Answer:
The population of interest is all the students at the University, to find their parking times.
Step-by-step explanation:
Sampling
This is a common statistics practice, when we want to study something from a population, we find a sample of this population.
For example:
I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected New York state residents wheter they are Buffalo Bills fans, and expand this to the entire population of New York State residents.
Here, the population of interest is all New York state residents.
An administrator inconspicuously followed 250 students and carefully recorded their parking times.
Sample of 250 students at the University.
So the population of interest is all the students at the University, to find their parking times.
According to the National Association of Theater Owners, the average price for a movie in the United States in 2012 was $7.96. Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.
Required:
a. Calculate the standard error of the mean.
b. What is the probability that the sample mean will be less than $7.75?
c. What is the probability that the sample mean will be less than $8.10?
d. What is the probability that the sample mean will be more than $8.20?
Answer:
(a) The standard error of the mean is 0.091.
(b) The probability that the sample mean will be less than $7.75 is 0.0107.
(c) The probability that the sample mean will be less than $8.10 is 0.9369.
(d) The probability that the sample mean will be more than $8.20 is 0.0043.
Step-by-step explanation:
We are given that the average price for a movie in the United States in 2012 was $7.96.
Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.
Let [tex]\bar X[/tex] = sample mean price for a movie in the United States
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean price for a movie = $7.96
[tex]\sigma[/tex] = population standard deviation = $0.50
n = sample of theaters = 30
(a) The standard error of the mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{0.50}{\sqrt{30} }[/tex]
= 0.091
(b) The probability that the sample mean will be less than $7.75 is given by = P([tex]\bar X[/tex] < $7.75)
P([tex]\bar X[/tex] < $7.75) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{7.75-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < -2.30) = 1 - P(Z [tex]\leq[/tex] 2.30)
= 1 - 0.9893 = 0.0107
The above probability is calculated by looking at the value of x = 2.30 in the z table which has an area of 0.9893.
(c) The probability that the sample mean will be less than $8.10 is given by = P([tex]\bar X[/tex] < $8.10)
P([tex]\bar X[/tex] < $8.10) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{8.10-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < 1.53) = 0.9369
The above probability is calculated by looking at the value of x = 1.53 in the z table which has an area of 0.9369.
(d) The probability that the sample mean will be more than $8.20 is given by = P([tex]\bar X[/tex] > $8.20)
P([tex]\bar X[/tex] > $8.20) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{8.20-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z > 2.63) = 1 - P(Z [tex]\leq[/tex] 2.63)
= 1 - 0.9957 = 0.0043
The above probability is calculated by looking at the value of x = 2.63 in the z table which has an area of 0.9957.
Margo borrows $1700, agreeing to pay it back with 4% annual interest after 6 months. How much interest
will she pay?
Round your answer to the nearest cent, if necessary.
Answer:
$1733.67
Step-by-step explanation:
Simple interest rate formula: A = P(1 + r)^t
Simply plug in your known variables
A = 1700(1 + 0.04)^0.5
A = 1733.67
Remember that t is time in years.
Factor completely 6x to the second power - 36xy + 12x
Answer:
6x(x - 6y +2)
Step-by-step explanation:
Step 1: Write out expression
6x² - 36xy + 12x
Step 2: Factor out x
x(6x - 36y + 12)
Step 3: Factor out 6
6x(x - 6y + 2)
That is the most we can do. We can only take GCF to factor. Since we don't have an y² term we do not have binomial factors.
An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students who are in both Spanish and French, 4 who are in both Spanish and German, and 6 who are in both French and German. In addition, there are 2 students taking all 3 classes. If two students are randomly chosen, what is the probability that at exactly one of them does exactly two language classes.
Answer:
The probability that at exactly one of them does exactly two language classes is 0.32.
Step-by-step explanation:
We can model this variable as a binomial random variable with sample size n=2.
The probability of success, meaning the probability that a student is in exactly two language classes can be calculated as the division between the number of students that are taking exactly two classes and the total number of students.
The number of students that are taking exactly two classes is equal to the sum of the number of students that are taking two classes, minus the number of students that are taking the three classes:
[tex]N_2=F\&S+S\&G+F\&G-F\&S\&G=12+4+6-2=20[/tex]
Then, the probabilty of success p is:
[tex]p=20/100=0.2[/tex]
The probability that k students are in exactly two classes can be calcualted as:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{2}{k} 0.2^{k} 0.8^{2-k}\\\\\\[/tex]
Then, the probability that at exactly one of them does exactly two language classes is:
[tex]P(x=1) = \dbinom{2}{1} p^{1}(1-p)^{1}=2*0.2*0.8=0.32\\\\\\[/tex]
100 pts You have a bag of 15 marbles: 5 blue, 3 red, 4 green, and 3 yellow. You draw 3 marbles without replacement. Which action, performed before the draws, increases the probability of drawing 3 green marbles in a row?
Answer:
see below
Step-by-step explanation:
You can remove one or more of the other color marbles to increase the probability of drawing a green marble
or
You can add one or more green marbles to have more green marbles in the bag
Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do homework regularly. Moreover, 95% of the students who do their homework regularly generally pass the course. She also knows that 85% of her students pass the course.
a. What is the probability that a student will do homework regularly and also pass the course?
b. What is the probability that a student will neither do homework regularly nor will pass the course?
c. Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.
d. Are the events "pass the course" and "do homework regularly" independent? Explain.
Answer:
a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57
b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12
c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.
d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.
Step-by-step explanation:
Let the event that a student does homework regularly be H.
The event that a student passes the course be P.
- 60% of her students do homework regularly
P(H) = 60% = 0.60
- 95% of the students who do their homework regularly generally pass the course
P(P|H) = 95% = 0.95
- She also knows that 85% of her students pass the course.
P(P) = 85% = 0.85
a) The probability that a student will do homework regularly and also pass the course = P(H n P)
The conditional probability of A occurring given that B has occurred, P(A|B), is given as
P(A|B) = P(A n B) ÷ P(B)
And we can write that
P(A n B) = P(A|B) × P(B)
Hence,
P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57
b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')
From Sets Theory,
P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1
P(H n P) = 0.57 (from (a))
Note also that
P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)
0.60 = P(H n P') + 0.57
P(H n P') = 0.60 - 0.57
Also
P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)
0.85 = P(H' n P) + 0.57
P(H' n P) = 0.85 - 0.57 = 0.28
So,
P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1
Becomes
0.03 + 0.28 + 0.57 + P(H' n P') = 1
P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12
c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.
Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,
P(A n B) = 0.
But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.
Hence, the two events aren't mutually exclusive.
d. Are the events "pass the course" and "do homework regularly" independent? Explain
Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when
P(A|B) = P(A)
P(B|A) = P(B)
P(A n B) = P(A) × P(B)
To check if the events pass the course and do homework regularly are mutually exclusive now.
P(P|H) = 0.95
P(P) = 0.85
P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671
P(H) = 0.60
P(H n P) = P(P n H)
P(P|H) = 0.95 ≠ 0.85 = P(P)
P(H|P) = 0.671 ≠ 0.60 = P(H)
P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)
None of the conditions is satisfied, hence, we can conclude that the two events are not independent.
Hope this Helps!!!
B
Round your answer to the nearest hundredth.
A
9
B
5
Answer:
56.25°
Step-by-step explanation:
The definition of the cosine function tells you that
cos(B) = BC/BA
B = arccos(BC/BA) = arccos(5/9)
B ≈ 56.25°
Among coffee drinkers, men drink a mean of 3.2 cups per day with a standard deviation of 0.8 cups. Assume the number of cups per day follows a normal distribution.
a. What proportion drink 2 cups per day or more?
b. What proportion drink no more than 4 cups per day?
c. If the top 5% of coffee drinkers are considered "heavy" coffee drinkers, what is the minimum number of cups consumed by a heavy coffee drinker?
d. If a sample of 20 men is selected, what is the probability that the mean number of cups per day is greater than 3?
Answer:
a) 0.9332 = 93.32% drink 2 cups per day or more.
b) 0.8413 = 84.13% drink no more than 4 cups per day
c) The minimum number of cups consumed by a heavy coffee drinker is 4.52.
d) 86.86% probability that the mean number of cups per day is greater than 3
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 3.2, \sigma = 0.8[/tex]
a. What proportion drink 2 cups per day or more?
This is 1 subtracted by the pvalue of Z when X = 2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2 - 3.2}{0.8}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
1 - 0.0668 = 0.9332
0.9332 = 93.32% drink 2 cups per day or more.
b. What proportion drink no more than 4 cups per day?
This is the pvalue of Z when X = 4.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 3.2}{0.8}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.8413 = 84.13% drink no more than 4 cups per day
c. If the top 5% of coffee drinkers are considered "heavy" coffee drinkers, what is the minimum number of cups consumed by a heavy coffee drinker?
This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 3.2}{0.8}[/tex]
[tex]X - 3.2 = 1.645*0.8[/tex]
[tex]X = 4.52[/tex]
The minimum number of cups consumed by a heavy coffee drinker is 4.52.
d. If a sample of 20 men is selected, what is the probability that the mean number of cups per day is greater than 3?
Sample of 20, so applying the central limit theore with n = 20, [tex]s = \frac{0.8}{\sqrt{20}} = 0.1789[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 3.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3 - 3.2}{0.1789}[/tex]
[tex]Z = -1.12[/tex]
[tex]Z = -1.12[/tex] has a pvalue of 0.1314
1 - 0.1314 = 0.8686
86.86% probability that the mean number of cups per day is greater than 3
What is the distance between (8, -3) and (4, - 7)?
Answer:
[tex]distance=\sqrt{32}[/tex] , which agrees with answer "c" in your list of possible options
Step-by-step explanation:
Use the formula for distance between two points [tex](x_1,y_1)[/tex], and [tex](x_2,y_2)[/tex] on the plane:
[tex]distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\distance= \sqrt{(4-8)^2+(-7-(-3))^2} \\distance= \sqrt{(-4)^2+(-4)^2} \\distance=\sqrt{16+16}\\distance=\sqrt{32}[/tex]
Pls help marking Brainliest! (x-1)(x-1)=?
Answer:
Brainleist!Step-by-step explanation:
x^2-2x+1thats the answer...
just: Expand the polynomial using the FOIL method.
Answer:
(x-1)(x-1)=(x-1)² because it's the same thing multiplied by itself
Using FOIL method:
(x-1)(x-1)=
x²-x-x+1=
x²-2x+1
Please help with this problem
Answer:
The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.
Step-by-step explanation:
The Pythagorean Theorem
If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
This relationship is represented by the formula:
[tex]a^2+b^2=c^2[/tex]
Applying the Pythagorean Theorem to find the lengths of the three sides we get:
[tex](x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0[/tex]
Solve with the quadratic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}[/tex]
Because a length can only be positive, the only solution is
[tex]x=5+3\sqrt{10}\approx 14.5[/tex]
The length of the short side is 14.5, the length of the other short side is [tex]14.5+4=18.5[/tex], and the length of the longest side is [tex]14.5+9=23.5[/tex].
I NEED HELP ASAP PLEASE!!! I REALLY NEED HELP!
Answer:
D.
Step-by-step explanation:
One slope is positive and one negative, so one line should go up and one down. B or D.
y = 1/2 x - 1 line goes up and y-int. = - 1. Answer D.
y = - 1/2 x + 3 line goes up and y-int. = 3. Answer D.
The answer to – 7x + y = -10
Step-by-step explanation:
y=7x-10
Answer:
[tex]\huge \boxed{y=7x-10}[/tex]
Step-by-step explanation:
[tex]-7x+y=-10[/tex]
[tex]\sf Add \ 7x \ on \ both \ sides.[/tex]
[tex]-7x+y+7x=-10+7x[/tex]
[tex]y=7x-10[/tex]
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. In a talent competition, half of the contestants are eliminated in each round. At the end of the nth round, 32 contestants remain. If there were 1,024 contestants at the start of the competition, what is the value of n? The value of n is .
Answer:
n =32
Step-by-step explanation:
If 1 contestant is eliminated each round
then of 1024contestants
32 left
1024/32=32
Answer:
n=32
Step-by-step explanation:
State the size of angle 'n' in the triangle illustrated below.
Answer:
Option B
Step-by-step explanation:
<r = 32 degrees (alternate angles )
<r = <n = 32 degrees (vertical angles)
Solve for x: −3x + 3 < 6
Answer:x>-1
Step-by-step explanation:
Step 1: Subtract 3 from both sides.
-3x+3-3<6-3
-3x<3
Step 2: Divide both sides by -3.
-3x/-3<3/3
X>-1
Where is my phone? I seem to have lost my phone. I know where I last saw it but it has been moved since then and I need help to locate it. It started at the following coordinates A (14, -12); B (14, -19); C (10, -19); D (10, -14); E (13, -14); F (13, -12). My Mom told me she translated it 6 units to the left Then my little brother said he had reflected it over the Y-axis My friend many found it and translated it 9 units up Dad said he tripped over it and reflected it over the X-axis My sister then rotated it 900 clockwise Uncle Jose translated it 5 units left and 4 units down Cousin Michelle then said she rotated it 900 clockwise Finally my dog picked it up and translated it 5 units down and 10 units to the right Where is my phone? Using the scenario on this page do the following. Graph the preimage using the given points. Label points (A, B, C, ...) Transform the objects using the information provided. Show each transformation and label. (A', B', C', ...) Determine the final location. Write a 2 to 3 sentence explain on how you found the phone location.
Answer:
see attached
Step-by-step explanation:
The attachments show the initial (brown) and final (blue) positions of the phone. The spreadsheet shows all the intermediate locations and the formulas used to determine them.
The two reflections cancel the total of 180° of CW rotation, so the net result is simply a translation. That translation is up by 9 units.
__
Translation up adds to the y-coefficient; translation right adds to the x-coefficient. Down or left use negative values.
90° CW does this: (x, y) ⇒ (y, -x)
Reflection across y does this: (x, y) ⇒ (-x, y)
Reflection across x does this: (x, y) ⇒ (x, -y)
The following questions pertain to the properties of the STANDARD NORMAL distribution. (a) True or False: The distribution is bell-shaped and symmetric. True False (b) True or False: The mean of the distribution is 0. True False (c) True or False The probability to the left of the mean is 0. True False (d) True or False The standard deviation of the distribution is 1. True False
Answer:
a. The distribution is bell-shaped and symmetric: True.
b. The distribution is bell-shaped and symmetric: True.
c. The probability to the left of the mean is 0: False.
d. The standard deviation of the distribution is 1: True.
Step-by-step explanation:
The Standard Normal distribution is a normal distribution with mean, [tex] \\ \mu = 0[/tex], and standard deviation, [tex] \\ \sigma = 1[/tex].
It is important to recall that the parameters of the Normal distributions, namely, [tex] \\ \mu[/tex] and [tex] \\ \sigma[/tex] characterized them.
We can use the Standard Normal distribution to find probabilities for any normally distributed data. All we have to do is normalized them through z-scores:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]
Where [tex] \\ x[/tex] is the raw score that we want to standardize.
Therefore, taking into account all this information, we can answer the following questions about the Standard Normal distribution:
(a) True or False: The distribution is bell-shaped and symmetric
Answer: True. As the normal distribution, the standard normal distribution is also bell-shape and it is symmetrical around the mean. The standardized values or z-scores, which represent the distance from the mean in standard deviations units, are the same but when it is above the mean, the z-score is positive, and negative when it is below the mean. This result is a consequence of the symmetry of this distribution respect to the mean of the distribution.
(b) True or False: The mean of the distribution is 0.
Answer: True. Since the Standard Normal uses standardized values, if we use [1], we have:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex]
If [tex] \\ x = \mu[/tex]
[tex] \\ z = \frac{\mu - \mu}{\sigma}[/tex]
[tex] \\ z = \frac{0}{\sigma}[/tex]
[tex] \\ z = 0[/tex]
Then, the value for the mean is where z = 0. A z-score is a linear transformation of the original data. For this reason, the transformed mean is equivalent to 0 in the standard normal distribution. We only need to find distances from this zero in standard normal deviations or z-scores to find probabilities.
(c) True or False: The probability to the left of the mean is 0.
Answer: False. The probability to the left of the mean is not 0. The cumulative probability from [tex] \\ -\infty[/tex] until the mean is 0.5000 or [tex] \\ P(-\infty < z < 0) = 0.5[/tex].
(d) True or False: The standard deviation of the distribution is 1.
Answer: True. The standard normal distribution is a convenient way of calculate probabilities for any normal distribution. The standardized variable, represented by [1], permits us to use one table (the standard normal table) for all normal distributions.
In this distribution, the z-score is always divided by the standard deviation of the population. Then, the standard deviation for the standard normal distribution are times or fractions of the standard deviation of the population, since we divide the distance of a raw score from the mean of the population, [tex] \\ x - \mu[/tex], by it. As a result, the standard deviation for the standard normal distribution will be times (1, 2, 3, 0.96, -1, -2, etc) the standard deviation of any normal distribution, [tex] \\ \sigma[/tex].
In this case, the linear transformation of the original data for one standard deviation from the mean is z = 1. Therefore, the standard deviation for the standard normal distribution is the unit.
Answer:
A: true
B: true
C: false
D: true
Please answer this correctly
Answer:
The number of employees classified into groups as shown below:
1 - 10: 3 6 (2companies)
11-20: 16 (1 company)
21-30: 25, 26, 27 (3 companies)
31-40: 34, 35, 35, 35, 36 (5 companies)
41-50: 41, 43, 48, 48 (4 companies)
Hope this helps!
Answer:
11-20 is 1
31-40 is 5
Step-by-step explanation:
Just count the amount
Hope that helps :D
At the U.S. Open Tennis Championship a statistician keeps track of every serve that a player hits during the tournament. The statistician reported that the mean serve speed was 100 miles per hour (mph) and the standard deviation of the serve speeds was 15 mph. Assume that the statistician also gave us the information that the distribution of serve speeds was mound- shaped and symmetric. What percentage of the player's serves were between 115 mph and 145 mph
Answer:
15.74% of the player's serves were between 115 mph and 145 mph
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 100, \sigma = 15[/tex]
What percentage of the player's serves were between 115 mph and 145 mph
This is the pvalue of Z when X = 145 subtracted by the pvalue of Z when X = 115.
X = 145
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{145 - 100}{15}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
X = 115
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{115 - 100}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.9987 - 0.8413 = 0.1574
15.74% of the player's serves were between 115 mph and 145 mph
2.
√3x + 7 + √x + 1 =2
Answer:
x = -1
Step-by-step explanation:
The usual approach to these is to square the radicals until they are gone.
[tex]\displaystyle\sqrt{3x+7}+\sqrt{x+1}=2\\\\(3x+7) +2\sqrt{(3x+7)(x+1)}+(x+1) = 4\qquad\text{square both sides}\\\\2\sqrt{(3x+7)(x+1)}=-4x-4\qquad\text{subtract $4x+8$}\\\\(3x+7)(x+1)=(-2x-2)^2\qquad\text{divide by 2, square again}\\\\3x^2+10x +7=4x^2+8x+4\qquad\text{simplify}\\\\x^2-2x-3=0\qquad\text{subtract the left expression}\\\\(x-3)(x+1)=0\qquad\text{factor}\\\\x=3,\ x=-1\qquad\text{solutions to the quadratic}[/tex]
Each time the equation is squared, the possibility of an extraneous root is introduced. Here, x=3 is extraneous: it does not satisfy the original equation.
The solution is x = -1.
_____
Using a graphing calculator to solve the original equation can avoid extraneous solutions. The attachment shows only the solution x = -1. Rather than use f(x) = 2, we have rewritten the equation to f(x)-2 = 0. The graphing calculator is really good at showing the function values at the x-intercepts.
y= -3/2x-6 x=15 plssssssssssssssssssssssss help
Answer:
-45/2 - 12/2 = -57/2
Step-by-step explanation:
Substitute 15 for x in the given equation: y = (-3/2)x - 6 becomes
y = (-3/2)(15) - 6 = -45/2 - 6 when x = 15. This is equivalent to -57/2
Justin spent $23 on fruit at grocery store. He spent a total of $25 at the store. What percentage of the total did he spend on fruit?
Step-by-step explanation:
In my opinion maybe he has spent 98%
Simplify the expression by combining like terms
15 + 12x – 5.2 + 4y - 7
9.8 +12x+y-7
2.8+12x+4y
Construct a boxplot for the given data. Include values of the 5-number summary in all boxplots. The test scores of 40 students are listed below. Construct a boxplot for the data set.
25 35 43 44 47 48 54 55 56 57
59 62 63 65 66 68 69 69 71 72
72 73 74 76 77 77 78 79 80 81
81 82 83 85 89 92 93 94 97 98
Answer:
Minimum = 25
First quartile = 58
Second quartile = 72
Third quartile = 80
Maximum = 98
Step-by-step explanation:
N
Write the rate as a unit rate
729 riders in 9 subway cars
А
The unit rate is
This
(Simplify your answ
riders/car
Answer:
Unit rate = 81 riders/ car.
Step-by-step explanation:
Given
729 riders in 9 cars
we have to find unit rate in terms of riders per car
let the the riders per car (i.e rate) be x.
If there are 9 cars then
total no. of riders in 9 cars = no. of cars * riders per car = 9*x = 9x
given that 729 riders in 9 cars
then
9x = 729
=> x = 729/9 = 81
Thus, riders per car = x = 81.
Unit rate is 81 riders per car.
Single adults: According to a Pew Research Center analysis of census data, in 2012, 20% of American adults ages 25 and older had never been married. Suppose that we select 3 random samples of 500 adults from this population. Which of the following is most likely to occur with the three samples?
A. The number that had never been married will equal 20% in each of the three samples.B. The number that had never been married will vary in each sample due to the random selection of adults.C. The average for the three samples of the number of adults that had never been married will equal 20%.D. The number of adults that had never been married will increase for each sample because the number is generally increasing over time.
Answer:
Option B
Step-by-step explanation:
The number that had never been married will vary in each sample due to the random selection of adults.
This number will vary in each sample to the random selection process but they might or might not be as close as possible to one another after sampling.
Can someone plz help me solved this problem! I’m giving you 10 points! I need help plz help me! Will mark you as brainiest!
Answer:
See the answers below.
Step-by-step explanation:
[tex]a.\:\frac{f\left(x\right)-f\left(a\right)}{x-a}=\frac{2x^2-x-5-\left(2a^2-a-5\right)}{x-a}\\\\=\frac{2x^2-x+a-2a^2}{x-a}\\\\=\frac{2\left(x+a\right)\left(x-a\right)-1\left(x-a\right)}{x-a}\\\\=\frac{\left(x-a\right)\left[2\left(x+a\right)-1\right]}{x-a}\\\\=2x+2a-1\\\\\\b.\:\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{2\left(x+h\right)^2-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\\\=\frac{2\left(x^2+2xh+h^2\right)-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\[/tex]
Expand and simplify to get:
[tex]=\frac{2h^2+4xh-h}{h}\\\\=\frac{h\left(2h+4x-1\right)}{h}\\\\=2h+4x-1[/tex]
Best Regards!
there are only red counters and blue counters in a bag. Jim takes at random a counter from a bag. the probability that the counter is red is 0.45 Jim puts the counter back into the bag. Molly takes at random a counter from the bag. She puts the counter back in the bag. What is the probability that Jim and Molly take counters of different colours? Give your answer as a decimal
Answer:
0.495 probability that Jim and Molly take counters of different colours
Step-by-step explanation:
For each trial, there are only two possible outcomes. Either a blue counter is picked, or a red counter is picked. The counter is put back in the bag after it is taken, which means that we can 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.
The probability that the counter is red is 0.45
This means that [tex]p = 0.45[/tex]
Jim taken a counter, then Molly:
Two trials, so [tex]n = 2[/tex]
What is the probability that Jim and Molly take counters of different colours?
One red and one blue. So this is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{2,1}.(0.45)^{1}.(0.55)^{1} = 0.495[/tex]
0.495 probability that Jim and Molly take counters of different colours
In general, shopping online is supposed to be more convenient than going to stores. However, according to a recent Harris Interactive poll, 87% of people have experienced problems with an online transaction (The Wall Street Journal, October 2, 2007). Forty-two percent of people who experienced a problem abandoned the transaction or switched to a competitor′s website. Fifty-three percent of people who experienced problems contacted customer-service representatives.
a. What percentage of people did not experience problems with an online transaction?
b. What percentage of people experienced problems with an online transaction and abandoned the transaction or switched to a competitor′s website?
c. What percentage of people experienced problems with an online transaction and contacted customer-service representatives?
Answer:
a) 13% of people did not experience problems with an online transaction.
b) 36.54% of people experienced problems with an online transaction and abandoned the transaction or switched to a competitor′s website
c) 46.11% of people experienced problems with an online transaction and contacted customer-service representatives.
Step-by-step explanation:
a. What percentage of people did not experience problems with an online transaction?
87% of people have experienced problems with an online transaction. So 100 - 87 = 13% of people did not experience problems with an online transaction.
b. What percentage of people experienced problems with an online transaction and abandoned the transaction or switched to a competitor′s website?
87% of people have experienced problems with an online transaction. Forty-two percent of people who experienced a problem abandoned the transaction or switched to a competitor′s website.
Then:
0.87*0.42 = 0.3654
36.54% of people experienced problems with an online transaction and abandoned the transaction or switched to a competitor′s website.
c. What percentage of people experienced problems with an online transaction and contacted customer-service representatives?
87% of people have experienced problems with an online transaction. Fifty-three percent of people who experienced problems contacted customer-service representatives.
Then:
0.87*0.53 = 0.4611
46.11% of people experienced problems with an online transaction and contacted customer-service representatives.