Answer:
The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
In a sample of 500 adults, 345 had children.
This means that [tex]n = 500, \pi = \frac{345}{500} = 0.69[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.69 - 2.575\sqrt{\frac{0.69*0.31}{500}} = 0.6367[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.69 + 2.575\sqrt{\frac{0.69*0.31}{500}} = 0.7433[/tex]
The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).
What is a1
of the arithmetic sequence for which a3=126
and a64=3,725
a
64
=
3
,
725
?
In an arithmetic sequence, every pair of consecutive terms differs by a fixed number c, so that the n-th term [tex]a_n[/tex] is given recursively by
[tex]a_n=a_{n-1}+c[/tex]
Then for n ≥ 2, we have
[tex]a_2=a_1+c[/tex]
[tex]a_3=a_2+c = (a_1+c)+c = a_1 + 2c[/tex]
[tex]a_4=a_3+c = (a_1 + 2c) + c = a_1 + 3c[/tex]
and so on, up to
[tex]a_n=a_1+(n-1)c[/tex]
Given that [tex]a_3=126[/tex] and [tex]a_{64}=3725[/tex], we can solve for [tex]a_1[/tex]:
[tex]\begin{cases}a_1+2c=126\\a_1+63c=3725\end{cases}[/tex]
[tex]\implies(a_1+63c)-(a_1+2c)=3725-126[/tex]
[tex]\implies 61c = 3599[/tex]
[tex]\implies c=59[/tex]
[tex]\implies a_1+2\times59=126[/tex]
[tex]\implies a_1+118 = 126[/tex]
[tex]\implies \boxed{a_1=8}[/tex]
HELP ASAP!!!!!!
Thank you so much
We know
[tex]\boxed{\sf P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]
[tex]\\ \sf\longmapsto p(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]
m:n=3:5
We know
[tex]\boxed{\sf M(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]
[tex]\\ \sf\longmapsto M(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]
the line that passes through the point (-4, 2) and has a
What is the equation of
slope of
2?
Answer:
y = 2x + 10
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m = 2 , then
y = 2x + c ← is the partial equation
To find c substitute (- 4, 2 ) into the partial equation
2 = - 8 + c ⇒ c = 2 + 8 = 10
y = 2x + 10 ← equation of line
Solve 5 to x-4 power =7
Answer:The value of x is 4.
Step-by-step explanation:
Given : Expression .
To find : Solve for x ?
Solution :
Applying exponential property,
Comparing the base,
Step-by-step explanation:
Please helpppppp me!!!!!!!!
Answer:
A --> y=cot(x)
Step-by-step explanation:
if you graph tan(x), it has a period of just PI, because tan(x) is just sin(x)/cos(), and cot(x) is the same because it is just sec(x)/csc(x).
A man owns 3/4 of the share of a business and sells 1/3 of his
shares for Birr 10,000. What is the value of the business in Bir?
Given:
A man owns 3/4 of the share of a business and sells 1/3 of his shares for Bir 10,000.
To find:
The value of the business in Bir.
Solution:
Let x be the value of the business.
It is given that a man owns 3/4 of the share of a business and sells 1/3 of his shares for Bir 10,000.
[tex]x\times \dfrac{3}{4}\times \dfrac{1}{3}=10000[/tex]
[tex]x\times \dfrac{1}{4}=10000[/tex]
Multiply both sides by 4.
[tex]x=4\times 10000[/tex]
[tex]x=40000[/tex]
Therefore, the value of the business is 40,000 Bir.
.
you invested $
7000
7000
between two accounts payin
6
%
6
%
and
8
%
8
%
annual interet, respectively. if the total interest earned for the year was $
480
,
480
,
how much was invested at each rate
9514 1404 393
Answer:
$3000 at 8%$4000 at 6%Step-by-step explanation:
Let x represent the amount invested at 8%. Then the total interest earned is ...
0.06(7000 -x) +0.08x = 480
420 +0.02x = 480 . . . . . . . . . eliminate parentheses
0.02x = 60 . . . . . . . . . . subtract 420
x = 60/0.02 = 3000 . . . . . divide by the coefficient of x
$3000 was invested at 8%; $4000 was invested at 6%.
A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 28 transects gave a sample variance s2 = 48.3 for the number of sites per transect.
Required:
Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3.
Answer:
There is significant evidence to conclude that carina e in New section is greater than 42.3
Step-by-step explanation:
Given :
Sample variance, s² = 48.3
Population variance, σ² = 42.3
Sample size, n = 28
α = 0.05
The hypothesis :
H0 : σ² = 42.3
H0 : σ² > 42.3
The test statistic, χ² : (n-1)*s²/σ²
χ² = [(28 - 1) * 48.3] / 42.3
χ² = (27 * 48.3) / 42.3
χ² = 1304.1 / 42.3
χ² = 30.829787
χ² = 30.830
The Critical value at α = 0.05 ; df = (28-1) = 27
Critical value(0.05, 27) = 27.587
Reject H0 if χ² statistic > Critical value
Since, 30.830 > 27.587 ; Reject H0 ; and conclude that variance in the new section is greater than 42.3
Open the graphing tool one last time. Compare the graphs of y=log (x-k) and y=log x+k in relation to their domain, range, and asymptotes. Describe what you see.
Answer:
sorry I don't know the answer
Answer:
For the equation y=log(x-k), the domain depends on the value of K. Sliding K moves the left bound of the domain interval. The range and the right end behavior stay the same. For the equation y=log x+k, the domain is fixed, starting at an x-value of 0. The vertical asymptote is also fixed. The range of the equation depends on K.
Step-by-step explanation:
https://www.drfrostmaths.com/util-generatekeyskillpic.php?name=AnglePoint4&width=400¶ms=%5B61%2C0%2C%22y%22%2C4%5D
Answer:
Complementary angle
<y+61° =90°
<y = 90°- 61°
<y =29°
I Hope this helps.
What is the measurement of N?
Answer:
the measurement of N is D, 81.
Step-by-step explanation:
The angle measurement of a Right Angled Triangle is 90 degrees. And based off the angle dimension given in the image above ( 9 degrees ), you need to subtract 90 ( the angle dimension of the triangle) with the angle dimension given (9 degrees) which gets you to an answer of 81 degrees.
A humanities professor assigns letter grades on a test according to the following scheme.
A: Top 8% of scores
B: Scores below the top 8% and above the bottom 62%
C: Scores below the top 38% and above the bottom 18%
D: Scores below the top 82% and above the bottom 9%
E: Bottom 9% of scores Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.
Find the numerical limits for a C grade.
Answer:
The numerical limits for a C grade are 60.6 and 69.1.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.
This means that [tex]\mu = 67, \sigma = 7.3[/tex]
Find the numerical limits for a C grade.
Below the 100 - 38 = 62th percentile and above the 18th percentile.
18th percentile:
X when Z has a p-value of 0.18, so X when Z = -0.915.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.915 = \frac{X - 67}{7.3}[/tex]
[tex]X - 67 = -0.915*7[/tex]
[tex]X = 60.6[/tex]
62th percentile:
X when Z has a p-value of 0.62, so X when Z = 0.305.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.305 = \frac{X - 67}{7.3}[/tex]
[tex]X - 67 = 0.305*7[/tex]
[tex]X = 69.1[/tex]
The numerical limits for a C grade are 60.6 and 69.1.
In a normally distributed data set with a mean of 19 and a standard deviation of 2.6, what percentage of the data would be between 13.8 and 24.2?a) 95% based on the Empirical Ruleb) 68% based on the histogramc) 68% based on the Empirical Ruled) 95% based on the histogram
Step-by-step explanation:
check the attachment above.
The work done by a machine in 2 minutes is 480J. Calculate the power of the machine
Answer:
I think the power is 4
Step-by-step explanation:
480J / 120 = 4
Put 2 mins into seconds which is 120 seconds
Sorry if it is wrong :)
Answer:
[tex]4\text{ watts}[/tex]
Step-by-step explanation:
In physics, the power of a machine is given by [tex]P=\frac{W}{\Delta t}[/tex], where [tex]W[/tex] is work in Joules and [tex]\Delta t[/tex] is time in seconds.
Convert 2 minutes into seconds:
2 minutes = 120 seconds.
Substitute [tex]W=480[/tex] and [tex]\Delta t=120[/tex] to solve for [tex]P[/tex]:
[tex]P=\frac{480}{120}=\boxed{4\text{ watts}}[/tex]
what is the area of this whole shape
Answer:
104 m²
Step-by-step explanation:
Area of the whole shape = area of the triangle + area of the rectangle
= ½*b*h + L*W
Where,
b = 8 m
h = 6 m
L = 10 m
W = 8 m
Plug in the values into the equation
Area of the whole shape = ½*8*6 + 10*8
= 24 + 80
= 104 m²
Help I don’t get this
Answer:
Step-by-step explanation:
5t² + 4t = 5t² + 4 if and only if t=1.
A zero coefficient makes the value of the term equal to zero.
Answer: 5t to the second power and 5t to the second power. I don't know the answer to b.
Step-by-step explanation:
5 = –6x2 + 24x
5 = –6(x2 – 4x)
inside the parentheses and
.
–19 = –6(x – 2)2
StartFraction 19 Over 6 EndFraction = (x – 2)2
Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot = x – 2
The two solutions are
Plus or minus StartRoot StartFraction 19 Over 6 EndFraction EndRoot.
Answer:
x = 2 - sqrt(19/6)
x = 2 + sqrt(19/6)
Step-by-step explanation:
Answer:
add 4
subtract 24 from 5
2
Step-by-step explanation:
А _______ equation can be written in the form ax2 + bx+c=0 where a, b, and c are real numbers, and a is a nonzero number.
Fill in the blank.
A) quadratic
B) quartic
C) linear
D) cubic
Wrong answers WILL be reported. Thanks!
Answer:
A) quadratic
Step-by-step explanation:
ax2 + bx+c=0
Since the highest power of the equation is 2
A) quadratic -2
B) quartic- 4
C) linear- 1
D) cubic-3
Which graph shows the solution to the given system of inequalities? [y<6x+1 y<-3.2x-4
Answer:
VERY NICE RACK U HAVE MAM
Step-by-step explanation:
Answer:
Its a
Step-by-step explanation:
found on another thing and im taking test
Use the procedures developed to find the general solution of the differential equation. (Let x be the independent variable.)
2y''' + 15y'' + 24y' + 11y= 0
Solution :
Given :
2y''' + 15y'' + 24y' + 11y= 0
Let x = independent variable
[tex](a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) = Q(x)[/tex] is a differential equation.
If [tex]Q(x) \neq 0[/tex]
It is non homogeneous then,
The general solution = complementary solution + particular integral
If Q(x) = 0
It is called the homogeneous then the general solution = complementary solution.
2y''' + 15y'' + 24y' + 11y= 0
[tex]$(2D^3+15D^2+24D+11)y=0$[/tex]
Auxiliary equation,
[tex]$2m^3+15m^2+24m +11 = 0$[/tex]
-1 | 2 15 24 11
| 0 -2 - 13 -11
2 13 11 0
∴ [tex]2m^2+13m+11=0[/tex]
The roots are
[tex]$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]
[tex]$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$[/tex]
[tex]$=\frac{-13\pm9}{4}$[/tex]
[tex]$=-5.5, -1$[/tex]
So, [tex]m_1, m_2, m_3 = -1, -1, -5.5[/tex]
Then the general solution is :
[tex]$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$[/tex]
A professor, transferred from Toronto to New York, needs to sell his house in Toronto quickly. Someone has offered to buy his house for $220,000, but the offer expires at the end of the week. The professor does not currently have a better offer but can afford to leave the house on the market for another month. From conversations with his realtor, the professor believes the price he will get by leaving the house on the market for another month is uniformly distributed between $210,000 and $235,000. If he leaves the house on the market for another month, what is the probability that he will get at least $225,000 for the house
If h(x) = −3x − 10, find h(−3). (1 point)
Answer:
h(- 3) = - 1
Step-by-step explanation:
Substitute x = - 3 into h(x) , that is
h(- 3) = - 3(- 3) - 10 = 9 - 10 = - 1
whats 2 plus 2
*just trying to help someone get points* :)
Answer:4 ma boi
Step-by-step explanation:
Answer:
14
Step-by-step explanation:
because I am god at meth and very smart
Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give two exams, one to a class of size 25 and the other to a class of size 64.
Required:
a. Approximate the probability that the average test score in the class of size 25 exceeds 80.
b. Repeat part (a) for the class of size 64.
c. Approximate the probability that the average test score in the larger class exceeds that of the other class by over 2.2 points.
Answer:
a) Hence the probability that the average test score in the class of size 25 exceeds 80.
P ( X > 74) = 0.9838
P ( Z > 2.14) = 0.0162
b) Hence the probability that the average test score for the class of size 64
P ( X > 74) = 0.9838
P ( Z > 2.14) = 0.0003
c) Probability of the difference exceeding 2.2 = 0.9936
P (Z < 2.49) = 0.0064
Step-by-step explanation:
Let's assume a normal distribution.
Now,
a) For a class of 25
[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(25) )\\\\= P (Z < 6 / 2.8)\\\\= P ( Z < 2.14)\\= 0.9838\\[/tex]
[tex]P ( Z > 2.14) = 1- 0.9838\\= 0.0162[/tex]
b)
Similarly:
For the class of 64
[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(64) )\\\\= P (Z < 6 / 1.75)= P ( Z < 3.428)\\= 0.9838\\[/tex]
[tex]P ( Z > 2.14) = 1- 0.9997\\= 0.0003[/tex]
c) Probability of the difference exceeding 2.2
[tex]= P (Z > 2.2/\sqrt{14 * {(1/25) + (1/64}\\[/tex]
[tex]= P (Z > 2.49)\\= 0.9936[/tex]
P (Z < 2.49)
= 1 - 0.9936
= 0.0064
please help me with this on the image
Answer:
ik only 1 of them and i dont even know if this is correct...
a) A
The sum of two binomials is 12x2 − 5x. If one of the binomials is x2 − 2x, the other binomial is:
1. 11x2 − 7x.
2. 12x2 − 3x.
3. 11x2 − 3x.
4. None of these choices are correct.
Answer:
C. 11x² - 3x
Step-by-step explanation:
(12x² - 5x) - (x² - 2x)
12x² - 5x - x² + 2x
12x - x² - 5x + 2x
11x² - 3x
What is the area of the figure
Answer:
72 in. sq.
Step-by-step explanation:
6 x 6 = 36
6 x 6 = 36 / 2 = 18
6 x 6 = 36 / 2 = 18
18 + 18 + 36 = 72.
Hope this helps!
If there is something wrong just let me know.
What is the numerical coefficient of the first term
Answer:
the number before the first variable (first term)
Step-by-step explanation:
this appears to be an incomplete question. The numerical coefficient of a term is the number before the variable.
the constant is the number without a variable.
Find the values of c such that the area of the region bounded by the parabolas y = 4x2 − c2 and y = c2 − 4x2 is 32/3. (Enter your answers as a comma-separated list.)
Answer:
-2,2
Step-by-step explanation:
Let
[tex]y_1=4x^2-c^2[/tex]
[tex]y_2=c^2-4x^2[/tex]
We have to find the value of c such that the are of the region bounded by the parabolas =32/3
[tex]y_1=y_2[/tex]
[tex]4x^2-c^2=c^2-4x^2[/tex]
[tex]4x^2+4x^2=c^2+c^2[/tex]
[tex]8x^2=2c^2[/tex]
[tex]x^2=c^2/4[/tex]
[tex]x=\pm \frac{c}{2}[/tex]
Now, the area bounded by two curves
[tex]A=\int_{a}^{b}(y_2-y_1)dx[/tex]
[tex]A=\int_{-c/2}^{c/2}(c^2-4x^2-4x^2+c^2)dx[/tex]
[tex]\frac{32}{3}=\int_{-c/2}^{c/2}(2c^2-8x^2)dx[/tex]
[tex]\frac{32}{3}=2\int_{-c/2}^{c/2}(c^2-4x^2)dx[/tex]
[tex]\frac{32}{3}=2[c^2x-\frac{4}{3}x^3]^{c/2}_{-c/2}[/tex]
[tex]\frac{32}{3}=2(c^2(c/2+c/2)-4/3(c^3/8+c^3/28))[/tex]
[tex]\frac{32}{3}=2(c^3-\frac{4}{3}(\frac{c^3}{4}))[/tex]
[tex]\frac{32}{3}=2(c^3-\frac{c^3}{3})[/tex]
[tex]\frac{32}{3}=2(\frac{2}{3}c^3)[/tex]
[tex]c^3=\frac{32\times 3}{4\times 3}[/tex]
[tex]c^3=8[/tex]
[tex]c=\sqrt[3]{8}=2[/tex]
When c=2 and when c=-2 then the given parabolas gives the same answer.
Therefore, value of c=-2, 2
Lesson Goals
Understand
relationships.
Define a function in
Identify the domain
Determine whether or
terms of its
and
of
not a
and output.
a relation.
is a function.
W
2K
Words to know
स्ज्द्ज्ज्ग्जेज्दिविफ्ज्र्ज्स्ज्स्ज्ग्झ्क्व्जेफिक
