making option a correct. This is in contrast to a discrete random variable, which can only take on a finite number of values. A key characteristic of continuous random variables is that they can take on an infinite number of values in any interval.
For example, the height of a person can be treated as a continuous variable, since it can theoretically take any value within a certain range, not just whole number values. The same applies to variables such as time, temperature, and weight in physics.
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