Answer:
a) 5+13k where k is integer
b) 20+13k where k is integer
c)12+13k where k is integer
Step-by-step explanation:
(a)
[tex]8x \equiv 1 (mod 13) \text{ means } 8x-1=13k[/tex].
8x-1=13k
Subtract 13k on both sides:
8x-13k-1=0
Add 1 on both sides:
8x-13k=1
I'm going to use Euclidean Algorithm.
13=8(1)+5
8=5(1)+3
5=3(1)+2
3=2(1)+1
Now backwards through the equations:
3-2=1
3-(5-3)=1
3-5+3=1
(8-5)-5+(8-5)=1
2(8)-3(5)=1
2(8)-3(13-8)=1
5(8)-3(13)=1
So compare this to:
8x-13k=1
We see that x is 5 while k is 3.
Anyways 5 is a solution or 5+13k is a solution where k is an integer.
b)
[tex]8x \equiv 4 (mod 13)[/tex]
8x-4=13k
Subtract 13k on both sides:
8x-13k-4=0
Add 4 on both sides:
8x-13k=4
We got this from above:
5(8)-3(13)=1
If we multiply both sides by 4 we get:
8(20)-13(12)=4
So x=20 and 20+13k is also a solution where k is an integer.
c)
[tex]99x \equiv 5 (mod 13)[/tex
99x-5=13k
Subtract 13k on both sides:
99x-13k-5=0
Add 5 on both sides:
99x-13k=5
Using Euclidean Algorithm:
99=13(7)+8
13=8(1)+5
Go back through the equations:
13-8=5
13-(99-13(7))=5
8(13)-99=5
99(-1)+8(13)=5
Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.
Answer:
a) x = 5 mod 13.
b) x = 7 mod 13.
Step-by-step explanation:
a) 8x = 1 mod 13
x = 2, 16 = 3 mod 13
x = 3, 24 = 11 mod 13
x = 4, 32 = 6 mod 13
x = 5 , 40 = 1 mod 13
8x = 40
x = 5 mod 13.
b) 8x = 4 mod 13
x = 7, 56 = 4 mod 13.
7 = 4 mod 13
x = 7 mod 13.
Raul received a score of 77 on a history test for which the class mean was 70 with a standard deviation of 6. He received a score of 79 on a biology test for which the class mean was 70 with standard deviation 4. On which test did he do better relative to the rest of the class?
Answer:
He did better on biology test.
Step-by-step explanation:
For comparing the scores we need to find z-scores for both subjects,
We know that,
z-score or standard score for x is,
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where, [tex]\mu[/tex] is mean,
[tex]\sigma[/tex] is standard deviation,
In history test,
[tex]x=77[/tex]
[tex]\mu=70[/tex]
[tex]\sigma = 6[/tex]
Thus, the z-score would be,
[tex]z_1=\frac{77-70}{6}[/tex]
[tex]\approx 1.167[/tex]
In biology test,
[tex]x=79[/tex]
[tex]\mu = 70[/tex]
[tex]\sigma = 4[/tex],
Thus, the z-score would be,
[tex]z_2=\frac{79-70}{4}[/tex]
[tex]= 2.25[/tex]
∵ [tex]z_2>z_1[/tex]
Hence, he did better on biology test.
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
Answer with explanation:
[tex]\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}[/tex]
--------------------------------------------------------Dividing both sides by 8 x
This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.
Integrating Factor
[tex]=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}[/tex]
Multiplying both sides by Integrating Factor
[tex]x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)[/tex]
When , x=1, gives , y=9.
Evaluate the value of C and substitute in the equation 1.
Assuming that in a box there are 10 black socks and 12 blue socks, calculate the maximum number of socks needed to be drawn from the box before a pair of the same color can be made. Using the pigeonhole principle
Answer:
Step-by-step explanation:
The pigeonhole principal states that if n items are put into m containers, with n > m then at least one container must contain more than one item.
In other words, to find the case where maximum attempts are required, we eliminate all the cases where our criterion is not met and we will be left with the desired result.
For this case, the box has 10 black socks and 12 blue socks.
10 black socks = 5 left foot ones and 5 right foot ones
12 blue socks = 6 left foot ones and 6 right foot ones
If we draw all left foot or right foot ones then we will not have a pair till 11 draws have been made.
The next socks drawn will be a right foot one of blue or black color and a pair will be made.
Therefore, the maximum number of socks needed to be drawn from the box are 12.
3. Let U and V be subspaces of a vector space W. Prove that their intersection UnV is also a subspace of W
Answer: The proof is done below.
Step-by-step explanation: Given that U and V are subspaces of a vector space W.
We are to prove that the intersection U ∩ V is also a subspace of W.
(a) Since U and V are subspaces of the vector space W, so we must have
0 ∈ U and 0 ∈ V.
Then, 0 ∈ U ∩ V.
That is, zero vector is in the intersection of U and V.
(b) Now, let x, y ∈ U ∩ V.
This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.
Since U and V are subspaces of U and V, so we get
x + y ∈ U and x + y ∈ V.
This implies that x + y ∈ U ∩ V.
(c) Also, for a ∈ R (a real number), we have
ax ∈ U and ax ∈ V (since U and V are subspaces of W).
So, ax ∈ U∩ V.
Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have
x + y and ax ∈ U ∩ V.
Thus, U ∩ V is also a subspace of W.
Hence proved.
Jayanta is raising money for the homeless, and discovers each church group requires 2 hr of letter writing and 1 hr of follow-up calls, while each labor union needs 2 hr of letter writing and 3 hr of follow-up. She can raise $125 from each church group and $175 from each union. She has a maximum of 20 hours of letter writing and 14 hours of follow-up available each month. Determine the most profitable mixture of groups she should contact and the most money she can raise in a month.
Answer:
8 churches, 2 unions; $1350 per monthStep-by-step explanation:
Let x and y represent the numbers of churches and unions contacted in the month, respectively. Then Jayanta's limit on letter writing hours is ...
2x +2y ≤ 20
and her limit on follow-up call hours is ...
x + 3y ≤ 14
Graphing these inequalities (see below) results in a feasible region with vertices at (x, y) = (0, 4 2/3), (8, 2), and (10, 0). Of these, the mixture of groups producing the most money is ...
8 churches and 2 unions.
The money she can raise from that mixture is ...
8×$125 +2×$175 = $1350 in a month
Answer:
Sqdancefan's answer is correct.
Step-by-step explanation:
I misread the question.
1. Solve the equation x = (2x+ 3)1/2
2. Consider f1(x) = ln(x + 1) + ln (x-1) and f2(x) = ln(x^2-1).
a) State domains and ranges of f1 and f2.
b) Sketch the curves y = f1(x) and y = f2(x).
Answer:
1. [tex]x=3[/tex]; 2. Domain: [tex]x>1[/tex] Range: all real numbers
Step-by-step explanation:
Let's find the solutions.
1. Solve the equation [tex]x=\sqrt{2x+3}[/tex] so:
[tex](x)^2=(\sqrt{2x+3})^2[/tex]
[tex]x^2=2x+3[/tex]
[tex]x^2-2x-3=0[/tex]
[tex]x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]x1=\frac{2+\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]
[tex]x1=3[/tex]
[tex]x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]x2=\frac{2-\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]
[tex]x2=-1[/tex]
Although we have two answers, remember that from the original equation the result of [tex]\sqrt{2x+3} > 0[/tex] is never negative. So -1 do not solve the equation.
In conlcusion, the equation is solved by x=3.
2A. Domains and ranges of f1(x) and f2(x)
[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]
Using logarithmic property [tex]ln(a)+ln(b)=ln(a*b)[/tex] we have:
[tex]f1(x)=ln(x^2-1)[/tex] because:
[tex]ln(x)[/tex] is defined by [tex]x>0[/tex] then:
[tex]x^2-1>0[/tex]
[tex]x>\sqrt{1}[/tex] so the domain of f1(x) is [tex]x>1[/tex]
Now for the range:
[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]
[tex]y=ln(x^2-1)[/tex]
[tex]e^y=x^2-1[/tex]
[tex]\sqrt{e^y+1}=x^2-1[/tex] notice that [tex]e^y+1[/tex] is always positive, so the range of f1(x) is all real numbers.
Be aware that although point number two of the problem mentioned two equations, f1(x)=f2(x) by logarithmic properties, so their domains and ranges are the same.
2B. Graph of f1(x) is attached. Because f1(x)=f2(x) both functions plot equal.
The range of for y = 4/5 sin x for pi [tex]\leq[/tex] x [tex]\leq[/tex] 3pi/2 is
Choices:
4/5 [tex]\leq[/tex] y [tex]\leq[/tex] 1
-1 [tex]\leq[/tex] y[tex]\leq[/tex] 4/5
-4/5[tex]\leq[/tex] y [tex]\leq[/tex] 0
-4/5 [tex]\leq[/tex] y[tex]\leq[/tex] 4/5
Answer:
see attachment
Step-by-step explanation:
The range of [tex]\(y = \frac{4}{5}\sin x\)[/tex] for [tex]\(\pi \leq x \leq \frac{3\pi}{2}\)[/tex] is [tex]\(-\frac{4}{5} \leq y \leq 0\)[/tex] ( Option C).
To find the range of [tex]\(y = \frac{4}{5}\sin x\)[/tex] for [tex]\(\pi \leq x \leq \frac{3\pi}{2}\)[/tex], we need to determine the minimum and maximum values of sin x in the given interval and then scale them using [tex]\(\frac{4}{5}\)[/tex].
In the interval [tex]\(\pi \leq x \leq \frac{3\pi}{2}\)[/tex], the sine function is negative since it corresponds to the third and fourth quadrants on the unit circle. The minimum value of sin x in this interval is -1, and the maximum value is 0.
Now, scale these values using [tex]\(\frac{4}{5}\)[/tex]:
[tex]\(-1 \times \frac{4}{5} = -\frac{4}{5}\) (minimum)\\\\\(0 \times \frac{4}{5} = 0\) (maximum)[/tex]
Therefore, the range of [tex]\(y = \frac{4}{5}\sin x\) for \(\pi \leq x \leq \frac{3\pi}{2}\)[/tex] is [tex]\(-\frac{4}{5} \leq y \leq 0\)[/tex]. The correct choice is:
-4/5 <= y <= 0
To know more about range, refer here:
https://brainly.com/question/17440903
#SPJ2
A brand name has a 40% recognition rate. If the owner of the brand wants to verify that rate by beginning with a small sample of 5 randomly selected consumers, find the probability that exactly 2 of the 5 consumers recognize the brand name. Also find the probability that the number who recognize the brand name is not 2.
Answer:
1) 0.3456
2) 0.6544.
Step-by-step explanation:
Let X represents the event of recognizing the brand,
Given,
The probability of recognizing the brand, p = 40% = 0.40,
Thus, the probability of not recognizing the brand, q = 1 - 0.40 = 0.60,
Since, the binomial distribution formula,
[tex]P(x) = ^nC_r (p)^r(q)^{n-r}[/tex]
Where,
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
1) Thus, the probability that exactly 2 of the 5 consumers recognize the brand name is,
[tex]P(X=2)=^5C_2 (0.40)^2 (0.60)^{5-2}[/tex]
[tex]=10 (0.40)^2 (0.60)^3[/tex]
[tex]=0.3456[/tex]
2) Also, the probability that the number who recognize the brand name is not 2 = 1 - P(X=2) = 1 - 0.3456 = 0.6544.
Prove that the set of all odd positive integers is countable.
Answer:
See below.
Step-by-step explanation:
The set of all positive integers N is countable so we need to show that there is a 1 to 1 correspondence between the elements in N with the set of all odd positive integers. This is the case as shown below:
1 2 3 4 5 6 ...
| | | | | | ....
1 3 5 7 9 11....
We are doing a study on shampoo buyers in Walmart stores across the US and we send them online surveys. Based on past studies, we know that of the people who start the survey, 80% qualify for the study, and only 40% of these people who do qualify actually complete the survey. Of the people who complete the survey, 10% of respondents are removed due to invalid or low quality answers. If we want 1000 valid responses for this particular study, how many people do we need to have start the survey?
Answer:
2500
Step-by-step explanation:
They tell you that only 40% of the total complete the quality test, and you want to know how many represent 100%.
If this 40% is 1000 valid responses use cross multiplication to know how many people are the total.
40%-----1000 valid responses
100%----x=
[tex]\frac{100*1000}{40} =X\\2500=X[/tex]
So, you can tell they need at least 2500 persons to have 1000 valid responses.
Answer:
Number of people = 3473
Step-by-step explanation:
Probability is typically the rate at which an event or occurrence is likely to happen. Whenever we're not so sure about an event outcome, we can then deliberate about the probabilities or possibility of certain outcomes that is how likely they are to happen.
Probability of success, =0.8×0.4×0.9 =0.288
Expected, E=1000
Expected = np
1000=n×0.288
n=0.2881000≈34
Find the equation of the following line and graph. Through (3,-10) perpendicular to 5x-y=9
bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of 5x - y = 9 then.
[tex]\bf 5x-y=9\implies -y=-5x+9\implies y=\stackrel{\stackrel{m}{\downarrow }}{5}x-9\leftarrow \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{5\implies \cfrac{5}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{5}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{5}}}[/tex]
so then, we're really looking for the equation of a line whose slope is -1/5 and runs through (3,-10).
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-10})~\hspace{10em} slope = m\implies -\cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-10)=-\cfrac{1}{5}(x-3)\implies y+10=-\cfrac{1}{5}x+\cfrac{3}{5} \\\\\\ y=-\cfrac{1}{5}x+\cfrac{3}{5}-10\implies y=-\cfrac{1}{5}x+\cfrac{53}{5}[/tex]
and it looks like the one in the picture below.
If f(x) = 2x - 1 and g(x) = x^2 - 2, find [g · f](x)
please show me how to do this
Answer:
(x² -7)/(2x + 1)
Step-by-step explanation:
f(x) = 2x+1 and g(x) = x² -7
thus: (g/f)(x) = g(x)/f(x) = x² -7/2x + 1
The answer is:
[tex](g \circ f)(x)=4x^{2}-4x-1[/tex]
Why?To solve the problem, we need to remember that composing functions means evaluate a function into another different function.
Also, we need to remember how to solve the following notable product:
[tex](a-b)^{2}=a^{2}-2ab+b^{2}[/tex]
We have that:
[tex](g \circ f)(x)=g(f(x))[/tex]
Now, we are given the equations:
[tex]f(x)=2x-1\\g(x)=x^{2}-2[/tex]
So, composing we have:
[tex](g \circ f)(x)=g(f(x))[/tex]
[tex](g \circ f)(x)=(2x-1)^{2}-2[/tex]
Now, we have to solve the notable product:
[tex](g \circ f)(x)=((2x)^{2}-2(2x*1)+1^{2})-2[/tex]
[tex](g \circ f)(x)=4x^{2}-4x+1-2[/tex]
Hence, we have that:
[tex](g \circ f)(x)=4x^{2}-4x-1[/tex]
Have a nice day!
For f(x) = 2|x+3| – 5, name the type of function and describe each of the three
transformations from the parent function f(x) = |x|
Answer:
There are three transformation i.e. vertically stretch, downward and towards left.
Step-by-step explanation:
Given : Parent function [tex]f(x)=|x|[/tex] and transformed function [tex]f(x)=2|x+3|-5[/tex]
To find : Name the type of function and describe each of the three transformations?
Solution :
Parent function [tex]f(x)=|x|[/tex] get transformed into [tex]f(x)=2|x+3|-5[/tex]
The transformations are as follow :
1) The function is multiplied by 2 which means there is a transformation vertically by a stretch factor '2'.
2) The function is subtracted by 5 which means there is a transformation downward by a factor '5'.
3) The function inside x value get added by 3 which means there is a transformation left by a factor '3'.
Therefore, There are three transformation i.e. vertically stretch, downward and towards left.
Suppose that the inflation rate is 2.5% and the real terminal value of an investment is expected to be $20,000 in 2 years. Calculate the nominal terminal value of the investment at the end of year 2.
Answer:
$21012.50
Step-by-step explanation:
$20,000 today will be inflated to $20,000·(1.025)^2 ≈ $21012.50 in 2 years. We presume this is the nominal terminal value you want.
A) In 2000, the population of a country was approximately 5.82 million and by 2040 it is projected to grow to 9 million. Use the exponential growth model A=A0e^kt, in which t is the number of years after 2000 and A0 is in millions, to find an exponential growth function that models the data.
B) By which year will the population be 15 million?
Answer:
By 2086
Step-by-step explanation:
The provided equation is:
[tex]A=A0*e^{k*t}[/tex] , where:
A=total of population after t years
A0=initial population
k= rate of growth
t= time in years
Given information:
The final population will be 15 million, then A=15.
We start in 2000 with a 5.82 million population, then A0=5.82.
Missing information:
Although k is not given, we can calculate k by using the following statement, from 2000 to 2040 (within 40 years) population is proyected to grow to 9 million, which means a passage from 5.8 to 9 million (3.2 million increament).
Then we can use the same expression to calculate k:
[tex]A=A0*e^{k*t}[/tex]
[tex]9=5.8*e^{40*k}[/tex]
[tex]ln(9/5.8)/40=k[/tex]
[tex]0.010984166494596147=k[/tex]
[tex]0.011=k[/tex]
Now that we have k=0.011, we can find the time (t) by which population will be 15 million:
[tex]A=A0*e^{k*t}[/tex]
[tex]15=5.8*e^{0.011t}[/tex]
[tex]ln(15/5.8)/0.011=t[/tex]
[tex]86.38111668634878=t[/tex]
[tex]86.38=t[/tex]
Because the starting year is 2000, and we need 86.38 years for increasing the population from 5.8 to 15 million, then by 2086 the population will be 15 million.
I need the answer to this question.
1) Divide 704 days 11 hours by 29.
There are 24 hours to a day, so you can write this as
704 days, 11 hours = 704 + 11/24 days
Then dividing by 29 gives
704/29 + 11/696 days
We have
704 = 24*29 + 8
so that the time is equal to
24 + 8/29 + 11/696 days
24 + (192 + 11)/696 days
24 + 7/24 days
which in terms of days and hours is
24 days, 7 hours
The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites.
The correct option is to reject the null hypothesis. The mean life spans for whites and non-whites born in 1900 in the certain county are not the same.
To conduct the hypothesis test, we will use a two-sample z-test for the difference in two means. The null hypothesis (H0) states that there is no difference in the mean life spans between whites and non-whites, while the alternative hypothesis (Ha) states that there is a difference.
The formula for the z-test statistic is:
[tex]\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \][/tex]
where:
- [tex]\(\bar{x}_1\)[/tex] and [tex]\(\bar{x}_2\)[/tex] are the sample means for whites and non-whites, respectively.
- [tex]\(\mu_1\)[/tex] and [tex]\(\mu_2\)[/tex] are the population means for whites and non-whites, respectively.
- [tex]\(\sigma_1\)[/tex] and [tex]\(\sigma_2\)[/tex] are the population standard deviations for whites and non-whites, respectively.
- [tex]\(n_1\)[/tex] and [tex]\(n_2\)[/tex] are the sample sizes for whites and non-whites, respectively.
Given:
- [tex]\(\bar{x}_1 = 45.3\) years, \(s_1 = 12.7\) years, \(n_1 = 124\)[/tex] (for whites)
- [tex]\(\bar{x}_2 = 34.1\)[/tex] years, [tex]\(s_2 = 15.6\) years, \(n_2 = 82\)[/tex] (for non-whites)
- [tex]\(\mu_1 = 47.6\)[/tex] years (for whites)
- [tex]\(\mu_2 = 33.0\)[/tex] years (for non-whites)
Since we do not have the population standard deviations, we will use the sample standard deviations as an estimate. This is appropriate given the sample sizes are large enough (generally [tex]\(n > 30\)[/tex] is considered sufficient).
First, we calculate the standard error (SE) of the difference in means:
[tex]\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{12.7^2}{124} + \frac{15.6^2}{82}} \][/tex]
[tex]\[ SE = \sqrt{\frac{161.29}{124} + \frac{243.36}{82}} \][/tex]
[tex]\[ SE = \sqrt{1.299 + 2.968} \][/tex]
[tex]\[ SE = \sqrt{4.267} \][/tex]
[tex]\[ SE \approx 2.066 \][/tex]
Now, we calculate the z-statistic:
[tex]\[ z = \frac{(45.3 - 34.1) - (47.6 - 33.0)}{2.066} \][/tex]
[tex]\[ z = \frac{11.2 - 14.6}{2.066} \][/tex]
[tex]\[ z = \frac{-3.4}{2.066} \][/tex]
[tex]\[ z \approx -1.646 \][/tex]
Next, we find the p-value for this z-statistic. Since we are conducting a two-tailed test, we will look at the probability of a z-score being less than -1.646 or greater than 1.646. Using a standard normal distribution table or a calculator, we find that the p-value is approximately 0.100.
Finally, we compare the p-value to our significance level (commonly denoted as [tex]\(\alpha\))[/tex]. If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis. If we choose [tex]\(\alpha = 0.05\)[/tex], then since [tex]\(0.100 > 0.05\)[/tex], we fail to reject the null hypothesis.
However, if we choose a significance level of [tex]\(\alpha = 0.10\)[/tex], then since [tex]\(0.100 \leq 0.10\)[/tex], we would reject the null hypothesis, indicating that there is a statistically significant difference in the mean life spans between whites and non-whites in the county.
Given the p-value is on the boundary of common significance levels, the conclusion may vary depending on the chosen level of significance. However, the correct option based on the provided conversation is to reject the null hypothesis, suggesting that the mean life spans are not the same for whites and non-whites in the county.
A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 40 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 42.3 milligrams. Assume the population is normally distributed and the population standard deviation is 7.1 milligrams. At alphaequals0.04, can you reject the company's claim
Answer:
Step-by-step explanation:
Put hypotheses as:
[tex]H_0: mu =40 mg\\H_a: mu \neq 40 mg.[/tex]
(Two tailed test at 4%)
Since population std deviation is given we can do Z test.
[tex]Sample size n =30\\Sample mean = 42.3 mg\\Mean diff = 2.3 mg\\Std error of sample = \frac{7.1}{\sqrt{30} } =1.296[/tex]
Test statistic t = mean diff/se = [tex]\frac{2.3}{1.296} =1.775[/tex]
p value=0.075
Since p >0.04 our alpha, we accept null hypothesis.
At alphaequals0.04, we cannot reject the company's claim
Step-by-step explanation:
Put hypotheses as:
\begin{gathered}H_0: mu =40 mg\\H_a: mu \neq 40 mg.\end{gathered}
H
0
:mu=40mg
H
a
:mu
=40mg.
(Two tailed test at 4%)
Since population std deviation is given we can do Z test.
\begin{gathered}Sample size n =30\\Sample mean = 42.3 mg\\Mean diff = 2.3 mg\\Std error of sample = \frac{7.1}{\sqrt{30} } =1.296\end{gathered}
Samplesizen=30
Samplemean=42.3mg
Meandiff=2.3mg
Stderrorofsample=
30
7.1
=1.296
Test statistic t = mean diff/se = \frac{2.3}{1.296} =1.775
1.296
2.3
=1.775
p value=0.075
Since p >0.04 our alpha, we accept null hypothesis.
At alphaequals0.04, we cannot reject the company's claim
Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, -1>, v = <7, -4>
Answer:
A
Step-by-step explanation:
Given
u = <6, -1>
u = 6i-j
and
v=<7,-4>
v=7i-4j
The formula for angle is:
Let x be the angle
[tex]cos\ x = \frac{u.v}{||u||.||v||}[/tex]
where ||u|| is the length and u.v is the dot product or scalar product of both vectors
So,
[tex]||u|| = \sqrt{(6)^2+(-1)^2}\\ = \sqrt{36+1}\\ = \sqrt{37}\\ ||v||=\sqrt{(7)^2+(-4)^2}\\ = \sqrt{49+16}\\ = \sqrt{65}\\[/tex]
[tex]u.v = u_1u_2+v_1v_2\\= (6)(7)+(-1)(-4)\\=42+4\\=46[/tex]
[tex]cos\ x=\frac{46}{\sqrt{37}\sqrt{65}} \\= \frac{46}{\sqrt{2405} }\\Can\ also\ be\ written\ as:\\= \frac{46}{\sqrt{2405} } * \frac{\sqrt{2405} }{\sqrt{2405}} \\=\frac{46\sqrt{2405} }{2405}[/tex]
The calculated angle will be in radians. To find the angle in degrees:
[tex]x = \frac{180}{\pi} cos^{-1} (\frac{46\sqrt{2405} }{2405})\\x = 20.282\\x= 20.3\\[/tex]
Hence Option A is correct ..
Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 1.1 years. Find the probability that 70 randomly selected washing machines will have a mean replacement time less than 9.1 years. Your answer should be a decimal rounded to the fourth decimal place.
Answer:
The probability is 0.0643
Step-by-step explanation:
* Lets revise some definition to solve the problem
- The standard deviation of the distribution of sample means is called σM
- σM = σ/√n , where σ is the standard deviation and n is the sample size
- z-score = (M - μ)/σM, where M is the mean of the sample , μ is the mean
of the population
* Lets solve the problem
- The mean of the washing machine is 9.3 years
∴ μ = 9.3
- The standard deviation is 1.1 years
∴ σ = 1.1
- There are 70 washing machines randomly selected
∴ n = 70
- The mean replacement time less than 9.1 years
∴ M = 9.1
- Lets calculate z-score
∵ σM = σ/√n
∴ σM = 1.1/√70 = 0.1315
∵ z-score = (M - μ)/σM
∴ z-score = (9.1 - 9.3)/0.1315 = - 1.5209
- Use the normal distribution table of z to find P(z < -1.5209)
∴ P(z < -1.5209) = 0.06426
∵ P(M < 9.1) = P(z < -1.5209)
∴ P(M < 9.1) = 0.0643
* The probability is 0.0643
Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector 15–√i+25–√j15i+25j. (Use symbolic notation and fractions where needed.)
[tex]f(x,y,z)=2z^2x+y^3[/tex]
[tex]f[/tex] has gradient
[tex]\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k[/tex]
which at the point (-1, 4, 3) has a value of
[tex]\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k[/tex]
I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say [tex]\vec u=15\,\vec\imath+25\,\vec\jmath[/tex], in which case we have
[tex]\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}[/tex]
Then the derivative of [tex]f[/tex] at (-1, 4, 3) in the direction of [tex]\vec u[/tex] is
[tex]D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}[/tex]
To find the directional derivative, we need to find the gradient of the function and dot product it with the given direction vector.
Explanation:To find the directional derivative, we need to find the gradient of the function and dot product it with the given direction vector. The gradient of the function f(x, y, z) = 2z^2x + y^3 is (∂f/∂x, ∂f/∂y, ∂f/∂z) = (4zx, 3y^2, 4xz). The directional derivative in the direction of the vector (15-√i + 25-√j) is given by the dot product of the gradient and the direction vector: (4(-1)(15-√) + 3(4^2)(25-√) + 4(3)(-1)(15-√))/√((15-√)^2 + (25-√)^2). Simplifying this expression gives the directional derivative.
Learn more about Directional derivative here:https://brainly.com/question/32589894
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Doing research for insurance rates, it is found that those aged 30 to 49 drive an average of 38.7 miles per day with a standard deviation of 6.7 miles. These distances are normally distributed. If a group of 60 drivers in that age group are randomly selected, what is the probability that the mean distance traveled each day is between 32.5 miles and 40.5 miles?
Answer: 0.4302
Step-by-step explanation:
Given : Mean : [tex]\mu=\text{38.7 miles }[/tex]
Standard deviation : [tex]\sigma=\text{6.7 miles }[/tex]
Sample size : [tex]n=60[/tex]
Also, these distances are normally distributed.
Then , the formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=32.5
[tex]\\\\ z=\dfrac{32.5-38.7}{6.7}=-0.925373134\approx-0.93[/tex]
For x=40.5
[tex]\\\\ z=\dfrac{40.5-38.7}{6.7}=0.268656\approx0.27[/tex]
The p-value = [tex]P(-0.93<z<0.27)[/tex]
[tex]=P(0.27)-P(-0.93)=0.6064198- 0.1761855=0.4302343\approx0.4302[/tex]
Hence, the required probability :-0.4302
The three sides of a triangle measure 9, 20, and n. What is the range of possible lengths of n?
9 < n < 20
6 < n < 29
11 < n < 29
11 < n < 20
Answer:
11 < n < 29
Step-by-step explanation:
The smallest the third side can be is just bigger than the difference of the other two sides
20-9 < n
11<n
The largest the third side can be is just smaller than the sum of the other two sides
20+9 > n
29 >n
Putting this together
11<n<29
You manage an ice cream factory that makes two flavors: Creamy Vanilla and Continental Mocha. Into each quart of Creamy Vanilla go 2 eggs and 3 cups of cream. Into each quart of Continental Mocha go 1 egg and 3 cups of cream. You have in stock 700 eggs and 1500 cups of cream. How many quarts of each flavor should you make in order to use up all the eggs and cream? HINT [See Example 5.]
Answer:
200 Creamy Vanilla
300 Continental Mocha
Step-by-step explanation:
3 cups cream for each:
500 Quarts total
Creamy Vanilla requires 400 eggs
Continental Mocha requres 300 eggs for a total of 700 eggs
The answer is:
[tex]\[ \boxed{V = 200, M = 300} \][/tex]
To solve this problem, we need to set up a system of equations based on the given information and then solve for the number of quarts of each flavour of ice cream that should be made.
Let [tex]\( V \)[/tex] represent the number of quarts of Creamy Vanilla ice cream and [tex]\( M \)[/tex] represent the number of quarts of Continental Mocha ice cream.
From the information given, we can derive the following equations:
For the eggs:
Each quart of Creamy Vanilla requires 2 eggs, so [tex]\( 2V \)[/tex] eggs are used for Creamy Vanilla.
Each quart of Continental Mocha requires 1 egg, so [tex]\( M \)[/tex] eggs are used for Continental Mocha.
Since there are 700 eggs in total, we have the equation:
[tex]\[ 2V + M = 700 \][/tex]
For the cream:
Each quart of Creamy Vanilla requires 3 cups of cream, so [tex]\( 3V \)[/tex] cups of cream are used for Creamy Vanilla.
Each quart of Continental Mocha also requires 3 cups of cream, so [tex]\( 3M \)[/tex] cups of cream are used for Continental Mocha.
Since there are 1500 cups of cream in total, we have the equation:
[tex]\[ 3V + 3M = 1500 \][/tex]
Now we have a system of two equations with two variables:
[tex]\[ \begin{cases} 2V + M = 700 \\ 3V + 3M = 1500 \end{cases} \][/tex]
To solve this system, we can simplify the second equation by dividing every term by 3:
[tex]\[ V + M = 500 \][/tex]
Now we have a simpler system:
[tex]\[ \begin{cases} 2V + M = 700 \\ V + M = 500 \end{cases} \][/tex]
We can subtract the second equation from the first to eliminate [tex]\( M \)[/tex] and solve for [tex]\( V \)[/tex] :
[tex]\[ (2V + M) - (V + M) = 700 - 500 \][/tex]
[tex]\[ V = 200 \][/tex]
Now that we have the value of [tex]\( V \)[/tex], we can substitute it back into the simplified second equation to find [tex]\( M \)[/tex]:
[tex]\[ 200 + M = 500 \][/tex]
[tex]\[ M = 500 - 200 \][/tex]
[tex]\[ M = 300 \][/tex]
Therefore, the factory should make 200 quarts of Creamy Vanilla and 300 quarts of Continental Mocha to use up all the eggs and cream.
The final answer is:
[tex]\[ \boxed{V = 200, M = 300} \][/tex]
Round each of the following, using front-end rounding. [1.1] 4. 50,987 5. 851,004
Answer:
4) 50,000
5) 900,000
Step-by-step explanation:
In Front end rounding what we do is we focus on first two digit of the number if the second digit number is greater than or equal to 5 we add the first digit by 1.
4) 50,987
here we can clearly see that the first two digit is 50 and second digit is not equal to 5 then rounding will be equal to 50,000.
5) 851,004
here we can clearly see that the first two digit is 85 and second digit is equal to 5 then we will round it 8+1=9
hence rounded number will be 900,000
Final answer:
Front-end rounding of 50,987 is 50,000 and for 851,004 it's 900,000, based on the most significant digits, using zeros as placeholders after rounding.
Explanation:
Front-end rounding involves rounding numbers based on the most significant digits. Let's apply this method to the numbers provided.
50,987 rounded to the nearest ten thousand would be 50,000.For 851,004, rounded to the nearest hundred thousand would be 900,000.When rounding, if the digit immediately after the place you are rounding to is 5 or greater, you round up. Otherwise, you round down. Placeholder zeros are used to maintain the value's place in the number system.
Consider the region satisfying the inequalities.y ≤ e−x, y ≥ 0, x ≥ 0a) Find area of regionb) Find the volume of the solid generated by revolving the region about the x-axis.c) Find the volume of the solid generated by revolving the region about the y-axis.
Using the disk method, the volume is
[tex]\displaystyle\pi\int_0^\infty e^{-2x}\,\mathrm dx=\boxed{\frac\pi2}[/tex]
Alternatively, using the shell method, the volume is
[tex]\displaystyle2\pi\int_0^1y(-\ln y)\,\mathrm dy=\frac\pi2[/tex]
Revolving about the [tex]y[/tex]-axis:Using the shell method, the volume is
[tex]\displaystyle2\pi\int_0^\infty xe^{-x}\,\mathrm dx=\boxed{2\pi}[/tex]
Alternatively, using the disk method, the volume is
[tex]\displaystyle\pi\int_0^1(-\ln x)^2\,\mathrm dx=2\pi[/tex]
The area of the region is 1 square unit. The volume of the solid generated by revolving the region about the x-axis can be found by integrating π(y^2) dx from x = 0 to x = ∞.
Explanation:To find the area of the region, we need to find the intersection points between the two curves. In this case, the curves are y = e^(-x) and y = 0. Since y ≥ 0, the region will lie between the x-axis and the curve y = e^(-x). The intersection point is where y = 0, which occurs at x = 0. To find the area, we integrate y = e^(-x) from x = 0 to x = ∞:
A = ∫0∞ e^(-x) dx = [-e^(-x)]0∞ = -[e^0 - 0]
= -[1 - 0] = 1
The area of the region is 1 square unit.
To find the volume of the solid generated by revolving the region about the x-axis, we use the disk method. The radius of each disk is given by y = e^(-x), and the height of each disk is given by dx. The volume can be found by integrating π(y^2) dx from x = 0 to x = ∞:
V = π∫0∞ (e^(-x))^2 dx = π∫0∞ e^(-2x) dx
A tank holds 300 gallons of water and 100 pounds of salt. A saline solution with concentration 1 lb salt/gal is added at a rate of 4 gal/min. Simultaneously, the tank is emptying at a rate of 1 gal/min. Find the specific solution Q(t) for the quantity of salt in the tank at a given time t.
The amount of salt in the tank changes with rate according to
[tex]Q'(t)=\left(1\dfrac{\rm lb}{\rm gal}\right)\left(4\dfrac{\rm gal}{\rm min}\right)-\left(\dfrac{Q(t)}{300+(4-1)t}\dfrac{\rm lb}{\rm gal}\right)\left(1\dfrac{\rm gal}{\rm min}\right)[/tex]
[tex]\implies Q'+\dfrac Q{300+3t}=4[/tex]
which is a linear ODE in [tex]Q(t)[/tex]. Multiplying both sides by [tex](300+3t)^{1/3}[/tex] gives
[tex](300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}[/tex]
so that the left side condenses into the derivative of a product,
[tex]\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}[/tex]
Integrate both sides and solve for [tex]Q(t)[/tex] to get
[tex](300+3t)^{1/3}Q=(300+3t)^{4/3}+C[/tex]
[tex]\implies Q(t)=300+3t+C(300+3t)^{-1/3}[/tex]
Given that [tex]Q(0)=100[/tex], we find
[tex]100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}[/tex]
and we get the particular solution
[tex]Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}[/tex]
[tex]\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}[/tex]
Approximate the number below using a calculator. Round your answer to three decimal places. el.3 a) 3.534 b) 3.669 c) 3.969 d) 2.040
Answer:
a. 3.53 b. 3.61 c. 3.91 d. 2.04
Step-by-step explanation:
a. 3.534 take 4 out
b. 3.669 nine is higher then 5 so, take six makes out of one
c. 3.969 nine is higher then 5 so, take six makes out of one
d. 2.040 take zero out
above that is the answer.... hope i helped this....
A father wishes to give his son P200, 000 ten years from now. What amount should he invest if it will earn interest at 10% compounded quarterly during the first five years and 12% compounded annually during the next five years? A. P68,757.82 B. P62,852.23 C. P69,256.82 D. P67,238.54
Answer:
C. P69,256.82
Step-by-step explanation:
We know that,
The amount formula in compound interest is,
[tex]A=P(1+\frac{r_1}{n_1})^{n_1t_1} (1+\frac{r_2}{n_2})^{n_2t_2}.......[/tex]
Where, P is the principal amount,
[tex]r_1, r_2....[/tex] are the annual rate for the different periods,
[tex]t_1, t_2,.....[/tex] are the number of year for different periods,
[tex]n_1, n_2, n_3...[/tex] are the number of periods,
Given,
A = P 200,000,
[tex]r_1=10%=0.1[/tex], [tex]n_1=4[/tex], [tex]t_1=5[/tex],[tex]r_2=12%=0.12[/tex], [tex]n_2=1[/tex], [tex]t_2=5[/tex]
Thus, by the above formula the final amount would be,
[tex]200000=P(1+\frac{0.1}{4})^{4\times 5}(1+\frac{0.12}{1})^{1\times 5}[/tex]
[tex]200000=P(1+0.025)^{20}(1+0.12)^5[/tex]
[tex]200000=P(1.025)^{20}(1.12)^5[/tex]
[tex]\implies P=69,256.824\approx 69,256.82[/tex]
Option C is correct.
The correct option is C. P69,256.82. The father should invest approximately P69,256.82 today to ensure his son receives P200,000 in ten years.
To determine the amount the father needs to invest today to give his son P200,000 ten years from now, we will break the problem into two phases due to different interest rates and compounding periods.
Phase 1: First Five Years (10% compounded quarterly)
→ Future Value (FV) needed after 10 years: P200,000
→ Future Value (FV) after first five years at 12% annual interest for the next five years:
Let's use the formula for compound interest to calculate amount required after the first five years.
Here,
→ n is the number of times the interest is compounded per year
→ t is time in years.
→ [tex]FV = PV*(1 + r/n)^{(nt)[/tex]
After the first five years, the amount needs to grow at 12% compounded annually for 5 years to reach P200,000.
We can calculate the present value (PV) at the end of the first five years needed to achieve P200,000 after next 5 years.
→ P200,000 = [tex]PV * (1 + 0.12/1)^{(1*5)[/tex]
→ P200,000 = [tex]PV * (1.12)^5[/tex]
Calculating PV:
→ PV = P200,000 / (1.7623)
≈ P113,477.57
Phase 2: First Five Years Investment Calculation
Now, we need to find out the amount the father should invest today to reach P113,477.57 after five years with 10% interest compounded quarterly.
Here,
→ r is the quarterly rate,
→ nt is the total number of quarters.
We use the same compound interest formula:
→ [tex]FV = PV * (1 + r/n)^{(nt)[/tex]
→ P113,477.57 = [tex]PV * (1 + 0.10/4)^{(4*5)[/tex]
→ P113,477.57 = [tex]PV * (1.025)^{20[/tex]
→ [tex](1.025)^{20} \approx 1.6386[/tex]
Calculating PV:
→ PV = P113,477.57 / 1.6386
≈ P69,256.82
So, the father needs to invest approximately P69,256.82 today.
You deposit $100 in an account earning 4% interest compounded annually. How much will you have in the account in 10 years? Round your answer to the nearest penny
Answer:
$148.02
Step-by-step explanation:
In the question
Principal = $100, rate(R) = 4% compounded annually, time(T)= 10 years
we know that the formula for compound interest
A=[tex]P\times_(1+\frac{R}{100} )^{T}[/tex] where A is amount
now putting values in the above formula we get
A=[tex]100\times_(1+\frac{4}{100} )^{10}[/tex]
therefore A= $148.024428
rounding off to the nearest penny we get amount as $148.02 and compound interest will be $48.02