Answer:
38.911≤p≤41.089
Step-by-step explanation:
The formula for calculating confidence interval for a population mean us as shown below;
CI = xbar ± Z×S/√N where;
xbar is the sample mean = 40
Z is the z score at 95% confidence interval = 1.96
S is the standard deviation = 5
N is the sample size = 81
Substituting this parameters in the formula we have;
CI = 40±1.96×5/√81
CI = 40±(1.96×5/9)
CI = 40±(1.96×0.556)
CI = 40±1.089
CI = (40-1.089, 40+1.089)
CI = (38.911, 41.089)
The 95% confidence interval for the population mean is 38.911≤p≤41.089
Answer:
38.9 ≤ U ≤ 41.1
Step-by-step explanation:
Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95
N = 81
The standard error, α(m) = α/√(N) = 5/√81 =5/9
Using table: 0.95 = 0.0379
Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96
Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}
But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1
(40 - 1.1 ≤ U ≤ 40 + 1.1)
∴ the confidence interval = 38.9 ≤ U ≤ 41.1
What is the solution to 5x - 15 = 5(-4x - 3) ? Group of answer choices -12 6 0 -16
Answer:
x = 0Step-by-step explanation:
5x - 15 = 5(-4x - 3)
Multiply the terms in the bracket
5x - 15 = - 20x - 15
Group like terms
Send the constants to the right side of the line and those with variables to the left side
That's
5x + 20x = - 15 + 15
Simplify
25x = 0
Divide both sides by 25
We have the final answer as
x = 0Hope this helps you
Answer:
x=0
Step-by-step explanation:
5x - 15 = 5(-4x - 3)
To find the solution to this equation, we have to get x by itself on one side of the equation.
First, distribute the 5 on the right side. Multiply each term by 5.
5x - 15= (5*-4x) + (5*-3)
5x-15 = -20x + (5*-3)
5x-15= -20x -15
Next, add 20x to both sides of the equation.
(5x+20x) -15 = (-20x+20x) -15
(5+20x) -15 = -15
25x -15=-15
Next, add 15 to both sides of the equation.
25x -15 +15 = -15+15
25x= -15+15
25x=0
Finally, divide both sides of the equation by 25.
25x/25=0/25
x= 0/25
x= 0
The solution to this equation is x=0
20 points!
Please help.
Suppose you have a bag with the following in it: 5 one dollar bills, 4 fives, 3 tens, 5 twenties, and 3 fifties. Assuming the experiment requires drawing one bill from the bag at random, complete the probability distribution for this experiment.
Required:
What is the probability of drawing 9 dollars or less in a single draw?
Answer:
(a) Probility Distribution
Outcome probability
$1 5/15 = 1/3
$5 4/15
$10 3/15 = 1/5
$20 5/15 = 1/3
(b) P($9 or less) = 3/5
Step-by-step explanation:
(a) Probility Distribution
Outcome probability
$1 5/15 = 1/3
$5 4/15
$10 3/15 = 1/5
$20 5/15 = 1/3
Any other denomination
0
(b)
ways to draw $9 or less in a single draw
P($1) = 1/3
P($5) = 4/15
P($9 or less) = P($1) + P($5) = 1/3 + 4/15 = 9/15 = 3/5
How do I solve these equations:
sin(2θ) + sin θ = 0
sin(2θ) = sqrt 3 cos θ
for intervals 0=< θ < 2pi
Answer:
Step-by-step explanation:
Given that
sin(2θ)+sinθ=0
We know that
sin(2θ)=2 sinθ x cosθ
Therefore
2 sinθ x cosθ + sinθ=0
sinθ(2 cosθ+1)=0
sinθ= 0
θ=0
2 cosθ+1=0
cosθ= - 1/2
θ=120°
_______________________________________________________
[tex]sin 2\theta=\sqrt{3cos\theta}[/tex]
By squaring both sides
[tex]sin^2 2\theta={3cos\theta}[/tex]
4 sin²θ x cos²θ=3 cosθ
4 sin²θ x cos²θ - 3 cosθ=0
cos θ = 0
θ= 90°
4 sin²θ=3
θ=60°
You meet with the financial aid office to discuss your costs for attending LSU next semester.Tuition is $113.67 per credit hour, and fees are a flat rate of $660. You have a grant of $350 and a scholarship of $400. If you are taking 15 credit hours what amount will you need go pay for your classes next semester?
Show you work
Answer:
$1615.05Step-by-step explanation:
Scholarship and grants are money given to the candidates to support his financial needs in school. It will serves as the means of revenue for the student.
Revenue generated = Grant + Scholarship amount
Revenue generated = $350 + $400
Revenue generated= $750
Total money needed to be spent in school = Tuition + fees
If tuition is $113.67 per credit hour and I used 15 credit hours, total amount of tuition paid = 15* $113.67 = $1705.05
Total fees = $660
Total money needed to be spent in school = $1705.05 + $660
Total money needed to be spent in school = $2365.05
Amount I will need to pay for classes next semester = Total money that will be spent - (grant+scholarship)
= $2365.05 - $750
= $1615.05
Hence, the amount I will need to pay for classes next semester is $1615.05
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you will lose $1, Suppose that in one evening you play this game n=2500 times in a row.
(a) Estimate the probability that by the end of the evening you will not have lost any money.
(b) Estimate the probability that the number of "even rolls" (roll a 2, 4, or 6) will fall between 1250 and 1300.
(c) Estimate the probability that you will win $100 or more.
Answer:
(a) 50%
(b) 47.5%
(c) 2.5%
Step-by-step explanation:
According to the honest coin principle, if the random variable X denotes the number of heads in n tosses of an honest coin (n ≥ 30), then X has an approximately normal distribution with mean, [tex]\mu=\frac{n}{2}[/tex] and standard deviation, [tex]\sigma=\frac{\sqrt{n}}{2}[/tex].
Here the number of tosses is, n = 2500.
Since n is too large, i.e. n = 2500 > 30, the random variable X follows a normal distribution.
The mean and standard deviation are:
[tex]\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25[/tex]
(a)
To not lose any money the even rolls has to be 1250 or more.
Since, μ = 1250 it implies that the 50th percentile is also 1250.
Thus, the probability that by the end of the evening you will not have lost any money is 50%.
(b)
If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.
Then for number of "even rolls" as 1300,
1300 = 1250 + 2 × 25
= μ + 2σ
Then P (μ + 2σ) for a normally distributed data is 0.975.
⇒ 1300 is at the 97.5th percentile.
Then the area between 1250 and 1300 is:
Area = 97.5% - 50%
= 47.5%
Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.
(c)
To win $100 or more the number of even rolls has to at least 1300.
From part (b) we now 1300 is the 97.5th percentile.
Then the probability that you will win $100 or more is:
P (Win $100 or more) = 100% - 97.5%
= 2.5%.
Thus, the probability that you will win $100 or more is 2.5%.
i need help quick!!!
Answer: A,C, and D
Step-by-step explanation:
Answer:
the answer to this question may be option B, C and D
One number is twice another. The sum of their reciprocals is 3/2 . Find the numbers.
Answer:
The two numbers are 1 and 2.
Step-by-step explanation:
Let the two numbers be a and b.
One number is twice another, so let's let b=2a.
Their reciprocals are 3/2. Thus:
[tex]\frac{1}{a}+\frac{1}{b} =\frac{3}{2}[/tex]
Substitute and solve for a:
[tex]\frac{1}{a}+\frac{1}{2a} =\frac{3}{2}\\[/tex]
Combine the fractions by forming a common denominator by multiplying the left term by 2:
[tex]\frac{2}{2a} +\frac{1}{2a}=\frac{3}{2}[/tex]
Combine and cross-multiply:
[tex]3/2a=3/2\\6a=6\\a=1\\b=2(1)=2[/tex]
Thus, the two numbers are 1 and 2.
Closing prices of two stocks are recorded for 50 trading days. The sample standard deviation of stock X is 4.665 and the sample standard deviation of stock Y is 8.427. The sample covariance is $35.826.
Calculate the sample correlation coefficient. (Round your answer to 4 decimal places.)
Answer:
The sample correlation coefficient r = 0.9113
Step-by-step explanation:
In this question, we are interested in calculating the sample correlation coefficient.
From the question, we are given;
We are given that,
The sample standard deviation of stock X = 4.665
The sample standard deviation of stock Y = 8.427
The sample covariance = 35.826
Mathematically, the Pearson correlation coefficient "r", it is given as;
r = (co-variance of X and Y)/(standard deviation of X * Standard deviation of Y)
Inputing these values, we have;
r = (35.826)/(4.665 * 8.427) = 0.9113
I really need help i will rate you branliest
Work Shown:
A = P*(1+r)^t
A = 21450*(1+(-0.08))^5
A = 21450*(1-0.08)^5
A = 21450*(0.92)^5
A = 21450*0.6590815232
A = 14137.29867264
A = 14,137.30
Notice how I used a negative r value to indicate depreciation rather than growth.
A sports club was formed in the month of May last year. The function below, M(t), models the number of club members for the first 10 months, where t represents the number of months since the club was formed in May. m(t)=t^2-6t+28 What was the minimum number of members during the first 10 months the club was open? A. 19 B. 28 C. 25 D. 30
Answer:
A: 19
Step-by-step explanation:
For this, we can complete the square. We first look at the first 2 terms,
t^2 and -6t.
We know that [tex](t-3)^2[/tex] will include terms.
[tex](t-3)^2 = t^2 - 6t + 9[/tex]
But [tex](t-3)^2[/tex] will also add 9, so we can subtract 9. Putting this into the equation, we get:
[tex]m(t) = (t-3)^2 - 9 +28[/tex]
[tex]m(t) = (t-3)^2 +19[/tex]
Using the trivial inequality, which states that a square of a real number must be positive, we can say that in order to have the minimum number of members, we need to make (t-3) = 0. Luckily, 3 months is in our domain, which means that the minimum amount of members is 19.
The entire graph of the function h is shown below write the domain and range of h using interval notation.
you can only see values of [tex] x[/tex] Ranging from $-3$ to $3$ and they're included, so domain is $[-3,3]$
and $y$ values ranging from $-2$ to $4$ but $-2$ is not included so range is $(-2,4]$
Q1) Two balls are randomly selected without replacement from a box containing three black balls numbered 1, 2, 3 and two white balls numbered 4 and 5. Assuming that all outcomes are equally likely. Find out the probabilities of following events. a) Probability that the color of second ball is white. b) Probability that the color of second ball is black. c) Probability that both balls are black. d) Probability that both balls are white.
[tex]|\Omega|=5\cdot4=20[/tex]
a)
[tex]|A|=3\cdot2+2\cdot1=8\\\\P(A)=\dfrac{8}{20}=\dfrac{2}{5}[/tex]
b)
[tex]|A|=3\cdot2+2\cdot3=12\\\\P(A)=\dfrac{12}{20}=\dfrac{3}{5}[/tex]
c)
[tex]|A|=3\cdot2=6\\\\P(A)=\dfrac{6}{20}=\dfrac{3}{10}[/tex]
d)
[tex]|A|=2\cdot1=2\\\\P(A)=\dfrac{2}{20}=\dfrac{1}{10}[/tex]
A chemical company makes two brands
of antifreeze. The first brand is 30% pure
antifreeze, and the second brand i$ 80% pure
antifreeze. In order to obtain 80 gallons of a
mixture that contains 70o£ pure antifreeze, hov
mabry gallons of each band ot antifneze must
bo used?
Answer:
16 bags for the first(30% pure) and 64 bags of the second(80% pure)
Step-by-step explanation:
If they are mixed in a ratio of x bags to y bags
(0.3x+0.8y)/(x+y) = 0.7
0.3x + 0.8y = 0.7(x+y)
Multiply both sides with 10
3x + 8y = 7(x+y)
4x = y ——(1)
x + y = 80 ——(2)
Solve simultaneously
x + 4x = 80
5x = 80
x = 16 bags
y = 4x = 64 bags
The margin of error ________ (increases or decreases) with an increased sample size and ________ (increases or decreases) with an increase in confidence level.
With such a larger sample size, the margin of error lowers (grows or decreases), whereas, at a higher confidence level, it increases (increases or declines).
The margin of error:
A margin of error increases as the confidence level rises, resulting in a larger interval. A margin of error is reduced as confidence is increased, resulting in a narrower interval.By increasing sample size and confidence level, the margin of error lowers and grows.
Margin of error [tex](E) = Z_c * {\frac{\sigma}{\sqrt{n}}}[/tex]
Here [tex]Z_c[/tex] denotes the confidence level's critical value.Whenever it comes to sample size, n is the number of people who take part in the study.As a result, the margin of error lowers as the sample size grows, and the margin of error increases as the confidence level grows.Find out more about the margin of error here:
brainly.com/question/10501147
Compute the flux of the vector field LaTeX: \vec{F}=F → =< y + z , x + z , x + y > though the unit cubed centered at origin.
Assuming the cube is closed, you can use the divergence theorem:
[tex]\displaystyle\iint_S\vec F\cdot\mathrm dS=\iiint_T\mathrm{div}\vec F\,\mathrm dV[/tex]
where [tex]S[/tex] is the surface of the cube and [tex]T[/tex] is the region bounded by [tex]S[/tex].
We have
[tex]\mathrm{div}\vec F=\dfrac{\partial(y+z)}{\partial x}+\dfrac{\partial(x+z)}{\partial y}+\dfrac{\partial(x+y)}{\partial z}=0[/tex]
so the flux is 0.
What is the 25th term in the following arithmetic sequence? -7, -2, 3, 8, ...
Answer:
108.
Step-by-step explanation:
-7, -2, 3, 8 is an arithmetic sequence with a1 (first term) = -7 and common difference (d) = 5.
The 24th term = a1 + (24 - 1)d
= -7 + 23 * 5
= -7 + 115
= 108.
16.50 and pays 20.00 in cash the change due is
Answer:
Change due is 3.50
Step-by-step explanation:
20.00-16.50 is 3.50
Answer: $3.50
Step-by-step explanation:
You subtract 20 from 16.50.
NEED HELP ASAP
Which point represents the center of the circle shown below?
Answer:
Point O represents the center of the circle
Step-by-step explanation:
HOPE IT HELPS. PLEASE MARK IT AS BRAINLIEST
Please answer this correctly without making mistakes
Answer:
1/8
Step-by-step explanation:
3/8-1/8-1/8=1/8
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is 0.500 in. A bearing is acceptable if its diameter is within 0.004 in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with a mean 0.499 in. and standard deviation 0.002 in. What percentage of bearings will now not be acceptable
Answer:
the percentage of bearings that will not be acceptable = 7.3%
Step-by-step explanation:
Given that:
Mean = 0.499
standard deviation = 0.002
if the true average diameter of the bearings it produces is 0.500 in and bearing is acceptable if its diameter is within 0.004 in.
Then the ball bearing acceptable range = (0.500 - 0.004, 0.500 + 0.004 )
= ( 0.496 , 0.504)
If x represents the diameter of the bearing , then the probability for the z value for the random variable x with a mean and standard deviation can be computed as follows:
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - \mu}{\sigma} \leq \dfrac{X -\mu}{\sigma} \leq \dfrac{0.504 - \mu}{\sigma})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - 0.499}{0.002} \leq \dfrac{X -0.499}{0.002} \leq \dfrac{0.504 - 0.499}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{-0.003}{0.002} \leq Z \leq \dfrac{0.005}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P(Z \leq 2.5) - P(Z \leq -1.5)[/tex]
From the standard normal tables
[tex]P(0.496\leq X \leq 0.504) = 0.9938-0.0668[/tex]
[tex]P(0.496\leq X \leq 0.504) = 0.927[/tex]
By applying the concept of probability of a complement , the percentage of bearings will now not be acceptable
P(not be acceptable) = 1 - P(acceptable)
P(not be acceptable) = 1 - 0.927
P(not be acceptable) = 0.073
Thus, the percentage of bearings that will not be acceptable = 7.3%
Determine whether the statement (p∧(p⟶q))⟶q is a tautology one time by using truth table and the other time without using truth table
It is.........................................................
One way to prove this without a truth table is to use a conditional proof. We assume the portion p ^ (p --> q). If that's true, then so is p and p-->q
Using p and p-->q, the modus ponens rule allows us to derive q. It says that if p is true and p --> q, then q must be true as well.
Since we arrive at q, we have found the conclusion we're after. The assumption (p ^ (p-->q)) leads to q, and therefore the entire statement (p ^ (p-->q)) --> q is true for any combination of p,q.
Policeman A and Policeman B hand out 70 speeding tickets in a month.
Policeman A hands out 4 times as many speeding tickets as Policeman B.
Policeman A handed out ? Speeding tickets.
Answer:
Policeman A = 56 tickets
Step-by-step explanation:
Policemen A + B = 70
If Policeman B hands out x no of tickets...
Then Policeman A hands out 4x no of tickets
meaning...
x + 4x = 70
5x = 70
x = 70/5
x = 14
Therefore Policeman A hands out..
4x = 4 × 14 = 56 tickets
Angela took a general aptitude test and scored in the 90th percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score? % (b) What percentage were above?
Answer:
Step-by-step explanation:
From the given information;
Angela took a general aptitude test and scored in the 90th percentile for aptitude in accounting.
What percentage of the scores were at or below her score?
The percentage of scores that were at or below her score is
Percentage P = 90%
This benchmark is more than average, so we can typically say more of the scores were at or below her score
What percentage were above?
Since The percentage of scores that were at or below her score is
Percentage P = 90%
Then the percentage of the scores that were above will be : (100 -90)%
= 10 %
We can see here that less percentage of score were above Angela's Score.
1) At AJ Welding Company they employ 253 people, 108 employees receive 2 weeks of paid 1) _______ vacation each year. Find the ratio of those who receive 2 weeks of paid vacation to those whose paid vacation is not 2 weeks.
Answer:
108 : 145
Step-by-step explanation:
253 - 108 = 145
Ratio of those who receive 2 weeks of paid vacation is 108
Ratio of those paid vacation is not 2 weeks is 145
108 : 145
Find the surface area of the triangular prism.
Answer:
169 [tex]cm^{2}[/tex]
Step-by-step explanation:
Surface area (SA) = 2B + PH
SA = 2 ([tex]\frac{1}{2}[/tex] x 9 x 6) + (7+7+9) 5
= 2 (27) + (23) 5
= 54 + 115
SA = 169 [tex]cm^{2}[/tex]
Students who score within 14 points of the number 88 will pass a particular test. Write this statement using absolute value notation and use the variable x for the score.
Answer:
|88-x| ≤ 14
Step-by-step explanation:
their score has to be within 14 points of 88.
if their score is above 88, the number will be negative, but the absolute value makes the number positive. if that number is still within 14 of 88, they pass.
if their score is below 88, the number will be negative, and the absolute value keeps the number positive. if that number is still within 14 of 88, they pass.
The distance that Sarah travels varies directly to how long she drives. She travels 440 miles in 8 hours. Write the equation that relates the distance, d, to the time, t. How many miles can Sarah travel in 6 hours?
Answer:
330
Step-by-step explanation:
If d = distance, t = time, and s = speed, then the relationship between the 3 is s * t = d.
Solve for speed by dividing the distance over the time, s = d/t. Then, plug in the speed which in this case is 55 mph and then multiply by the time of 6 hours.
Can I have help with 43 and 44 I need to see how to do them thanks.
Answer:
see explanation
Step-by-step explanation:
(43)
3[tex]x^{5}[/tex] - 75x³ ← factor out 3x³ from each term
= 3x³(x² - 25) ← this is a difference of squares and factors in general as
a² - b² = (a - b)(a + b) , thus
x² - 25 = x² - 5² = (x - 5)(x + 5)
Thus
3[tex]x^{5}[/tex] - 75x³ = 3x³(x - 5)(x + 5)
(44)
81c² + 72c + 16 ← is a perfect square of the form
(ac + b)² = a²c² + 2abc + b²
Compare coefficients of like terms
a² = 81 ⇒ a = [tex]\sqrt{81}[/tex] = 9
b² = 16 ⇒ b = [tex]\sqrt{16}[/tex] = 4
and 2ab = 2 × 9 × 4 = 72
Thus
81c² + 72c + 16 = (9c + 4)²
1. 3x^5 -75x³
=3x³(x²-25)
=3x³(x²-5²)
=3x³(x-5)(x+5)
2. 81c²+72c+16
=81c²+36c+36c+16
=9c(9c+4)+4(9c+4)
=(9c+4)(9c+4)
=(9c+4)²
A test is being conducted to test the difference between two population means using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the:
Answer:
Student t-distribution.
Step-by-step explanation:
In this scenario, a test is being conducted to test the difference between two population "means" using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the student t-distribution.
In Statistics and probability, a student t-distribution can be defined as the probability distribution which can be used to estimate population parameters when the population variance is not known (unknown) and the sample population is relatively small. The student t-distribution is a statistical distribution which was published in 1908 by William Sealy Gosset.
A student t-distribution has a similar curve with the normal distribution curve, except that it is fatter and a little bit shorter.
