Answer:
33.33%
Step-by-step explanation:
Add them together 8 plus 8 plus 8 which is 24 there are 8 tp so its 8/24 which is equal to 1/3 so the percent is
What’s the correct answer for this question?
Answer:
The last option is the correct choice 33.5
Step-by-step explanation:
[tex]V=\pi r^2\frac{h}{3} \\=\pi 2^2\frac{8}{3} \\=33.51\\=33.5[/tex]
Answer:
D
Step-by-step explanation:
In the attached file
What is the range of the function y = -x ^2 + 1?
A. y ≤ -1
B. y ≥ -1
C. y ≤ 1
D. y ≥ 1
Answer:
C. y ≤ 1
Step-by-step explanation:
The maximum value of the function is 1. So, the range is all values of y less than or equal to that.
y ≤ 1
Please help!!! Which of the following is equal to the rational expression when x ≠ -2 or 3? x^2+5x+6/x^2-x-6
Answer:
see below
Step-by-step explanation:
These are always simplified by cancelling common factors from numerator and denominator. In order to do that, you have to factor the expressions. The restrictions on x give a clue as to the factors of the denominator.
[tex]\dfrac{x^2+5x+6}{x^2-x-6}=\dfrac{(x+3)(x+2)}{(x-3)(x+2)}=\boxed{\dfrac{x+3}{x-3}}[/tex]
The best possible statement to your question is x+3 / x-3
Please answer this correctly as soon as possible.I have to finish this today. A triangular prism is 19 yards long and has a triangular face with a base of 12 yards and a height of 8 yards. The other two sides of the triangle are each 10 yards. What is the surface area of the triangular prism?
total SA = 764 yd²
A triangular prism is 13 yards long and has a triangular face with a base of 12 yards and a height of 8 yards. The other two sides of the triangle are each 10 yards. What is the surface area of the triangular prism?
See attachment.
if length = 13 yards then total SA = 512 yd²
if length = 19 yards then total SA = 764 yd²
Find the slope and y-intercept of this linear function:
2x + x = 4(y - 1)
Answer:
slope: 3/4y-intercept: 1Step-by-step explanation:
Solve for y to put the equation in slope-intercept form.
3x = 4y -4 . . . . . eliminate parentheses, collect terms
3x +4 = 4y . . . . . add 4
y = 3/4x +1 . . . . . divide by 4
The slope is the x-coefficient: 3/4.
The y-intercept is the constant: 1.
write eight hundred and seven thousand, two hundred and five in figures
Answer:
807,205
Step-by-step explanation:
Take the eight hundred and seven thousand and express that has 807,000. Then, add the two hundred and five at the end to get 807,205
The given statement is written in figures 807,205.
The given statement is eight hundred and seven thousand, two hundred and five in figures.
We need to write the given statement as the number.
What are numbers?A number is a mathematical object used to count, measure, and label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. Numbers can be represented in language with number words.
Now, eight hundred and seven thousand, two hundred and five=807,205.
Therefore, the given statement is written in figures 807,205.
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What’s the correct answer for this question?
Answer:
B:
Step-by-step explanation:
According to theorem, "the angle in a semi-circle is a right angle" So,
<O = 90°
<M = 54
<K = 180-90-54
<OKM = 36°
If g(x) = 2x - 4), find the value of xf g(x) = 20. 12 points)
Answer:
x = 12
Step-by-step explanation:
g(x)= 2x-4
g(x)= 20
Therefore,
2x-4 = 20
Bringing -4 to the other side it becomes positive,so..
2x= 20+4
= 24
x =24/2
= 12
f(x)=x^3+10x^2-25x-250
Answer:
-16x^5
Step-by-step explanation:
f(x)=x^3+10x^2-25x-250
f(x) = x^3-15x+x^2-250
f(x) = x^5-15x-250
f(x) = x^5 -x + 16
f(x) = -x^5+16
f(x) = -16x^5
// have a great day //
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 9 ppm and standard deviation 1.5 ppm. 38 randomly selected large cities are studied. Round all answers to 4 decimal places where possible.
1. What is the distribution of XX? XX ~ N(,)
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
3. What is the probability that one randomly selected city's waterway will have more than 9.6 ppm pollutants?
4. For the 37 cities, find the probability that the average amount of pollutants is more than 9.6 ppm.
5. For part d), is the assumption that the distribution is normal necessary? YesNo
6. Find the IQR for the average of 37 cities.
Q1 = ppm
Q3 = ppm
IQR: ppm
Answer:
Step-by-step explanation:
Hello!
There are two values of n in the text, I'll use the one that appears in all the questions.
The variable of interest is
X: pollutants found in waterways near large cities. (ppm)
This variable has a normal distribution with parameters μ= 9ppm and σ= 1.5ppm
1) X~N(μ;σ²)
X~N(9;2.25)
2) The distribution of the sample mean is X~N(μ;σ²/n)
σ²/n= 2.25/37= 0.06
X~N(9;0.06)
3) P(X>9.6)
To calculate this probability you have to use the standard normal distribution. Using the population parameters, you can calculate the corresponding Z value:
Z= (X-μ)/σ= (9.6-9)/1.5= 0.4
P(Z>0.4)= 1-P(Z≤0.4)= 1 - 0.65542= 0.34458
The probability of selecting a city at random and finding 9.6ppm pollutants.
4) In this item, instead of calculating the probability of one value of the variable you have to calculate the probability of the sample average taking a determined value. Because of this, you have to work using the distribution of the sample mean, instead of the distribution of the variable.
P(X[bar]>9.6)
Z= (X[bar]-μ)/(σ/√n)= (9.6-9)/√0.06= 2.45
P(Z>2.45)= 1 - P(Z≤2.45)= 1 - 0.99286= 0.00714
5) The assumption of a normal distribution is not necessary for item 4. Since the sample size is large enough (greater than 30) you can apply the central limit theorem and approximate the distribution of the sample mean to normal, regarding the distribution of the original variable.
6)
In this case, you have to work starting with the standard normal distribution and then "translate" the Z values into values of the average amount of pollutants.
The first quartile divides the bottom 25% of the distribution from the top 75%, symbolically:
P(Z≤z₁)= 0.25
z₁= -0.674
z₁= (X[bar]-μ)/(σ/√n)
z₁*(√n/σ)=X[bar]-μ
X[bar]=z₁*(√n/σ)+μ
X[bar]=(-0.674)*(√37/1.5)+9= 6.27ppm
The third quartile divides the bottom 75% of the distribution from the top 25%, symbolically:
P(Z≤z₂)= 0.75
z₂= 0.674
z₂= (X[bar]-μ)/(σ/√n)
z₂*(√n/σ)=X[bar]-μ
X[bar]=z₂*(√n/σ)+μ
X[bar]=(0.674)*(√37/1.5)+9= 11.7.3ppm
IQR= Q₃-Q₁= 11.73-6.27= 5.46ppm
I hope this helps!
A survey was sent out to compare the proportion of adults who use their car horns when driving for two age populations (1=younger adults, defined as between 20 and 39 years old and 2 =older adults, defined as over 60 years old). The following data was obtained from those who responded.
Calculate the 90% confidence interval using the standard normal distribution. Note that 1 =0.52. P2= 0.35, and s.e.(P1-P2) =0.0338. Round to the fourth decimal point. Please enter you answer in the following format: (lower value, upper value)
Use the horn Use the horn
Group Yes No Total
1= younger adults 261 240 501
2= older adults 123 229 352
Answer:
The 90% confidence interval for the difference between proportions is (0.115, 0.228).
As the value 0 is not included in the interval, we can conclude that there is significant difference in the proportion of youger adults that use the horn and older adults that use the horn.
Step-by-step explanation:
We want to calculate the bounds of a 90% confidence interval.
For a 90% CI, the critical value for z is z=1.645.
The sample 1 (younger adults) , of size n1=501 has a proportion of p1=0.521.
[tex]p_1=X_1/n_1=261/501=0.5210[/tex]
The sample 2 (older adults), of size n2=352 has a proportion of p2=0.3494.
[tex]p_2=X_2/n_2=123/352=0.3494[/tex]
The difference between proportions is (p1-p2)=0.1715.
[tex]p_d=p_1-p_2=0.5210-0.3494=0.1715[/tex]
The pooled proportion, needed to calculate the standard error, is:
[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{261+123}{501+352}=\dfrac{384}{853}=0.4502[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.4502*0.5498}{501}+\dfrac{0.4502*0.5498}{352}}\\\\\\s_{p1-p2}=\sqrt{0.0005+0.0007}=\sqrt{0.0012}=0.0346[/tex]
Then, the margin of error is:
[tex]MOE=z \cdot s_{p1-p2}=1.645\cdot 0.0346=0.0569[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.1715-0.0569=0.115\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.1715+0.0569=0.228[/tex]
The 90% confidence interval for the difference between proportions is (0.115, 0.228).
Please answer this correctly
Answer:
0-4: Make it 2 units tall
5-9: Make it 5 units tall
10-14: Make it 1 unit tall
15-19: Make it 4 units tall
20-24: Make it 4 units tall
Step-by-step explanation:
0-4: 2, 2 (2 numbers)
5-9: 6, 7, 7, 8, 9 (5 numbers)
10-14: 14 (1 number)
15-19: 15, 16, 16, 18 (4 numbers)
20-24: 21, 23, 23, 24 (4 numbers)
3. Find the mean and range of the following data.
14, 14, 15, 15, 16, 15, 15, 16
A 15; 15
B 12; 15
C 12; 2
D 15; 2
Answer:
D: 15 and 2
Step-by-step explanation:
Mean
To find the mean, or average, add up all the values in the data set,then divide by the number of values in the data set.
1. Add up all the values
Values: 14, 14, 15, 15, 16, 15, 15, 16
Add them :14+14+15+ 15+16+15+15+16=120
120
2. Divide by the number of values
Count how many numbers are in the data set. In this case there are 8. Divide 120 by 8.
120/8=15
The mean is 15
Range
To find the range, subtract the smallest number in the set from the biggest number in the set.
14, 14, 15, 15, 16, 15, 15, 16
Biggest number: 16
Smallest number: 14
biggest-smallest
16-14=2
The range is 2
Therefore, the answer is D: 15 and 2
observation means number.
mean= sum of all observation ÷ number of observation
= 14+ 14+ 15+ 15+ 16+ 15+ 16
7
= 105
7
= 15
range= the highest observation - lowest observation
= highest number- 16
lowest number- 14
= 16-14
= 2
therefore the answer is
OPTION- D 15;2
Use the compound interest formulas A = Pert and A = P(1 + ) to solve. Suppose that you have $11,000 to invest. Which investment yields the greater return over 10 years: 6.25% compounded continuously or 6.3% compounded semiannually? Show your work
Answer:
Continuously
Step-by-step explanation:
Compounded continuously:
A = Pe^(rt)
A = 11,000 e^(0.0625 × 10)
A = 20,550.71
Compounded semiannually (twice per year):
A = P(1 + r)^t
A = 11,000 (1 + 0.063/2)^(2×10)
A = 11,000 (1 + 0.0315)^20
A = 20,453.96
helpppppppppppppppppppppppppppppppppp
Answer:
answer is 2/3
Step-by-step explanation:
probability it is an eclair is 1/15=3/(3+2x+6+x)= 1/(x+3)
so x+3=15 and then x = 12
so the probability it is a humbug is (2*12+6)/(3*12+9) = 30/45 = 2/3
Plz help! U will get full points!
Answer:
2 wild cards
Step-by-step explanation:
Typical would mean most often
2 wild cards shows up 6 times which is most often
Write the point slope form of an equation of the line through the points (-2,6) and (3,-3)
Answer:
A.
Step-by-step explanation:
So first you need to find the slope:
[tex]\frac{-2-6}{3+2} =-\frac{8}{5}[/tex]
Since it's point slope, you have to use a point:
It's either:
[tex](y - 6)=-\frac{8}{5}(x+2)[/tex]
or
[tex](y+2)=-\frac{8}{5}(x-3)[/tex]
Check which answer has those:
A.
The solution is Option A.
The equation of line is y - 6 = ( -8/5 ) ( x + 2 ) where the slope is -8/5
What is an Equation of a line?The equation of a line is expressed as y = mx + b where m is the slope and b is the y-intercept
And y - y₁ = m ( x - x₁ )
y = y-coordinate of second point
y₁ = y-coordinate of point one
m = slope
x = x-coordinate of second point
x₁ = x-coordinate of point one
The slope m = ( y₂ - y₁ ) / ( x₂ - x₁ )
Given data ,
Let the equation of line be represented as A
Now , the value of A is
Let the first point be P ( -2 , 6 )
Let the second point be Q ( 3 , -2 )
The slope of the line between the point is given by m = ( y₂ - y₁ ) / ( x₂ - x₁ )
Substituting the values in the equation , we get
Slope m = ( 6 - ( - 2 ) ) / ( -2 - 3 )
On simplifying the equation , we get
Slope m = ( 8 / -5 ) = -8/5
Now , the equation of line is y - y₁ = m ( x - x₁ )
Substituting the values in the equation , we get
y - 6 = ( -8/5 ) ( x - ( -2 ) )
On simplifying the equation , we get
y - 6 = ( -8/5 ) ( x + 2 )
Hence , the equation of line is y - 6 = ( -8/5 ) ( x + 2 )
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Angle θ is in standard position. If (8, -15) is on the terminal ray of angle θ, find the values of the trigonometric functions.
sin(θ) =
cos(θ) =
tan(θ) =
csc(θ) =
sec(θ) =
✔ 17/8
cot(θ) =
✔ -8/15
i have only gotten the last two right and i need help with the others.
Answer:
cos =1/ sec
=8/17
tan =1/cot
= -15/8
sin = 15/17 or -15/17
cosec = 1/ sin
= 17/15 or -17/15
Answer:
Did the same assignment. lol can see how that went but here's the answers. hope it helps.
Any help would be great
Answer:
88/57
Step-by-step explanation:
Answer: 88:57
Step-by-step explanation:
Length is 88 and width is 57
So the ratio is 88:57
Please answer this correctly
Answer:
10-19 ⇒ 4
40-49 ⇒ 3
Answer:
10-19: 4 numbers
40-49: 3 numbers
Step-by-step explanation:
10-19: 11, 13, 17, 18 (4 numbers)
40-49: 41, 44, 47 (3 numbers)
Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty
Answer:
26.68% probability that exactly three will end up being replaced under warranty
Step-by-step explanation:
For each telephone under warranty, there are only two possible outcomes. Either they need to be replaced, or they do not need to be replaced. Each telephone is independent of other telephones. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
30% must be replaced with new units
This means that [tex]p = 0.3[/tex]
If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty
This is [tex]P(X = 3)[/tex] when [tex]n = 10[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668[/tex]
26.68% probability that exactly three will end up being replaced under warranty
Type your answers into the boxes.
There are 36 students in a class. The pie chart shows the colour of their hair.
Students' Hair Colours
40°
Red
Blonde
Dark
240°
How many students have blonde hair?
How many students have dark hair?
How many students have red hair?
Answer:
(a)24
(b)8
(c)4
Step-by-step explanation:
Number of STudents in the Class = 36
Angle representing Students with Red Hair =40 degrees
Angle representing Students with Blonde Hair =240 degrees
Therefore:
(a)Number of Students with Blonde Hair
[tex]=\dfrac{240^\circ}{360^\circ} \times 36\\\\ =24$ students[/tex]
(b)Number of Students with Dark Hair
Angle representing students with dark hair = 360-(240+40)=80 degrees
Therefore:
Number of Students with Dark Hair
[tex]=\dfrac{80^\circ}{360^\circ} \times 36\\\\ =8$ students[/tex]
(c)Number of Students with Blonde Hair
[tex]=\dfrac{40^\circ}{360^\circ} \times 36\\\\ =4$ students[/tex]
There are 8 students that have blond hair
There are 24 students that have dark hair
There are 4 students that have red hair
Please find attached the pie chart used in answering this question
A pie chart is a graph that displays information in a circle. The circle is divided into slices which represent a numerical proportion. The sum of angles in a pie chart is 360 degrees
To determine the number of students with a type of hair, use this formula :
(degree of the slice that represents the hair type / 360) x total number of students in the class
Degree of the slice that represents blond hair = 360 - (240 + 40) = 80
Students that have blonde hair = [tex]\frac{80}{360}[/tex] x 36 = 8
Students that have dark hair = [tex]\frac{240}{360}[/tex] x 36 = 24
Students that have red hair = [tex]\frac{40}{360}[/tex] x 36 = 4
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What is the area of the obtuse triangle below?
A. 90 sq units
B. 23 sq units
C. 18 sq units
D. 45 sq units
Answer:
A. 90 sq. units
Step-by-step explanation:
5(18) = 90
Some college professors make bound lecture notes available to their classes in an effort to improve teaching effectiveness. A study of business student's opinions of lecture notes. Two groups of students were surveyed - 86 students enrolled in a promotional strategy class that required the purchase of lecture notes, and 35 students enrolled in a sales/retailing elective that did not offer lecture notes. At the end of the semester :"Having a copy of the lecture notes was helpful in understanding the material." Responses were measured on a nine-point semantic difference scale, where 1="strongly disagree" and 9=" strongly agree." A summary of the results is reported in the follow:
Classes Buying Lecture Notes Classes Not Buying Lecture Notes
n1=86 n2=35
X1=8.48 X2=7.80
S21=.94 S22=2.99
a. Describe the two populations involved in the comparison.
b. Do the samples provides sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students? Test using α=.01
c. Construct a 99% confidence interval for (μ1-μ2). Interpret the result.
d. Would a 95% confidence interval for (μ1-μ2) be narrow or wider than the one you found in part c? Why?
Answer:
Step-by-step explanation:
a) The number of students sampled in both populations are large. We can assume that the populations are normally distributed. The populations are also independent.
b) This is a test of 2 independent groups. Let μ1 be the mean responses of students buying lecture notes and μ2 be the mean responses of students not buying lecture notes.
The random variable is μ1 - μ2 = difference in the mean responses of students buying lecture notes and the mean responses of students not buying lecture notes.
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 = μ2 H0 : μ1 - μ2 = 0
The alternative hypothesis is
H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0
This is a two tailed test.
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
From the information given,
x1 = 8.48
x2 = 7.8
s1 = 0.94
s2 = 2.99
n1 = 86
n2 = 35
t = (8.48 - 7.8)/√(0.94²/86 + 2.99²/35)
t = 1.32
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [0.94²/86 + 2.99²/35]²/[(1/86 - 1)(0.94²/86)² + (1/35 - 1)(2.99²/35)²] = 0.0706/0.00192021883
df = 37
We would determine the probability value from the t test calculator. It becomes
p value = 0.195
c) Since alpha, 0.01 < than the p value, 0.195, then we would fail to reject the null hypothesis. Therefore, at 5% significance level, the samples do not provide sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students.
d) The formula for determining the confidence interval for the difference of two population means is expressed as
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
For a 99% confidence interval, the z score is 1.2.58. This is determined from the normal distribution table.
x1 - x2 = 8.48 - 7.8 = 0.68
z√(s1²/n1 + s2²/n2) = 2.58√(0.94²/86 + 2.99²/35) = 1.33
The confidence interval is
0.68 ± 1.33
The upper boundary for the confidence interval is
0.68 + 1.01 = 2.01
The lower boundary for the confidence interval is
0.68 - 1.33 = - 0.65
We are confident that the difference in population means responses between the students buying lecture notes and the students not buying lecture notes is between - 0.65 and 2.01
d) For a 95% confidence interval, the z score is 1.96.
z√(s1²/n1 + s2²/n2) = 1.96√(0.94²/86 + 2.99²/35) = 1.01
The confidence interval is
0.68 ± 1.01
The upper boundary for the confidence interval is
0.68 + 1.01 = 1.69
The lower boundary for the confidence interval is
0.68 - 1.01 = - 0.33
Therefore, a 95% confidence interval for (μ1-μ2) would be narrower. This is seen in the values in both scenarios.
A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to the treatment group and received the new experimental drug. Based on these data, the computed two-sample t statistic is:
Answer:
I think the complete question should be:
A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to the treatment group and received the new experimental drug. The other 23 subjects were assigned to the control group and received a standard well known treatment. After a suitable period of time, the reduction in blood pressure for each subject was recorded.
Treatment group n = 21, x1 mean = 23.48, sd = 8.01
Control group n = 23, x2 = 18.52, sd = 7.15
Based on these data, the computed two-sample t statistic is:
Step-by-step explanation:
Since the variances to be calculated from the sd are unequal we use this formula:
t statistics = (x1 - x2) / [(sd1²/n1) + (sd2²/n2) where n1 = 21, x1 mean = 23.48, sd1 = 8.01, n2 = 23, x2 = 18.52, sd2 = 7.15
Thus, we have
test statistic= (23.48-18.52) / [(8.01²/21) + (7.15²/23)]
Test statistics = 4.96 / (324.36/21)+(51.12/23)]
Test statistics = 4.96/ (15.45+2.43)
t statistic = 4.96 / 17.88
t statistics = 0.2774
I hope that helps, you can use this to solve for tours if the values are not the same
Captain Jessica has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Michael and his merciless band of thieves.
The Captain has probability \dfrac{1}{2}
2
1
start fraction, 1, divided by, 2, end fraction of hitting the pirate ship. The pirate only has one good eye, so he hits the Captain's ship with probability \dfrac{1}{6}
6
1
start fraction, 1, divided by, 6, end fraction.
If both fire their cannons at the same time, what is the probability that both the pirate and the Captain hit each other's ships?
Answer:
[tex]\dfrac{1}{12}[/tex]
Step-by-step explanation:
Probability of the captain hitting the pirate ship [tex]=\dfrac{1}{2}[/tex]
Probability of the pirate hitting the captain's ship [tex]=\dfrac{1}{6}[/tex]
If both fire cannons at the same time, the probability that both the pirate and the captain hit each other's ship
=P(Captain Hits AND Pirate Hits)
=P(Captain Hits) X P(Pirate Hits)
[tex]=\dfrac{1}{2} X \dfrac{1}{6}\\\\=\dfrac{1}{12}[/tex]
uppose the correlation between two variables, math attitude (x) and math achievement (y) was found to be .78. Based on this statistic, we know that the proportion of the variability seen in math achievement that can be predicted by math attitude is:
Answer:
The proportion of the variability seen in math achievement that can be predicted by math attitude is 0.78, the same value as the correlation coefficient.
Step-by-step explanation:
The correlation coefficient r between this two variables is found to be 0.78.
This coefficient can be calculated as:
[tex]r=\dfrac{SSY'}{SSY}[/tex]
where SSY' is the sum of the squares deviation from the mean for the predicted value and SSY is the sum of the squares deviation from the mean for the criterion variable.
Then, the value of the coefficient r is giving the proportion of the variability seen in the criterion value Y that can be explained by the predictor variable X.
Answer:
r=SSY'/SSY
Step-by-step explanation:
Problem 3.3.9 • (a) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner with probability 0.1. Find the PMF of K, the number of tickets you buy up to and including your fifth winning ticket. (b) L is the number of flips of a fair coin up to and including the 33rd occurrence of tails. What is the PMF of L? (c) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner with probability 0.01. Let M equal the number of tickets you buy up to and including your first winning ticket. What is the PMF of M?
Answer:
a) The probability mass function of K = [tex]P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...[/tex]
b)
c)
Step-by-step explanation:
a) Let p be the probability of winning each ticket be = 0.1
Then q which is the probability of failing each ticket = 1 - p = 1 - 0.1 = 0.9
Assume X represents the number of failure preceding the 5th success in x + 5 trials.
The last trial must be success whose probability is p = 0.1 and in the remaining (x + r- 1) ( x+ 4 ) trials we must have have (4) successes whose probability is given by:
[tex]\binom{x+r-1}{r-1}*p^{r-1}*q^{x} = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........[/tex]
Then, the probability distribution of random variable X is
[tex]P(X=x) = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........[/tex]
where;
X represents the negative binomial random variable.
K= X + 5 = number of ticket buy up to and including fifth winning ticket.
Since K =X+5 this signifies that X = K-5
as X takes value 0, 1 ,2,...
K takes value 5, 6 ,...
Therefore:
The probability mass function of K = [tex]P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...[/tex]
b)
Let p represent the probability of getting a tail on a flip of the coin
Thus p = 0.5 since it is a fair coin
where L = number of flips of the coin including 33rd occurrence of tails
Thus; the negative binomial distribution of L can be illustrated as:
[tex]P(X=x) = \binom{x-1}{r-1}(1-p)^{x-r}p^r[/tex]
where
X= L
r = 33 &
p = 0.5
Since we are looking at the 33rd success; L is likely to be : L = 33,34,35...
Thus; the PMF of L = [tex]P(L=l) = \binom{l-1}{33-1}(1-0.5)^{l}(0.5)^{33} \\ \\ \\ \mathbf{P(L=l) = \binom{l-1}{33-1}(0.5)^{l} }[/tex]
c)
Given that:
Let M be the random variable which represents the number of tickets need to be bought to get the first success,
also success probability is 0.01.
Therefore, M ~ Geo(0.01).
Thus, the PMF of M is given by:
[tex]P(M = m) = (1-0.01)^{m-1} * 0.01 , \ \ \ since \ \ \ (m = 1,2,3,4,....)[/tex]
[tex]P(M=m) = (0.99)^{m-1} * 0.01 , m = 1,2,3,4,....[/tex]
In a recent​ year, the total scores for a certain standardized test were normally​ distributed, with a mean of 500 and a standard deviation of 10.4. A) Find the probability that a randomly selected medical student who took the test had a total score that was less than 484. The probability that a randomly selected medical student who took the test had a total score that was less than 484 is:_______.B) Find the probability that a randomly selected study participant's response was between 4 and 6 The probability that a randomly selected study participant's response was between 4 and 6 is:_______.C) Find the probability that a randomly selected study participant's response was more than 8. The probability that a randomly selected study participant's response was more than 8 is:________.
Answer:
A) The probability that a randomly selected medical student who took the test had a total score that was less than 484 = 0.06178
B) The probability that a randomly selected study participant's response was between 504 and 516 = 0.29019
C) The probability that a randomly selected study participant's response was more than 528 = 0.00357
D) Option D is correct.
Only the event in (c) is unusual as its probability is less than 0.05.
Step-by-step explanation:
The b and c parts of the question are not complete.
B) Find the probability that a randomly selected study participant's response was between 504 and 516
C) Find the probability that a randomly selected study participant's response was more than 528.
D) Identify any unusual event amongst the three events in A, B and C. Explain the reasoning.
a) None.
b) Events A and B.
C) Event A
D) Event C
Solution
This is a normal distribution problem with
Mean = μ = 500
Standard deviation = σ = 10.4
A) Probability that a randomly selected medical student who took the test had a total score that was less than 484 = P(x < 484)
We first normalize or standardize 484
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (484 - 500)/10.4 = - 1.54
To determine the required probability
P(x < 484) = P(z < -1.54)
We'll use data from the normal distribution table for these probabilities
P(x < 484) = P(z < -1.54) = 0.06178
B) Probability that a randomly selected study participant's response was between 504 and 516 = P(504 ≤ x ≤ 516)
We normalize or standardize 504 and 516
For 504
z = (x - μ)/σ = (504 - 500)/10.4 = 0.38
For 516
z = (x - μ)/σ = (516 - 500)/10.4 = 1.54
To determine the required probability
P(504 ≤ x ≤ 516) = P(0.38 ≤ z ≤ 1.54)
We'll use data from the normal distribution table for these probabilities
P(504 ≤ x ≤ 516) = P(0.38 ≤ z ≤ 1.54)
= P(z ≤ 1.54) - P(z ≤ 0.38)
= 0.93822 - 0.64803
= 0.29019
C) Probability that a randomly selected study participant's response was more than 528 = P(x > 528)
We first normalize or standardize 528
z = (x - μ)/σ = (528 - 500)/10.4 = 2.69
To determine the required probability
P(x > 528) = P(z > 2.69)
We'll use data from the normal distribution table for these probabilities
PP(x > 528) = P(z > 2.69) = 1 - P(z ≤ 2.69)
= 1 - 0.99643
= 0.00357
D) Only the event in (c) is unusual as its probability is less than 0.05.
Hope this Helps!!!
Which of the following is an arithmetic sequence?
Answer:
D
Step-by-step explanation:
An arithmetic sequence is a series of numbers that increases or decreases by a certain quantity every step. A is not an arithmetic sequence, since it alternates between 2 and -2. B is not an arithmetic sequence, since it does not grow constantly in one direction. C is not an arithmetic sequence, but rather a geometric one. D is an arithmetic sequence, decreasing by 3 with each step. Hope this helps!
