Answer:
$[58543.42; 73456.58]
Step-by-step explanation:
Hello!
For the variable
X: salary of a college graduate that took a statistics course
Out of n= 43 students, the calculated mean is [tex]\frac{}{X}[/tex]= $66000
The population standard deviation is δ= $18908
There is no information about the variable distribution, but since the sample size is big enough (n≥30), you can apply the CLT and approximate the distribution of the sample mean to normal [tex]\frac{}{X}[/tex]≈N(μ;σ²/n)
Then you can apply the approximation of the standard normal distribution to calculate the 99% CI
[tex]\frac{}{X}[/tex] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{Singma}{\sqrt{n} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.995}= 2.586[/tex]
[tex]\frac{Singma}{\sqrt{n} }= \frac{18908}{\sqrt{43} }= 2883.44[/tex]
[66000±2.586*2883.44]
$[58543.42; 73456.58]
With a 99% confidence level you'd expect that the interval $[58543.42; 73456.58] will include the average salary of college graduates that took a course of statistics.
I hope this helps!
Please help. I’ll mark you as brainliest if correct!
Answer:
8lb of the cheaper Candy
17.5lb of the expensive candy
Step-by-step explanation:
Let the cheaper candy be x
let the costly candy be y
X+y = 25.5....equation one
2.2x +7.3y = 25.5(5.7)
2.2x +7.3y = 145.35.....equation two
X+y = 25.5
2.2x +7.3y = 145.35
Solving simultaneously
X= 25.5-y
Substituting value of X into equation two
2.2(25.5-y) + 7.3y = 145.35
56.1 -2.2y +7.3y = 145.35
5.1y = 145.35-56.1
5.1y = 89.25
Y= 89.25/5.1
Y= 17.5
X= 25.5-y
X= 25.5-17.5
X= 8
Crime and Punishment: In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
(A) If one of the study subjects is randomly selected, find the probability of getting someone who was not sent to prison.
(B) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, find the probability that this person was not sent to prison.
Answer:
(a) The probability of getting someone who was not sent to prison is 0.55.
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.
Step-by-step explanation:
We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
Let the probability that subjects studied were sent to prison = P(A) = 0.45
Let G = event that subject chose to plead guilty
So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40
and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55
(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison
P(A') = 1 - P(A)
= 1 - 0.45 = 0.55
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)
We will use Bayes' Theorem here to calculate the above probability;
P(A'/G) = [tex]\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}[/tex]
= [tex]\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}[/tex]
= [tex]\frac{0.3025}{0.4825}[/tex]
= 0.63
A drawer contains 3 white shirts, 2 blue shirts, and 5 gray shirts. A shirt is randomly
selected from the drawer and set aside. Then another shirt is randomly selected from the
drawer.
What is the probability that the first shirt is white and the second shirt is gray?
Answer:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]
Step-by-step explanation:
Given that
3 white, 2 blue and 5 gray shirts are there.
To find:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?
Solution:
Here, total number of shirts = 3+2+5 = 10
First of all, let us learn about the formula of an event E:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
[tex]P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}[/tex]
[tex]P(First\ White) = \dfrac{3}{10}[/tex]
Now, this shirt is set aside.
So, total number of shirts left are 9 now.
[tex]P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }[/tex]
So, the answer is:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]
ASAP PLEASE HELP!!!!!! Find the y-intercept of the rational function. A rational function is graphed in the first quadrant, and in the second, third and fourth quadrants are other pieces of the graph. The graph crosses the x axis at negative 10 and crosses the y axis at negative 2.
Answer:
(0,-2)
Step-by-step explanation:
The y-intercept is simply when the function touches or crosses the y-axis.
We're told that the graph crosses the y-axis at -2. In other words, the y-intercept is at -2.
The ordered pair would be (0,-2)
Find the area of the figure. Round to the nearest tenth if necessary. 386.3m^2 194.3m^2 193.1m^2 201.9m^2
Add the top and bottom numbers together, divide that by 2 then multiply by the height.
15.3 + 19.5 = 34.8
34.8/2 = 17.4
17.4 x 11.1 = 193.14
Answer is 193.1 m^2
plzzzzz helpp j + 9 - 3 < 8
Answer:
j < 2
Step-by-step explanation:
Simplify both sides of the inequality and isolating the variable would get you the answer
what is the answer to 100×338
Answer:
33800
Step-by-step explanation:
100 x 338 = 33800
Answer:
33800
Step-by-step explanation:
338x10=3380 then 3380x10=33800
-------------------------------------------------------
Good luck with your assignment...
15 points + brainliest if you can figure this out!
Answer:
(H1, T1)
Step-by-step explanation:
Since we know that the only number option is 1, we can cancel out the first 3 options. and obviously, there are only heads, and tails. So, using only the # 1 and heads and tails, we can conclude that the answer is (H1, T1).
Answer:
D. (H1, T1)
Step-by-step explanation:
Since all outcomes require card #1 is chosen, so any answer with 2 or 3 can be rejected, therefore the answer is
D. (H1, T1)
Mia agreed to borrow a 3 year loan with 4 percent interest to buy a motorcycle if Mia will pay a total of $444 in interest how much money did she borrow how much interest would Mia pay if the simple interest rate was 5 percent
Answer:
a) $3700
b) $555
Step-by-step explanation:
The length of the loan is 3 years.
The interest after 3 years is $444.
The rate of the Simple Interest is 4%.
Simple Interest is given as:
I = (P * R * T) / 100
where P = principal (amount borrowed)
R = rate
T = length of years
Therefore:
[tex]444 = (P * 3 * 4) / 100\\\\444 = 12P / 100\\\\12P = 444 * 100\\\\12P = 44400\\\\P = 44400 / 12\\[/tex]
P = $3700
She borrowed $3700
b) If the simple interest was 5%, then:
I = (3700 * 5 * 3) / 100 = $555
The interest would be $555.
In which table does y vary inversely with x? A. x y 1 3 2 9 3 27 B. x y 1 -5 2 5 3 15 C. x y 1 18 2 9 3 6 D. x y 1 4 2 8 3 12
Answer:
In Table C, y vary inversely with x.
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Step-by-step explanation:
We are given four tables and asked to find out in which table y vary inversely with x.
We know that an inverse relation has a form given by
y = k/x
xy = k
where k must be a constant
Table A:
x | y
1 | 3
2 | 9
3 | 27
1×3 = 3
2×9 = 18
3×27 = 81
3 ≠ 18 ≠ 81
Hence y does not vary inversely with x.
Table B:
x | y
1 | -5
2 | 5
3 | 15
1×-5 = -5
2×5 = 10
3×15 = 45
-5 ≠ 10 ≠ 45
Hence y does not vary inversely with x.
Table C:
x | y
1 | 18
2 | 9
3 | 6
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Hence y vary inversely with x.
Table D:
x | y
1 | 4
2 | 8
3 | 12
1×4 = 4
2×8 = 16
3×12 = 36
4 ≠ 16 ≠ 36
Hence y does not vary inversely with x.
Find the surface area of the attached figure and round your answer to the nearest tenth, if necessary.
Answer:
[tex] S.A = 246.6 in^2 [/tex]
Step-by-step explanation:
The figure given above is a square pyramid, having a square base and 4 triangular faces on the sides that are of the same dimensions.
Surface area of the square pyramid is given as: [tex] B.A + \frac{1}{2}*P*L [/tex]
Where,
B.A = Base Area of the pyramid = 9*9 = 81 in²
P = perimeter of the base = 4(9) = 36 in
L = slant height of pyramid = 9.2 in
Plug in the values into the given formula to find the surface area
[tex] S.A = 81 + \frac{1}{2}*36*9.2 [/tex]
[tex] = 81 + 18*9.2 [/tex]
[tex] = 81 + 165.6 [/tex]
[tex] S.A = 246.6 in^2 [/tex]
Scores made on a certain aptitude test by nursing students are approximately normally distributed with a mean of 500 and a variance of 10,000. If a person is about to take the test what is the probability that he or she will make a score of 650 or more?
Answer:
0.0668 or 6.68%
Step-by-step explanation:
Variance (V) = 10,000
Standard deviation (σ) = √V= 100
Mean score (μ) = 500
The z-score for any test score X is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X = 650:
[tex]z=\frac{650-500}{100}\\z=1.5[/tex]
A z-score of 1.5 is equivalent to the 93.32nd percentile of a normal distribution. Therefore, the probability that he or she will make a score of 650 or more is:
[tex]P(X\geq 650)=1-P(X\leq 650)\\P(X\geq 650)=1-0.9332\\P(X\geq 650)=0.0668=6.68\%[/tex]
The probability is 0.0668 or 6.68%
The probability that he or she will make a score of 650 or more is 0.0668.
Let X = Scores made on a certain aptitude test by nursing students
X follows normal distribution with mean = 500 and variance of 10,000.
So, standard deviation = [tex]\sqrt{10000}=100[/tex].
z score of 650 is = [tex]\frac{\left(650-500\right)}{100}=1.5[/tex].
The probability that he or she will make a score of 650 or more is:
[tex]P(X\geq 650)\\=P(z\geq 1.5)\\=1-P(z<1.5)\\=1-0.9332\\=0.0668[/tex]
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You are selling your product at a three-day event. Each day, there is a 60% chance that you will make money. What is the probability that you will make money on the first two days and lose money on the third day
Answer:
The required probability = 0.144
Step-by-step explanation:
Since the probability of making money is 60%, then the probability of losing money will be 100-60% = 40%
Now the probability we want to calculate is the probability of making money in the first two days and losing money on the third day.
That would be;
P(making money) * P(making money) * P(losing money)
Kindly recollect;
P(making money) = 60% = 60/100 = 0.6
P(losing money) = 40% = 40/100 = 0.4
The probability we want to calculate is thus;
0.6 * 0.6 * 0.4 = 0.144
An experiment involves 17 participants. From these, a group of 3 participants is to be tested under a special condition. How many groups of 3 participants can
be chosen, assuming that the order in which the participants are chosen is irrelevant?
Answer: 680
Step-by-step explanation:
When order doesn't matter,then the number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
Given: Total participants = 17
From these, a group of 3 participants is to be tested under a special condition.
Number of groups of 3 participants chosen = [tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\[/tex]
[tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\\\\=\dfrac{17\times16\times15\times14!}{3\times2\times14!}\\\\=680[/tex]
Hence, there are 680 groups of 3 participants can be chosen,.
Refer to the following wage breakdown for a garment factory:
Hourly Wages Number of employees
$4 up to $7 18
7 up to 10 36
10 up to 13 20
13 up to 16 6
What is the class interval for the preceding table of wages?
A. $4
B. $2
C. $5
D. $3
Answer:
The class interval is $3Step-by-step explanation:
The class interval is simply the difference between the lower or upper class boundary or limit of a class and the lower or upper class boundary or limit of the next class.
In this case for the class
$4 up to $7 18 and
$7 up to $10 36
The lower class boundary of the first class is $4 and the lower class boundary of the second class is $7
Hence the class interval = $7-$4= $3the mean monthly income of trainees at a local mill is 1100 with a standard deviation of 150. find rthe probability that a trainee earns less than 900 a month g
Answer:
The probability is [tex]P(X < 900 ) = 0.0918[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 1100[/tex]
The standard deviation is [tex]\sigma = 150[/tex]
The random number value is x =900
The probability that a trainee earn less than 900 a month is mathematically represented as
[tex]P(X < x) = P(\frac{X -\= x}{\sigma} < \frac{x -\= x}{\sigma} )[/tex]
Generally the z-value for the normal distribution is mathematically represented as
[tex]z = \frac{x -\mu }{\sigma }[/tex]
So From above we have
[tex]P(X < 900 ) = P(Z < \frac{900 -1100}{150} )[/tex]
[tex]P(X < 900 ) = P( Z <-1.33)[/tex]
Now from the z-table
[tex]P(X < 900 ) = 0.0918[/tex]
Find the area of the surface given by z = f(x, y) that lies above the region R. f(x, y) = 64 + x2 − y2 R = {(x, y): x2 + y2 ≤ 64}
The area of the surface above the region R is 4096π square units.
Given that:
The function: [tex]f(x, y) = 64 + x^2 - y^2[/tex]
The region R is the disk with a radius of 8 units [tex]x^2 + y^2 \le 64[/tex].
To find the area of the surface given by z = f(x, y) that lies above the region R, to calculate the double integral over the region R of the function f(x, y) with respect to dA.
The integral for the area is given by:
[tex]Area = \int\int_R f(x, y) dA[/tex]
To evaluate this integral, we need to set up the limits of integration for x and y over the region R, which is the disk cantered at the origin with a radius of 8 units.
Using polar coordinates, we can parameterize the region R as follows:
x = rcos(θ)
y = rsin(θ)
where r goes from 0 to 8, and θ goes from 0 to 2π.
Now, rewrite the integral in polar coordinates:
[tex]Area =\int\int_R f(x, y) dA\\Area = \int_0 ^{2\pi} \int_0^8(64 + r^2cos^2(\theta) - r^2sin^2(\theta)) \times r dr d \theta[/tex]
Now, we can integrate with respect to r first and then with respect to θ:
[tex]Area = \int_0^{2\pi} \int_0^8] (64r + r^3cos^2(\theta) - r^3sin^2(\theta)) dr d \theta[/tex]
Integrate with respect to r:
[tex]Area = \int_0^{2\pi}[(32r^2 + (1/4)r^4cos^2(\theta) - (1/4)r^4sin^2(\theta))]_0^8 d \theta\\Area = \int_0^{2\pi} (2048 + 256cos^2(\theta) - 256sin^2(\theta)) d \theta[/tex]
Now, we can integrate with respect to θ:
[tex]Area = [2048\theta + 128(sin(2\theta) + \theta)]_0 ^{2\pi}[/tex]
Area = 2048(2π) + 128(sin(4π) + 2π) - (2048(0) + 128(sin(0) + 0))
Area = 4096π + 128(0) - 0
Area = 4096π square units
So, the area of the surface above the region R is 4096π square units.
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A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90. its price-earnings ratio equals:
Answer: Price-earnings ratio= 22.0
Step-by-step explanation:
Given: A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90
To find: price-earnings ratio
Required formula: [tex]\text{price-earnings ratio }=\dfrac{\text{ Market Price per Share}}{\text{Earnings Per Share}}[/tex]
Then, Price-earnings ratio = [tex]\dfrac{\$38.50}{\$1.75}[/tex]
⇒Price-earnings ratio = [tex]\dfrac{22}{1}[/tex]
Hence, the price-earnings ratio= 22.0
TRIANGLE ABC IS DILATED BY A SCALE FACTOR OF 0.5 WITH THE ORIGIN AS THE CENTER OF DILATION, RESULTING IN THE IMAGE TRIANGLE A'B'C. IF A=(2,2). IF A (2,2), B= (4,3) AND C=(6,3), WHAT IS THE LENGTH OF LINE B'C'?
Answer: The length of the line B'C" is 1 unit.
Step-by-step explanation:
Given: Triangle ABC is dilated by a scale factor of 0.5 with the origin as the center of dilation , resulting in the image Triangle A'B'C'.
If A (2,2), B= (4,3) and C=(6,3).
Distance between (a,b) and (c,d): [tex]D=\sqrt{(d-b)^2+(c-b)^2}[/tex]
Then, BC [tex]=\sqrt{(3-3)^2+(6-4)^2}[/tex]
[tex]\\\\=\sqrt{0+2^2}\\\\=\sqrt{4}\\\\=2\text{ units}[/tex]
Length of image = scale factor x length in original figure
B'C' = 0.5 × BC
= 0.5 × 2
= 1 unit
Hence, the length of the line B'C" is 1 unit.
A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else. A random sample of 600 18-29 year-olds is obtained today. What is the probability that no more than 70% would prefer to start their own business?
Answer:
The probability that no more than 70% would prefer to start their own business is 0.1423.
Step-by-step explanation:
We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.
Let [tex]\hat p[/tex] = sample proportion of people who prefer to start their own business
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, p = population proportion who would prefer to start their own business = 72%
n = sample of 18-29 year-olds = 600
Now, the probability that no more than 70% would prefer to start their own business is given by = P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%)
P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.07) = 1 - P(Z < 1.07)
= 1 - 0.8577 = 0.1423
The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.
please I need help with this question!
The weight of adult males in Boston are normally distributed with mean 69 kilograms and variance 25 kilograms.
I. what percentage of adult male in Boston weigh more than 72 kilograms?
ii. what must an adult male weigh in order to be among the heaviest 10% of the population?
Thank you in advance!
Answer:
lmkjhvjgcfnhjkhbmgnc gfghh
Step-by-step explanation:
A car travels 133 mi averaging a certain speed. If the car had gone 30 mph faster, the trip would have taken 1 hr less. Find the car's average speed.
Answer:
49.923 mph
Step-by-step explanation:
we know that the car traveled 133 miles in h hours at an average speed of x mph.
That is, xh = 133.
We can also write this in terms of hours driven: h = 133/x.
If x was 30 mph faster, then h would be one hour less.
That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).
We can rewrite the latter equation as h = 133/(x + 30) + 1
We can then make a system of equations using the formulas in terms of h to find x:
h = 133/x = 133/(x + 30) + 1
133/x = 133/(x + 30) + (x + 30)/(x + 30)
133/x = (133 + x + 30)/(x + 30)
133 = x*(133 + x + 30)/(x + 30)
133*(x + 30) = x*(133 + x + 30)
133x + 3990 = 133x + x^2 + 30x
3990 = x^2 + 30x
x^2 + 30x - 3990 = 0
Using the quadratic formula:
x = [-b ± √(b^2 - 4ac)]/2a
= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)
= [-30 ± √(900 + 15,960)]/2
= [-30 ± √(16,860)]/2
= [-30 ± 129.846]/2
= 99.846/2 ----------- x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)
= 49.923
Check if the answer is correct:
h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.
If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.
Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster
Use all the information below to find the missing x-value for the point that is on this line. m = - 1 / 3 b = 7 ( x, 4 )
Answer:
[tex]\boxed{x = 9}[/tex]
Step-by-step explanation:
m = -1/3
b = 7
And y = 4 (Given)
Putting all of the givens in [tex]y = mx+b[/tex] to solve for x
=> 4 = (-1/3) x + 7
Subtracting 7 to both sides
=> 4-7 = (-1/3) x
=> -3 = (-1/3) x
Multiplying both sides by -3
=> -3 * -3 = x
=> 9 = x
OR
=> x = 9
Answer:
x = 9
Step-by-step explanation:
m = -1/3
b = 7
Using slope-intercept form:
y = mx + b
m is slope, b is y-intercept.
y = -1/3x + 7
Solve for x:
Plug y as 4
4 = 1/3x + 7
Subtract 7 on both sides.
-3 = -1/3x
Multiply both sides by -3.
9 = x
given sin theta=3/5 and 180°<theta<270°, find the following: a. cos(2theta) b. sin(2theta) c. tan(2theta)
I hope this will help uh.....
a 12- inch ruler is duvided into 3 parts. the large part is 3 times longer than the small. the meddium part is times longer than then small, the medium part is 2 times long as the smallest .how long is the smallest part?
Answer:
2 inches
Step-by-step explanation:
x= smallest
3x=largest
2x=medium
x+3x+2x=12
6x=12
x=2
so smallest is 2
largest is 6 (3x)
medium is 4 (2x)
2+6+4=12
The automatic opening device of a military cargo parachute has been designed to open when the parachute is 155 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 155 and standard deviation 30 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes
Answer:
the probability that one parachute of the five parachute is damaged is 0.156
Step-by-step explanation:
From the given information;
Let consider X to be the altitude above the ground that a parachute opens
Then; we can posit that the probability that the parachute is damaged is:
P(X ≤ 100 )
Given that the population mean μ = 155
the standard deviation σ = 30
Then;
[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]
[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]
[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]
From standard normal tables
[tex]P(X \leq 100 ) = 0.0334[/tex]
Hence; the probability of the given parachute damaged is 0.0334
Let consider Q to be the dropped parachute
Given that the number of parachute be n= 5
The probability that the parachute opens in each trail be p = 0.0334
Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334
The probability mass function is:
Q [tex]\sim[/tex] B(5, 0.0334)
Similarly; the event that one parachute is damaged is :
Q ≥ 1
P( Q ≥ 1 ) = 1 - P( Q < 1 )
P( Q ≥ 1 ) = 1 - P( Y = 0 )
P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )
P( Q ≥ 1 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = 1 - 0.8437891838
P( Q ≥ 1 ) = 0.1562108162
P( Q ≥ 1 ) [tex]\approx[/tex] 0.156
Therefore; the probability that one parachute of the five parachute is damaged is 0.156
If w'(t) is the rate of growth of a child in pounds per year, what does 7 w'(t)dt 4 represent? The change in the child's weight (in pounds) between the ages of 4 and 7. The change in the child's age (in years) between the ages of 4 and 7. The child's weight at age 7. The child's weight at age 4. The child's initial weight at birth.
Complete Question
If w'(t) is the rate of growth of a child in pounds per year, what does
[tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex] represent?
a) The change in the child's weight (in pounds) between the ages of 4 and 7.
b) The change in the child's age (in years) between the ages of 4 and 7.
c) The child's weight at age 7.
d) The child's weight at age 4. The child's initial weight at birth.
Answer:
The correct option is option a
Step-by-step explanation:
From the question we are told that
[tex]w'(t)[/tex] represents the rate of growth of a child in [tex]\frac{pounds}{year}[/tex]
So [tex]{w'(t)} \, dt[/tex] will be in [tex]pounds[/tex]
Which then mean that this [tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex] the change in the weight of the child between the ages of [tex]4 \to 7[/tex] years
Historically, the proportion of students entering a university who finished in 4 years or less was 63%. To test whether this proportion has decreased, 114 students were examined and 51% had finished in 4 years or less. To determine whether the proportion of students who finish in 4 year or less has statistically significantly decreased (at the 5% level of signficance), what is the critical value
Answer:
z(c) = - 1,64
We reject the null hypothesis
Step-by-step explanation:
We need to solve a proportion test ( one tail-test ) left test
Normal distribution
p₀ = 63 %
proportion size p = 51 %
sample size n = 114
At 5% level of significance α = 0,05, and with this value we find in z- table z score of z(c) = 1,64 ( critical value )
Test of proportion:
H₀ Null Hypothesis p = p₀
Hₐ Alternate Hypothesis p < p₀
We now compute z(s) as:
z(s) = ( p - p₀ ) / √ p₀q₀/n
z(s) =( 0,51 - 0,63) / √0,63*0,37/114
z(s) = - 0,12 / 0,045
z(s) = - 2,66
We compare z(s) and z(c)
z(s) < z(c) - 2,66 < -1,64
Therefore as z(s) < z(c) z(s) is in the rejection zone we reject the null hypothesis
1000 randomly selected Americans were asked if they believed the minimum wage should be raised. 600 said yes. Construct a 95% confidence interval for the proportion of Americans who believe that the minimum wage should be raised.
a. Write down the formula you intend to use with variable notation).
b. Write down the above formula with numeric values replacing the symbols.
c. Write down the confidence interval in interval notation.
Answer:
a. p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]
b.0.6 ± 1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]
c. { -1.96 ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ 1.96} = 0.95
Step-by-step explanation:
Here the total number of trials is n= 1000
The number of successes is p` = 600/1000 = 0.6. The q` is 1 - p`= 1- 0.6 = 0.4
The degree of confidence is 95 % therefore z₀.₀₂₅ = 1.96 ( α/2 = 0.025)
a. The formula used will be
p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ( z with the base alpha by 2 (α/2 = 0.025))
b. Putting the values
0.6 ± 1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]
c. Confidence Interval in Interval Notation.
{ -1.96 ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ 1.96} = 0.95
{ -z( base alpha by 2) ≤ p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex] ≥ z( base alpha by 2) } = 1- α
given g(x)=3/x^2+2x find g^-1(x)
Answer:
A
Step-by-step explanation:
[tex]g(x) = \frac{3}{{x}^{2} + 2x} \\ {x}^{2} + 2x - \frac{3}{g(x)} = 0 \\ x = \frac{1}{2} \Big( - 2 + \sqrt{12 + \frac{12}{g(x)} }\Big) \\ x = - 1 + \sqrt{1 \pm \frac{3}{g(x)} } [/tex]
Now replace $x$ by $g^{-1}(x)$ and $g(x)$ by $x$ and you have your answer.