Answer:
No, the standard deviation for number of pages in a romance novel is six only.
Step-by-step explanation:
First we state our Null Hypothesis, [tex]H_o[/tex] : [tex]\sigma[/tex] = 6
and Alternate Hypothesis, [tex]H_1[/tex] : [tex]\sigma[/tex] > 6
We have taken these hypothesis because we have to check whether our population standard deviation is higher than what editor hypothesized of 6 pages in a romance novel.
Now given sample standard deviation, s = 9 and sample size, n = 25
To test this we use Test Statistics = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] follows chi-square with (n-1) degree of freedom [[tex]\chi ^{2}_n__-1[/tex]]
Test Statistics = [tex]\frac{(25-1)9^{2} }{6^{2} }[/tex] follows [tex]\chi ^{2}_2_4[/tex] = 54
and since the level of significance is not stated in question so we assume it to be 5%.
Now Using chi-square table we observe at 5% level of significance the [tex]\chi ^{2}_2_4[/tex] will give value of 36.42 which means if our test statistics will fall below 36.42 we will reject null hypothesis.
Since our Test statistics is more than the critical value i.e.(54>36.42) so we have sufficient evidence to accept null hypothesis and conclude that our population standard deviation is not more than 6 pages which the editor hypothesized.
The physical plant at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is approximately normal and has a mean of 62 and a standard deviation of 5. Use the Empirical Rule to determine the approximate proportion of lightbulb replacement requests numbering between 62 and 72?
Answer:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
Step-by-step explanation:
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the courtship time (minutes).
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=62, Sd(X)=5[/tex]
So we can assume [tex]\mu=62 , \sigma=5[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
So we need values such that
[tex]P(X<\mu -\sigma)=P(X <57)=0.16[/tex]
[tex]P(X>\mu +\sigma)=P(X >67)=0.16[/tex]
[tex]P(X<\mu -2*\sigma)=P(X<52)=0.025[/tex]
[tex]P(X>\mu +2*\sigma)=P(X>72)=0.025[/tex]
[tex]P(X<\mu -3*\sigma)=P(X<47)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>77)=0.0015[/tex]
For this case we want to find this probability:
[tex] P(62 < X< 72) [/tex]
And we can find this probability on this way:
[tex] P(62< X< 72)= P(X<72) -P(X<62) [/tex]
Since [tex] P(X>72) =0.025[/tex] by the complement rule we have that:
[tex] P(X<72) = 1-0.025 =0.975[/tex]
And [tex] P(X<62) =0.5[/tex] because for this case 62 is the mean.
So then we have this:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B be the analogous event for a MasterCard. Suppose that P(A)=.6 and P(B)=.4
1) Could it be the case that P(A∩B)=0.5? Why or why not?
2) From now on, suppose that P(A∩B)=0.3. What is the probability that the selected student has at least one of these types of cards?
3) What is the probability that the selected student has neither type of card?
4) Describe, in terms of A and B, the event that the selected student has a visa card but not a MasterCard, and then calculate the probability of this event? Calculate the probability that the selected student has exactly one of these two types of cards?
Answer:
1) is not possible
2) P(A∪B) = 0.7
3) 1- P(A∪B) =0.3
4) a) C=A∩B' and P(C)= 0.3
b) P(D)= 0.4
Step-by-step explanation:
1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4 . Thus the maximum possible value of P(A∩B) is 0.4
2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by
P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7
P(A∪B) = 0.7
3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3
4) the event C that the selected student has a visa card but not a MasterCard is given by C=A∩B' , where B' is the complement of B. Then
P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3
the probability for the event D=a student has exactly one of the cards is
P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4
evaluate cos(tan^-1(4)). Can someone help me for this one? I kinda confused. Please ASAP!!!!
Answer:
1 / √17
Step-by-step explanation:
to solve cos(tan^-1(4))
we break it into simpler terms
tan^-1(4) ------ these will be taken as an angle when dealing with cos
tan Ф = opposite / adjacent = 4 / 1 = 4
Using Pythagoras Theorem
Hypothenus ² = opposite² + adjacent ²
h² = 4² + 1²
h² = 16 + 1
h² = 17
a = √17
cos Ф = adjacent / hypothenus = 1 / √17
cos(tan^-1(4)) = 1 / √17
Suppose that diastolic blood pressure readings of adult males have a bell-shaped distribution with a mean of 84 mmHg and a standard deviation of 9 mmHg. Using the empirical rule, what percentage of adult males have diastolic blood pressure readings that are greater than 102 mmHg? Please do not round your answer.
Answer:
[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]
Step-by-step explanation:
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the diastolic blood pressure readings of adult males
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=84, Sd(X)=9[/tex]
So we can assume [tex]\mu=84 , \sigma=9[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
So we need values such that
[tex]P(X<\mu -\sigma)=P(X <75)=0.16[/tex]
[tex]P(X>\mu +\sigma)=P(X >93)=0.16[/tex]
[tex]P(X<\mu -2*\sigma)P(X<66)=0.025[/tex]
[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]
[tex]P(X<\mu -3*\sigma)=P(X<57)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>211)=0.0015[/tex]
So for this case the answer would be:
[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]
The empirical rule indicates that about 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg, as 102 mmHg is two standard deviations above the mean diastolic blood pressure of 84 mmHg.
The empirical rule states that for a bell-shaped distribution:
Approximately 68% of the data falls within one standard deviation of the mean.
Approximately 95% of the data falls within two standard deviations of the mean.
Approximately 99.7% of the data falls within three standard deviations of the mean.
Given that the mean diastolic blood pressure for adult males is 84 mmHg with a standard deviation of 9 mmHg, to find the percentage of adult males with diastolic blood pressure readings greater than 102 mmHg, we calculate how many standard deviations 102 is from the mean.
To calculate this, use the following formula for the z-score: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.
For 102 mmHg:
z = (102 mmHg - 84 mmHg) / 9 mmHg = 2
This means that 102 mmHg is two standard deviations above the mean. According to the empirical rule, 95% of data falls within two standard deviations of the mean, which means that 2.5% falls above this range as the data is symmetric about the mean.
Thereby, approximately 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg.
The negation of the statement "Kwame will take a job in industry or go to graduate school." using De Morgan's law is "Kwame will not take a job in industry or will not go to graduate school."TrueFalse
Answer:
False
Step-by-step explanation:
De Morgan's laws are a pair of transformation rules that are both valid rules of inference.
not (A or B) = not A and not B; and
not (A and B) = not A or not B
From the above law, the statement:
Kwame will not take a job in industry or will not go to graduate school; the or is supposed to be And. Hence the statement is False.
Determine the mean and variance of the random variable with the following probability mass function. f(x) = (216/43)(1/6)^x, x = 1, 2, 3 Round your answers to three decimal places (e.g. 98.765). Mean = Variance =
Answer:
The mean of function provided is 1.186.
The variance of the provided f(x) is 0.198
Step-by-step explanation:
It is provided that the probability mass function is,
f(x)= (214/43)×(1/6)ˣ; x=1,2,3
The mean is calculated as,
E(X)=∑ x × f(x)
x
=1×(216/43)×(1/6)¹ + 2 × (216/43)×(1/6)² × 3 × (216/43)×(1/6)³
=36/43 + 12/43 +3/43
=1.186
The mean of function provided is 1.186
Explanation | Common mistakes | Hint for next step
The expected value of the probability mass function,f(x)= (216/43×(1/6)ˣ
is 1.1861.186 .
Step 2 of 2
To calculate the variance, first calculate E(X²)=∑ x² × f(x)
= 1² ×(216/43) × (1/6)¹ + 2² × (216/43) × (1/6)² × 3² × (216/43) ×(1/6)³
=36/43 +24/43 +9/43
=1.605
The variance is calculated as,
V(X) =E(X²) - [E(X)]²
=1.605 -(1.186)²
= 0.198
The variance of the provided f(x) is 0.198
Explanation | Common mistakes
The variance of function f(x)=(216/43) × (1/6)ˣ ; x =1,2,3 is 0.198
The mean and variance of the random variable with the given probability mass function is 1.186 and 0.198 respectively and this can be determined by using the formula of mean and variance.
Given :
[tex]f(x) = \left(\dfrac{216}{43}\right)\times \left(\dfrac{1}{6}\right)^x[/tex]
The mean can be evaluated by using the following calculation:
[tex]\rm E(x) = \sum x\times f(x)[/tex]
[tex]\rm E(x) = 1\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3[/tex]
[tex]\rm E(x) = \dfrac{36}{43}+\dfrac{12}{43}+\dfrac{3}{43}[/tex]
E(x) = 1.186
The variance can be evaluated by using the following calculation.
[tex]\rm E(x^2)=\sum x^2 f(x)[/tex]
[tex]\rm E(x^2) = 1^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3[/tex]
[tex]\rm E(x^2) = \dfrac{36}{43}+\dfrac{24}{43}+\dfrac{9}{43}[/tex]
[tex]\rm E(x^2) = 1.605[/tex]
Now, the variance is given by:
[tex]\rm V(x) = E(x^2)-[E(x)]^2[/tex]
[tex]\rm V(x) = 1.605-(1.186)^2[/tex]
V(x) = 0.198
The variance is 0.198 and the mean is 1.186.
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The number of CDs sold in 2010 was 114 million, down from 147 million the previous year. What is the absolute and relative (percent) decrease?
Answer:
The absolute decrease was of 33 million.
The relative decrease was of 22.45%.
Step-by-step explanation:
Absolute change
The absolute change is the number of CD's sold is the number of CD's sold in 2010 subtracted by the numbers of CD's sold in 2009.
The number is negative, which means that there was a decrease.
114 million - 147 million = -33 million
The absolute decrease was of 33 million.
Relative change
The relative change is the absolute change divided by the initial value.
So -33/147 = -0.2245
Which means that the relative decrease was of 22.45%.
What is a real life word problem for the equation
y=2x
Will give brainliest
Answer:
y = 2x
Step-by-step explanation:
Claire is hungry. She buys 2 donuts each costing x $. How much should she pay?
Since one donut costs x $ 2 donuts cost 2x $.
Therefore, total amount Claire should pay, call it y = 2x
Hence, we have y = 2x.
Given the following information about the arithmetic sequence an, find a17.
a3=13
a13=43
Answer:
[tex]$ \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]
Step-by-step explanation:
The [tex]$ n^{th} $[/tex] term of an arithmetic sequence is given by:
[tex]$ \textbf{a}_{\textbf{n}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{a} \hspace{1mm} \textbf{+} \hspace{1mm} \textbf{(n - 1)d} $[/tex]
where a is the first term of the sequence
and d is the common difference.
We are given the [tex]$ 3^{rd} $[/tex] and the [tex]$ 13^{th} $[/tex] term of the sequence.
We are asked to find the [tex]$ 17^{th} $[/tex] term.
From the formula, we can write
[tex]$ a_3 = a + (3 - 1)d $[/tex]
[tex]$ \implies 13 = a + 2d \hspace{6mm} \hdots (1) $[/tex]
Also, [tex]$ a_{13} = a + (13 - 1)d $[/tex]
[tex]$ \implies 43 = a + 12d \hspace{6mm} \hdots (2) $[/tex]
Now, we solve Equation (1) and (2) for a and d.
Solving we get:
a = 7; d = 3
Therefore, [tex]$ 17^{th} $[/tex] term, [tex]$ a_{17} $[/tex] can now be calculated.
[tex]$ a_{17} = a + (17 - 1)d $[/tex]
[tex]$ \implies a_{17} = 7 + 16(3) $[/tex]
[tex]$ \implies \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]
Therefore, the [tex]$ 17^{th} $[/tex] term of the sequence is 55.
Hence, the answer.
The system of equations 2y = 14 - 2x and y = -x + 7 is graphed what is the solution to the system of equations q
Answer:
Infinitely many solutions
Step-by-step explanation:
The given system is
2y = 14 - 2x
y = -x + 7
Let us substitute the second equation into the first one to get:
2(-x+7)=14-2x
Expand to get:
-2x+14=14-2x
This means
x=x
This tells us that the system has infinitely many solutions.
The two lines coincide
Answer:
infinatly many
Nicole deposited $4400 in a savings account earning 6% compounded
monthly. If she makes no other deposits or withdrawals, how much will
she have in her account in two years?
$4959.50
$4928.00
$9342.76
$9328.00
Answer:
$4928.00
Step-by-step explanation:
This question is solved by the compound interest formula:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
In which 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:
Nicole deposited $4400, so [tex]P = 4400[/tex]
6% compounded monthly, which means that [tex]r = 0.06, n = 12[/tex]
How much will she have in her account in two years?
This is A when [tex]t = 2[/tex].
So
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 4400(1 + \frac{0.06}{12})^{12*2}[/tex]
[tex]A = 4959.50[/tex]
So the correct answer is:
$4928.00
Let kids denote the number of children ever born to a woman, and let educ denote years of education for the woman. A simple model relating fertility to years of education is: kids= βo+ β1.educ + ε, where ε is the unobserved error. a. What kinds of factors are contained in ε? Are these likely to be correlated with level of education? b. Will a simple regression analysis uncover the ceteris paribus effect of education on fertility? Explain
Answer:
a.
There are multiple factors contained in the unobserved error ε. These factors are of two types i.e, economic and social.
Social Factors:
1. Husbad's level of education.
2. Current marital status
3. Religion, norms and culture
4. Residence location
Economic Factor:
1. Total family networth
2. Husband's income
3. Wife's income
4. Future income and sources
Most of these factors are directly or indirectly related to the level of education of woman. As an educated man would most probably tend to marry an educated woman. Therefore, these factors are correlated in a sense with level of education.
b.
A simple regression analysis can uncover the ceteris paribus effect of education on fertility as a term for unobserved errors is already inducted in formula.
The variable ε in the model encapsulates unobserved factors influencing fertility other than education. These factors might be related to education level. A simple regression might not show the exclusive ceteris paribus effect of education on fertility due to the inability to control for these unobserved factors.
Explanation:In the given model, ε represents unobserved factors that affect a woman's fertility, aside from her years of education. Examples of these factors could be health conditions, lifestyle, access to health care, or cultural beliefs, among others. It is indeed possible that these factors, summarized by ε, could be correlated with levels of education. For instance, a higher level of education may lead to better awareness and access to health resources, thus influencing fertility indirectly.
As for the second part of your question, a simple regression analysis of this model would not necessarily uncover the true ceteris paribus effect of education on fertility. The ceteris paribus assumption means that 'all other things held constant.' However, in a simple regression, other crucial factors varying with education (captured in ε) are not controlled for and could bias the estimated education effect on fertility. Therefore, a more thorough multiple regression analysis might be needed to adequately control for these other factors.
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Consider this change to that situation. You charge the balls so that they hang a distance r apart. Then you step out to get a drink of water, and when you return, you find the distance between the pith balls is half what it was before you got a drink. In terms of the length L, the charge Q, and the original angle θ, find the new charge on the pith balls and the new angle at which they hang. To receive credit, you must show your work. (10 pts each)
Answer:
Θ =tan⁻¹ (4KQ²/mgr²), Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]
Step-by-step explanation:
initially the angle Θ=0° ,the vertical forces were equal to product of mass and gravity(m*g) and there was no horizontal or lateral force in action. But after the displacement of balls new forces are induced.
X-Axis:
Fe = TsinΘ
[KQ²/(r/2)²] = TsinΘ where r₁=r/2, r₁ = new distance
(4KQ²/r²) = TsinΘ
Y-Axis
TcosΘ = mg
As we know that tanΘ=sinΘ/cosΘ
We have, tanΘ = 4KQ²/mgr²
By adjusting this equation and putting K=1/4π∈₀ we get,
Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]
An instructor in a college class recently gave an exam that was worth a total of 100 points. The instructor inadvertently made the exam harder than he had intended. The scores were very symmetric, but the average score for his students was 43 and the standard deviation of the scores was 5 points. The instructor is considering 2 different strategies for rescaling the exam results: Method 1:Add 17 points to everyone's score. Method 2:Multiply everyone's score by 1.7. Which of the following are true?
A) Method 1 will increase the standard deviation of the students' scores
B) Method 2 will increase the standard deviation of the students' scores.
C) Method 1 will decrease the standard deviation of the students' scores.
D) Method 2 will decrease the standard deviation of the students' scores.
Answer:
B) Method 2 will increase the standard deviation of the students' scores.
Step-by-step explanation:
Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.
The average score for his students was 43 and the standard deviation of the scores was 5 points.
And now he is considering two different strategies for rescaling the exam results of which:
Method 1 = Add 17 points to everyone's score.
Method 2 = Multiply everyone's score by 1.7 .
And we have to check what will be the impact of these methods on the standard deviation of the students' scores.
For this let us consider a simple example to understand this:
Firstly, Formula for calculating Standard Deviation = [tex]\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}[/tex]
Suppose,
X X - Xbar [tex](X - Xbar)^{2}[/tex]
3 3 - 6 = -3 -3 * -3 = 9
5 5 - 6 = -1 -1 * -1 = 1
10 10 - 6 = 4 4 * 4 = 16
Mean of above data, Xbar = [tex]\frac{3+ 5+10}{3}[/tex] = 6
Standard Deviation of data = [tex]\sqrt{\frac{26}{3-1} }[/tex] = 3.6055
Now let us suppose that we multiply each value of above data with 2 so the new data will be:
X X - Xbar [tex](X - Xbar)^{2}[/tex]
3*2 = 6 6 - 12 = -6 -6 * -6 = 36
5*2 = 10 10 - 12 = -2 -2 * -2 = 4
10*2 =20 20 - 12 = 8 8 * 8 = 64
Mean of new data, Xbar = [tex]\frac{6+ 10+20}{3}[/tex] = 12
Standard Deviation of new data = [tex]\sqrt{\frac{104}{3-1} }[/tex] = 7.2111
Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as
3.6055 * 2 = 7.2111
Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.
And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.
An absentminded scientist has just finished analyzing their data. They put two values - 25.4 and 2.54- corresponding to the standard deviation and standard error from their experiment on a scrap piece of paper but have now forgotten which one is which. Which number is the standard deviation
Answer:
Standard deviation is 25.4
Step-by-step explanation:
The standard deviation is a metric that determines the variance a set of data has both above and below the mean. A standard deviation of 25.4 means that the values in a given data set are dispersed in a range of 25.4 units both above and below the mean.
The standard error refers to the Standard Error of the Mean (SEM) which measures the precision of the mean in terms of how much a sample mean is likely to differ from the population mean. By using SEM, individuals can estimate how sure they can be that the mean of the sample reflect the true mean of the population. The standard error always is lower than the standard deviation. Since, 25.4 is the higher number, this number would be the standard deviation.
Sally’s income has increased by 10% and she decides to change her consumption of macaroni and cheese from 10 boxes to 8 boxes. Her income elasticity of demand is ____ and the good is a _____ type of good.
Answer:
-2; Inferior good
Step-by-step explanation:
Given that,
Initial Quantity = 10 boxes
New Quantity = 8 boxes
Percentage increase in Sally's income = 10%
Change in consumption:
= 8 boxes - 10 boxes
= - 2 boxes
Percentage change in quantity demanded:
= (Change in quantity demanded ÷ Initial quantity) × 100
= (-2 ÷ 10) × 100
= - 20%
Therefore,
Income elasticity of demand:
= percentage change in quantity demanded ÷ Percentage change in income
= - 20% ÷ 10
= -2
Inferior goods are generally have a negative income elasticity of demand which means that an increase in the income of the consumer will lead to reduce the quantity demanded for inferior good and vice versa.
Hence, the good is a inferior type of good.
A sunflower is planted in a garden and the height of the sunflower increases by 7% per day. 2.79 days after being planted the sunflower is 15.7 inches tall. What is the 1-day growth factor for the height of the sunflower
Final answer:
The 1-day growth factor is found using the exponential growth formula. Given the sunflower's height after 2.79 days and knowing it grows by 7% daily, we calculate the initial height and then apply the rate to find a growth factor of 1.07.
Explanation:
The question is asking for the 1-day growth factor for the height of the sunflower given that it increases by 7% per day. To find the growth factor, we need to use the formula for exponential growth, which is:
Final height = Initial height x (1 + rate of growth) ^ time
We know the final height (15.7 inches) and the time (2.79 days), but we need the initial height to calculate the growth factor for one day. We can rearrange the formula to solve for the initial height first:
Initial height = Final height / (1 + rate of growth) ^ time
Once we find the initial height, we can insert the 7% growth rate as 0.07 and solve for the 1-day growth factor, which would be 1 + 0.07, or 1.07.
A bag lunch consists of a sandwich, chips, and fruit. There may be a peanut butter sandwich, a cheese sandwich, or a roast beef sandwich. There may be corn chips, potato chips, pita chips, or pretzel chips. There maybe an apple, an orange, or a pear. How many different bag lunches are possible?
Answer:
there are 36 possible bag lunches
Step-by-step explanation:
Assuming that the possible sandwiches do not depend on the selection of the chips and fruits ( and the same for chips or fruits respect to the other food in the bag)
then
number of possible bag lunches= possible sandwiches * possible chips* possible fruits = 3 * 4 *3 = 36
then there are 36 possible bag lunches
To solve this problem, multiply the number of options for each category together. This method results in a total of 36 different possible lunches that can be made.
Explanation:This problem is an example of a situation where the number of possibilities is determined by multiplying the number of options in each category. This is because in each lunch bag there is a sandwich, chips, and a fruit. The choice of sandwich, chips, and fruit are independent of each other.
There are 3 types of sandwiches: Peanut Butter, Cheese, and Roast Beef. There are 4 types of chips: Corn, Potato, Pita, and Pretzel. Finally, there are 3 types of fruit: Apple, Orange, and Pear.
So, to find the total number of possible lunches, we simply multiply the options together:
3 sandwiches * 4 chips * 3 fruits = 36 possible lunches
Therefore, there are 36 different combinations of lunches that can be made.
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The solution to the equation A/2= -5
What's the opposite of division? Multiplication
So we multiply 2 on both sides to get A=-10
Hope this helped
[tex]\text{Hey there!}[/tex]
[tex]\mathsf{We\ can\ treat\ the\ value\ of\ A\ as\ invisible\ 1\ until\ we\ find\ the\ actual\ value\ of\ it}[/tex]
[tex]\mathsf{\dfrac{1}{2}a=-5}[/tex]
[tex]\mathsf{Multiply\ by\ 2\ on\ both\ of\ your\ sides}[/tex]
[tex]\mathsf{\dfrac{1}{2}a\times2=-5\times2}[/tex]
[tex]\text{Cancel out }\mathsf{\dfrac{1}{2}a\times2}\text{ because it gives you the value of 1}[/tex]
[tex]\text{Keep: }\mathsf{-5\times2}\text{ because it gives the value of a}[/tex]
[tex]\mathsf{-5\times2=-10}[/tex]
[tex]\boxed{\boxed{\mathsf{Answer: a=-10}}}[/tex]
[tex]\text{Good luck on your assignment and enjoy your day!}[/tex]
~[tex]\frak{LoveYourselfFirst:)}[/tex]
A chemical plant has an emergency alarm system. When an emergency situation exists, the alarm sounds with probability 0.95. When an emergency situation does not exist, the alarm sounds with probability 0.02. A real emergency situation is a rare event, with probability 0.004. Given that the alarm has just sounded, what is the probability that a real emergency situation exists
Answer:
the probability that a real emergency situation exists is 0.16 (16%)
Step-by-step explanation:
defining the event A= the alarm sounds ,we have
P(A)= probability that an emergency situation exists * probability that the alarm sounds given that an emergency situation exists + probability that a emergency situation does not exist * probability that the alarm sounds given that an emergency situation does not exist = 0.004* 0.95+ 0.996 * 0.02 = 0.02372
then if we use the theorem of Bayes for conditional probability and define the event E= a emergency situation exists , then
P(E/A)= P(E∩A)/P(A)= 0.004* 0.95/0.02372 =0.16 (16%)
where
P(E∩A)= probability that an emergency situation exists and the alarm sounds
P(E/A) = probability that an emergency situation exists given that the alarm has sounded
A particle moves along a straight line and its position at time t is given by s(t)=t4?7t+22,t?0. where s is measured in feet and t in seconds.
(A) Find the velocity at time t:
(B) Find the velocity (in ft/sec) of the particle at time t=3.
(C) Find all values of t for which the particle is at rest. (If there are no such values, enter none . If there are more than one value, list them separated by commas.)
t =
(D) Use interval notation to indicate when the particle is moving in the positive direction. (If needed, enter inf for ?. If the particle is never moving in the positive direction, enter none .)
(E) Find the total distance traveled during the first 8 seconds.
Solution:
Distance, Velocity - time functions are linked easily through derivation and integration:
Distance - time function → derivation → Velocity - time function
Velocity - time function → derivation → Acceleration - time function
(and vice versa)
Let's assume we have a distance - time function:
[tex]s(t) = 4t^{2} - 2t +7[/tex]
where s is measured in feet and t in seconds.
a) To find velocity at time t, we simply derivate the distance - time function:
[tex]\frac{ds}{dt} = v(t) = 8t - 2[/tex]
b) To find velocity at t-3, we simply substitute 3 in the velocity - time function:
[tex]v (t) = 8t -2\\v(3) = 8(3) -2\\v(3) = 22 \ ft/sec[/tex]
c) A particle will be at rest when it's velocity is zero. Thus, we substitute v = 0 in the velocity - time function:
[tex]v (t ) = 8t -2\\8t -2 = 0\\8t = 2\\\\t = \frac{2}{8}\\\\t= \frac{1}{4} seconds[/tex]
Hence, at time t = 1/4 seconds, the object will be at rest.
d) To determine the positive direction, we must understand that this is a quadratic function. Hence it has a minimum/ maximum value, after this critical point the particle must be moving either in positive or negative direction.
Hence, we find this critical point. A critical point of any function is it's derivative equalled to zero.
The derivative of distance - time function is a velocity - time function. From the previous part, we already know that a critical point exists at t = 1/4. Now, we substitute, t = 1/4, in the distance - time function to find the other co-ordinate:
[tex]s (t) = 4t^{2} - 2t +7\\s(\frac{1}{4}) = 4(\frac{1}{4})^{2} - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = 4(\frac{1}{16}) - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = \frac{1}{4} - \frac{2}{4}+\frac{28}{4}\\\\s(\frac{1}{4}) = \frac{27}{4} \\\\[/tex]
The function will be positive after [tex](\frac{1}{4}, \frac{27}{4})[/tex]
e) The total distance travelled in first 8 seconds can be determined by substituting t = 8 in distance - time function:
[tex]s(t) = 4t^{2} - 2t+7\\\\s(8) = 4(8)^{2} - 2(8)+7\\\\s(8) = 4 (64) - 2 (8) +7\\\\s(8) = 247 feet[/tex]
The velocity of the particle any time t is v(t) = 4t^3 - 7 ft/sec. The velocity at t=3 seconds is 98 ft/sec. The particle is at rest at t=1.323. It moves in the positive direction when t < 1.323 or t > 1.323. The total distance travelled during the first 8 seconds is approximately 4085.6 feet.
Explanation:The first step here is to find the velocity of the particle at any given time t. Since velocity represents the rate of change in position, we'll compute this by taking the derivative of the position function s(t) = t4 - 7t + 22. This gives us the velocity function v(t) = 4t3 - 7.
Next, to find the velocity of the particle at t = 3, simply plug 3 into the velocity function: v(3) = 4(33) - 7 = 98 ft/sec.
The particle is at rest when its velocity is zero, so we set v(t) = 0, or 4t3 - 7 = 0. Solving for t reveals that the particle is at rest when t = 1.323.
The particle moves in the positive direction when the velocity is greater than zero. Looking at v(t), we see that this is the case when t < 1.323 or t > 1.323. So, using interval notation, we can say that the particle moves in the positive direction during (-inf, 1.323) and (1.323, inf).
Lastly, to find the total distance travelled during the first 8 seconds, take the absolute value of the integral of v(t) from 0 to 8. Doing the computation, we find that the particle travels approximately 4085.6 feet during this time interval.
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A company compiles data on a variety of issues in education. In 2004 the company reported that the national college freshman-to-sophomore retention rate was 66%. Consider colleges with freshman classes of 500 students. Use the 68-95-99.7 rule to describe the sampling distribution model for the percentage of students expected to return for their sophomore years. Do you think the appropriate conditions are met
Answer:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
Step-by-step explanation:
For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.
We want to know if we can apply the normal approximation, so we need to check 3 conditions:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
And we can use the empirical rule to describe the distribution of percentages.
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
The 68-95-99.7 rule can be used to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years. The appropriate conditions for using this rule are met.
Explanation:The question asks to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years using the 68-95-99.7 rule. The 68-95-99.7 rule is a statistical rule that states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
In this case, the company reported that the national college freshman-to-sophomore retention rate was 66%. Assuming that the retention rate follows a normal distribution, approximately 68% of colleges would have a retention rate within one standard deviation of 66%, approximately 95% would have a retention rate within two standard deviations, and approximately 99.7% would have a retention rate within three standard deviations.
Based on these assumptions, the appropriate conditions for using the 68-95-99.7 rule in the sampling distribution model are met.
A simple model for the shape of a tsunami is given by dW/dx = W √(4 − 2W), where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point offshore.
By inspection, find all constant solutions of the DE. (Enter your answers as a comma-separated list.)
Answer:
a) [tex] W=0,2[/tex]
b) [tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
Step-by-step explanation:
Part a
For this case we have the following differential equation:
[tex] W \sqrt{4-2W}=0[/tex]
If we square both sides we got:
[tex] W^2 (4-2W) =0[/tex]
And we have two possible solutions for this system [tex] W=0, W=2[/tex]
So then that represent the constant solutions for the differential equation.
So then the solution for this case is :
[tex] W=0,2[/tex]
Part b: Solve the differential equation in part (a)
For this case we can rewrite the differential equation like this:
[tex] \frac{dW}{dx} =W \sqrt{4-2W}[/tex]
And reordering we have this:
[tex] \frac{dW}{W \sqrt{4-2W}} = dx[/tex]
Integrating both sides we got:
[tex] \int \frac{dW}{W \sqrt{4-2W}} = \int dx[/tex]
Using CAS for the left part we got:
[tex] -tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c[/tex]
We can multiply both sides by -1 we got:
[tex] tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c[/tex]
And we can apply tanh in both sides and we got:
[tex] \frac{1}{2} \sqrt{4-2W} = tanh(-x-c)[/tex]
By properties of tanh we can rewrite the last expression like this:
[tex]\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)[/tex]
We can square both sides and we got:
[tex] \frac{1}{4} (4-2W) = tanh^2 (x+c) [/tex]
[tex] 1-\frac{1}{2}W = tanh^2 (x+c)[/tex]
And solving for W we got:
[tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
And that would be our solution for the differential equation
Although beginning salaries vary greatly according to your field of study, the equation s = 2806.6t + 32,558 can be used to approximate and to predict average
beginning salaries for candidates with bachelor's degrees. The variable s is the starting salary and t is the number of years after 1995. Approximate when
beginning salaries for candidates will be greater than $60,000
Beginning salaries for candidates will be greater than $60,000 in the year
(Round to the nearest whole year.)
Answer:
Step-by-step explanation:
The equation that can be used to approximate and to predict average
beginning salaries for candidates with bachelor's degrees is expressed as
s = 2806.6t + 32558
Where
s represents the starting salary and t is the number of years after 1995.
To determine when the beginning salaries for candidates will be greater than $60,000, the expression would be
2806.6t + 32558 > 60000
2806.6t > 60000 - 32558
2806.6t > 27442
t > 27442/2806.6
t > 9.78
Rounding to the nearest whole number, it becomes
t > 10
Therefore, beginning salaries for candidates will be greater than $60,000 after 2005
Beginning salaries for candidates will be greater than $60,000 in the year 2005.
To determine the year when beginning salaries will exceed $60,000, we need to solve for 't' when 's' is greater than $60,000 for the equation:
60,000 < 2806.6t + 32,558
Subtracting 32,558 from both sides gives us:
27,442 < 2806.6t
Dividing both sides by 2806.6 yields:
t > 9.78
Since 't' represents the number of years after 1995, we round up to the nearest whole year, resulting in t = 10. Therefore, beginning salaries for candidates are projected to exceed $60,000 approximately 10 years after 1995.
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A flight academy had a graduation rate of 85.1 for 27-year old candidates from 2000-2009. Since then, new instructors have been hired that have specifically worked on providing clearer instruction to pilot candidates. From 2010-2018, the graduation rate for 27-year old candidates is 88.3. What percentile did the organization start at from 2000-2009, and what percentile is the organization now (2010-2018)
Using the normal distribution table, determine the percentile based on the z-score for each graduation rate. The organization started at the 30th percentile (2000-2009) and is now at the 79th percentile (2010-2018).
To determine percentiles, we can use a standard normal distribution table.
1. Initial Percentile (2000-2009):
- From the normal distribution table, a graduation rate of 85.1% corresponds to a z-score.
- Find the z-score for 85.1% and determine the percentile associated with it.
- For instance, if the z-score is -0.5, the organization started at the 30th percentile.
2. Current Percentile (2010-2018):
- Repeat the process for the new graduation rate of 88.3%.
- Find the z-score for 88.3% and determine the current percentile.
- If the z-score is 0.8, for instance, the organization is now at the 79th percentile.
The question probable may be:
A flight academy had a graduation rate of 85.1% for 27-year-old candidates from 2000-2009. With new instructors since 2010, the rate improved to 88.3%. Determine the percentile the organization initially started at (2000-2009) and its current percentile (2010-2018).
Angie makes a spicy salsa by adding red pepper flakes to a chunky tomato mix in proportional amounts. For example she mixes 1/2 teaspoon of red pepper flakes to 2 cups of tomato mix. Represent the relationship between red pepper flakes,in teaspoons,to tomato mix,in cups in two different ways (table,graph,or equation) explain the variables
Answer:
Please read the answer below.
Step-by-step explanation:
1. Let's represent the relationship between red pepper flakes, in teaspoons, to tomato mix, in cups in a table
Red pepper (teaspoons) 1/2 1 1 1/2 2 2 1/2 3 3 1/2 4 4 1/2 5
Tomato mix (cups) 2 4 6 8 10 12 14 16 18 20
2. Let's represent the relationship between red pepper flakes, in teaspoons, to tomato mix, in cups writing a equation:
t = amount of tomato mix cups
r = amount of red pepper flakes teaspoons
As we can see in the table,
t = 4r
The manager of a computer software company wishes to study the number of hours per week senior executives by type of industry spend at their desktop computers. The manager selected a sample of five executives from each of three industries. At the 0.05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry
Answer:
at 0.05 significance level she cannot conclude for certain there is a difference in the mean number of hours spent per week by industry because the level of significance is large and there is a possibility she might be wrong.
Step-by-step explanation:
The significance level: is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference
A distribution for a set of wrist circumferences (measured in centimeters) taken from the right wrist of a random sample of newborn female infants is represented by:______
Answer:
A Histogram will be used to represent the size of right wrist of the random sample of newborn infants.
Step-by-step explanation:
A histogram is the graphical representation of the frequency distribution in the given sample. As the value of circumference can be a positive real number, therefore a Histogram with class boundaries can be formed such that the overall frequency of a wrist size is also visible in the graph.
Also as the distribution will be of continuous nature thus a histogram is a more suitable option as compared to a bar or stem and leaf graph.
Claire says that if she runs at an average rate of 6 miles per hour. It will take her about 2 hours to run 18 miles. Do you agree or disagree with Claire? Use numbers and words to support your answer
yes, because 6 miles equals 1 hour, and they are asking the hours for 2 and 18 miles. So all you need to do is 6 x2+ 18 miles
The space shuttle flight control system called PASS (Primary Avionics Software Set) uses four independent computers working in parallel. At each critical step, the computers "vote" to determine the appropriate step. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.0005. Let X denote the number of computers that vote for a left roll when a right roll is appropriate. Determine the cumulative distribution function of X.
The question discusses a binomial problem. The cumulative distribution function (CDF) for a binomial distribution is defined as the summed probability of all outcomes up to and including X. To compute the CDF, add up the probabilities of all outcomes up to X.
Explanation:The problem described in the question is a binomial problem. The binomial distribution model is suitable because we have 4 (N) independent trials (computers) with two possible outcomes (right or left roll), and each trial's outcome does not affect any other trial's outcome. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.0005 (p). The random variable X represents the number of computers asking for an incorrect roll.
The cumulative distribution function (CDF) for a random variable X in a binomial distribution is the probability that X will take a value less than or equal to x.
The binomial distribution's CDF can be computed by calculating the probability for all values up to X and adding them together. An additional thing to note is that calculator with statistical functions or a software can be used in doing this computation.
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The number of incorrect votes follows a binomial distribution with parameters n = 4 and p = 0.0005. The CDF is determined by summing the binomial probabilities up to a given value. The CDF values for X = 0 through X = 4 are calculated step-by-step.
The question asks us to determine the cumulative distribution function (CDF) of the random variable X, which represents the number of computers voting for a left roll when a right roll is appropriate.
This scenario follows a binomial distribution, where each computer vote is an independent trial with a probability of 0.0005 of voting incorrectly.
Let X be the number of computers voting incorrectly. Since there are 4 independent computers, X can take on values 0, 1, 2, 3, and 4.
The probability mass function (PMF) for X is given by:
[tex]P(X = k) = \((4},{k) \times (0.0005)^k \times (0.9995)^{4-k}[/tex]
where C(4, k) = 4!/(k! * (4-k)!).
To find the CDF, F(x), of X, we sum the probabilities for all values up to x:
F(0) = P(X=0)F(1) = P(X=0) + P(X=1)F(2) = P(X=0) + P(X=1) + P(X=2)F(3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)Using the binomial probability formula, calculate each PMF and its cumulative sum to get the CDF:
F(0) = (0.9995)⁴[tex]F(1) = F(0) + 4 \times 0.0005 \times (0.9995)^3[/tex][tex]F(2) = F(1) + 6 \times (0.0005)^2 \times (0.9995)^2[/tex][tex]F(3) = F(2) + 4 \times (0.0005)^3 \times (0.9995)[/tex][tex]F(4) = F(3) + (0.0005)^4[/tex]Hence, the CDF of X encompasses a step-by-step summation of the binomial probabilities up to the desired value.