Answer:$462.50
Step-by-step explanation:multiply $1,850 by .25 to get your answear
An electronics company wants to compare the quality of their cell phones to the cell phones from three of their competitors. They sample 10 phones from their own company and 10 phones from each of the other three companies and count the number of defects for each phone. If ANOVA was used to compare the average number of defects, then the treatments would be defined as:
Answer:
The treatment should be stated by the four companies,since it more interested in the quality among each of the companies to be compared.
Step-by-step explanation:
From the example given, Since an electronic company wants to differentiate their cell phones quality to the cell phones from their three main competitors.
If ANOVA is used to determine the average number of defects, then the treatment should be defined for the four companies because it is more interested in comparing the quality among the different companies.
In ANOVA analysis for the given question, the treatments would be defined as the cell phones from the electronics company itself and three of its competitors. ANOVA evaluates if there is any significant difference in the average number of defects amongst these four groups.
Explanation:In the situation described, an electronics company testing the quality of their cell phones against three competitors, the treatments in the context of ANOVA (Analysis of Variance) would be the four different companies' cell phones. Specifically, a one-way ANOVA is being used here since there is only one factor or variable (the company) affecting the outcome variable (the number of defects).
Each sample of phones (10 from each company) represents a level within the treatment. Therefore, the treatments here are the cell phones from the electronics company, and Cell Phone Company 1, Cell Phone Company 2, and Cell Phone Company 3.
This comparison is made by analyzing the variance within each group's data and between the groups. The essential goal of ANOVA is to test if there is any significant difference between these means. In this case, it is used to ascertain whether the mean number of defects is significantly different across the four companies.
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Jade says that 8 divided by ½ means the same thing as ½ of 8, so the answer is 4. Do you agree with her? Why or why not?
Answer:
Jade's statement is incorrect because "8 divided by ½" does not mean the same thing as "½ of 8".
Explanation:
No, I don't agree with Jade's statement.
When we say "8 divided by ½", it means we are dividing 8 into two equal parts, each representing one-half. Mathematically, [tex]\(8 \div \frac{1}{2}\)[/tex] means we are dividing 8 by [tex]\(\frac{1}{2}\)[/tex] , which is equivalent to multiplying 8 by the reciprocal of [tex]\(\frac{1}{2}\)[/tex], which is 2. So:
[tex]\[ 8 \div \frac{1}{2} = 8 \times 2 = 16 \][/tex]
On the other hand, "½ of 8" means finding half of 8, which is indeed 4. However, this is not the same as "8 divided by ½".
Find the mean. Round to the nearest tenth. Help help?????
Answer:
(2). 26 (4). 547 (6). 3,132 (8). 46.1 (10). 10.6
Step-by-step explanation:
Which of the following statement is true? A. For any discrete random variable X and constants a and b, E(aX+b) = (a + b). E(X) B. For any discrete random variable X and constants a and b, V(aX+b) = . C. If a constant c is added to each possible value of a discrete random variable X, then the variance of X will be shifted by that same constant amount. D. If a constant c is added to each possible value of a discrete random variable X, then the expected value of X will be shifted by that same constant amount.
Answer:
A. False
B. False
C. False
D. True
Step-by-step explanation:
A. False
Remember that E[aX+b] = aE[X] + b, therefore the equality would be true only if E[X] = 1.
B. False
Remember that [tex]V[aX+b] = a^2 V[X][/tex].
C. False
In other words they are telling you in words that [tex]V[X+b] = V[X]+b[/tex], that is false because [tex]V[X+b] = V[X][/tex]
D. True
That is a property of the expected value E[X+b] = E[X]+b
Option D is correct in stating that adding a constant to each value of a random variable shifts the expected value of that variable by that same constant. Adding a constant does not alter the variance.
Explanation:The correct answer among the given options is D. This statement says that if a constant c is added to each possible value of a discrete random variable X, then the expected value of X will be shifted by that same constant amount. This is in accordance with the properties of expected value, where the expected value of (X + c) is E(X) + c. On the other hand, adding a constant does not affect the variance of a random variable. Variance measures the dispersion of a variable's possible values from its expected value, and adding a constant shifts every possible value, including the expected value, by the same amount, leaving dispersion unchanged.
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A researcher focusing on birth weights of babies found that the mean birth weight is 3370 grams (7 pounds, 6.9 ounces) with a standard deviation of 582 grams. Complete parts (a) through (c) below. a. Identify the population and variable. Choose the correct population below. A. The group of all babies ever born B. The group of researchers C. The number of babies born the year the research was performed D. The group of all of the babies whose recorded weights were examined Choose the correct variable below. A. The weights of the babies at birth B. The accuracy of the measurements of baby birth weights C. The number of babies that were born D. The number of births per capita b. For samples of size 175, find the mean mu Subscript x overbar and standard deviation sigma Subscript x overbar of all possible sample mean weights. mu Subscript x overbarequals nothing (Type an integer or a decimal. Do not round.) sigma Subscript x overbarequals nothing (Round to two decimal places as needed.) c. Repeat part (b) for samples of size 350. mu Subscript x overbarequals nothing (Type an integer or a decimal. Do not round.) sigma Subscript x overbarequals nothing (Round to two decimal places as needed.)
Answer:
a) D. The group of all of the babies whose recorded weights were examined
b) A. The weights of the babies at birth
c) i) Ux' = u = 3370;
s.d = 43.95
ii) Ux' = u = 3370
s.d = 35.109
Step-by-step explanation:
Given:
Mean, u = 3370
Standard deviation = 582
a) Population in statistics involves every member of the group in study.
Here, the correct population is the group of all of the babies whose recorded weights were examined.
b) In statistics, variable involves data collected.
The correct variable is the weights of the babies at birth
c)
i) at X' = 175
Sample mean:
Ux' = u = 3370
Standard deviation:
[tex] \sigma x' = \frac{\sigma}{\sqrt{n}} = \frac{582}{\sqrt{175}} = 43.995 [/tex]
ii) at X' = 175
Sample mean:
Ux' = u = 3370
Stanard deviation:
[tex] \sigma x' = \frac{\sigma}{\sqrt{n}} = \frac{582}{\sqrt{350}} = 35.109 [/tex]
a. The population includes all babies whose recorded weights were examined, and the variable is the birth weights of the babies.
b. For sample sizes of 175, the mean birth weight remains 3370 grams, with a standard deviation of approximately 43.96 grams.
c. For sample sizes of 350, the mean birth weight remains 3370 grams, with a standard deviation of approximately 31.15 grams.
Let's address each part of the question step-by-step to ensure clarity and accuracy.
Part (a): Identifying Population and Variable
To identify the correct population:
Population: The group of all of the babies whose recorded weights were examined.To identify the correct variable:
Variable: The weights of the babies at birth.Hence, the correct answers are:
Population: D. The group of all of the babies whose recorded weights were examined.Variable: A. The weights of the babies at birth.Part (b): Calculating Mean ([tex]\mu_\bar{x}[/tex]) and Standard Deviation ([tex]\sigma_\bar{x}[/tex]) for Sample Size 175
For samples of size 175, we need to find the mean and standard deviation of all possible sample means:
Mean ([tex]\mu_\bar{x}[/tex]) is the same as the population mean ([tex]\mu[/tex]): [tex]\mu_\bar{x}[/tex] = 3370 grams.Standard Deviation ([tex]\sigma_\bar{x}[/tex]):Part (c): Calculating Mean ([tex]\mu_\bar{x}[/tex]) and Standard Deviation ([tex]\sigma_\bar{x}[/tex]) for Sample Size 350
For samples of size 350, we perform similar calculations:
Mean ([tex]\mu_\bar{x}[/tex]) is the same as the population mean ([tex]\mu[/tex]): [tex]\mu_\bar{x}[/tex] = 3370 grams.Standard Deviation ([tex]\sigma_\bar{x}[/tex]):In summary, we have identified the population and variable, and calculated the mean and standard deviation for different sample sizes of 175 and 350.
a car drives 195 miles in 3 hours and 15 mins.
what is the average speed of the car?
Answer:
one mile per hour
Step-by-step explanation:
Answer:
It is 60 miles per hour
Step-by-step explanation: average speed = total distance / total time
We have 50 video gamers who share information about their favorite games. 20 people say that Borderlands II is their favorite game. 33 say their favorite game is Fortnite, while 19 declare Minecraft as their favorite game (note that a single person can have several favorite games). 15 people say that Borderlands and Fortnite are both favorites, while 10 say that both Borderlands and Minecraft are their favorites. Finally, 5 people declare all 3 of the games as their favorites. How many people declared Minecraft and Fortnite (exclusively) as their favorites?
Answer:
2 people
Step-by-step explanation:
Total number of people (n) = 50
Borderlands II (B) = 20
Fortnite (F) = 33
Minecraft (M) = 19
Borderlands and Fortnite (B&F) = 15
Borderlands and Minecraft (B&M)= 10
All games (B&M&F) = 5
The number of people that claimed more than one game as their favorite (X) is given by:
[tex]X=M+B+F-n\\X=19+20+33-50\\X=22\ people[/tex]
Those people can be divided in B&F only, B&M only, M&F only and B&M&F:
[tex]X=(B\&M-B\&M\&F)+(B\&F-B\&M\&F)+(M\&F-B\&M\&F)+B\&M\&F\\X=B\&M+B\&F+M\&F-2*B\&M\&F\\22=10+15+M\&F-2*5\\M\&F=7\ people[/tex]
7 people claimed that Minecraft and Fortnite were their favorites however, 5 of those people claimed all of the games as their favorite. Therefore, only 2 people declared Minecraft and Fortnite (exclusively) as their favorites.
Answer:
2
Step-by-step explanation:
Aditi downloads ten paid apps and sixteen free apps on her tablet. Fourteen of them are game apps, and she paid for five of the game apps.
Complete the statements to determine if the events “paid” and “game” are independent.
P(paid) =
P(paid | game) =
The events “paid” and “game” are
Answer:
-P(paid) = 10/26 and P(paid|game) = 5/14.
-The events "paid" and "game" are not independent.
Step-by-step explanation:
Number of paid apps downloaded = 10
Number of free apps downloaded = 16
Total number of apps = 10 + 16 = 26
Thus;
P(paid) = 10/26
Now, it says she paid for 5 out if 14 which were game apps. Thus;
P(paid|game) = 5/14
Now, Two events are independent if the result of the second event is not affected by the result of the first event. If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events.
In this question, paid and game are not affected by each other and the probability of of P(paid) and P(paid|game) occurring are not products of each individual event paid and game. Thus, they are not independent.
Answer:
10/26 , 5/14 , not independent
Step-by-step explanation:
Brad bought a 1/3 pound of bag of beans he divided all the beans into two equal size pile how much did each of the piles weigh
Answer: the answer is 2/3
Answer:
3 ounces
Step-by-step explanation:
1/3 of a pound is 6 ounces.
6 divided by 2 is 3.
Each pile of beans weighs 3 ounces.
Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean μ=6
ml and standard deviation σ=0.2 ml is a reasonable model for the distribution of the variable x = amount of red dye in the paint mixture. Use the normal distribution model to calculate the following probabilities. (Round all answers to four decimal places.)
(a) P(x > 6) =
(b) P(x < 6.2)=
(c) P(x ≤ 6.2) =
(d) P(5.8 < x < 6.2) =
(e) P(x > 5.7) =
(f) P(x > 5) =
Answer:
a) 0.5.
b) 0.8413
c) 0.8413
d) 0.6826
e) 0.9332
f) 1
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 problem, we have that:
[tex]\mu = 6, \sigma = 0.2[/tex]
(a) P(x > 6) =
This is 1 subtracted by the pvalue of Z when X = 6. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6-6}{0.2}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5.
1 - 0.5 = 0.5.
(b) P(x < 6.2)=
This is the pvalue of Z when X = 6.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6.2-6}{0.2}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
(c) P(x ≤ 6.2) =
In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.
(d) P(5.8 < x < 6.2) =
This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.
X = 6.2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6.2-6}{0.2}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
X = 5.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.8-6}{0.2}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
(e) P(x > 5.7) =
This is 1 subtracted by the pvalue of Z when X = 5.7.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.8-6}{0.2}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
1 - 0.0668 = 0.9332
(f) P(x > 5) =
This is 1 subtracted by the pvalue of Z when X = 5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5-6}{0.2}[/tex]
[tex]Z = -5[/tex]
[tex]Z = -5[/tex] has a pvalue of 0.
1 - 0 = 1
Summary of probabilities
[tex](a)\ P(x > 6) = 0.5\\(b)\ P(x < 6.2) = 0.8413\\(c)\ P(x \leq 6.2) = 0.8413\\(d)\ P(5.8 < x < 6.2) = 0.6826\\(e)\ P(x > 5.7) = 0.9332\\(f)\ P(x > 5) \approx 1\\[/tex]
(a) [tex]\(P(x > 6)\)\\[/tex]
1. Calculate the z-score for [tex]\(x = 6\)[/tex] :
[tex]\[ z = \frac{6 - 6}{0.2} = \frac{0}{0.2} = 0 \][/tex]2. Find [tex]\(P(Z > 0)\)[/tex] :
Since the standard normal distribution is symmetric, [tex]\(P(Z > 0) = 0.5\)[/tex]So, [tex]\(P(x > 6) = 0.5\)[/tex](b) [tex]\(P(x < 6.2)\)[/tex]
1. Calculate the z-score for [tex]\(x = 6.2\)[/tex] :
[tex]\[ z = \frac{6.2 - 6}{0.2} = \frac{0.2}{0.2} = 1 \][/tex]2. Find [tex]\(P(Z < 1)\)[/tex]:
Using the Z-table, [tex]\(P(Z < 1) = 0.8413\)[/tex]So, [tex]\(P(x < 6.2) = 0.8413\)[/tex](c) [tex]\(P(x \leq 6.2)\)[/tex]
For continuous distributions, [tex]\(P(x \leq 6.2) = P(x < 6.2)\)[/tex]So, [tex]\(P(x \leq 6.2) = 0.8413\)[/tex](d) [tex]\(P(5.8 < x < 6.2)\)[/tex]
1. Calculate the [tex]\(z\)[/tex]-score for [tex]\(x = 5.8\):[/tex]
[tex]\[ z = \frac{5.8 - 6}{0.2} = \frac{-0.2}{0.2} = -1 \][/tex]2. Calculate the [tex]\(z\)[/tex]-score for [tex]\(x = 6.2\):[/tex]
[tex]\[ z = \frac{6.2 - 6}{0.2} = \frac{0.2}{0.2} = 1 \][/tex]3. Find [tex]\(P(Z < 1)\)[/tex] and [tex]\(P(Z < -1)\):[/tex]
[tex]\[ P(Z < 1) = 0.8413 \][/tex][tex]\[ P(Z < -1) = 0.1587 \][/tex]4. Calculate [tex]\(P(-1 < Z < 1)\):[/tex]
[tex]\[ P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826 \][/tex]So, [tex]\(P(5.8 < x < 6.2) = 0.6826\)[/tex](e) [tex]\(P(x > 5.7)\)[/tex]
1. Calculate the [tex]\(z\)[/tex]-score for [tex]\(x = 5.7\):[/tex]
[tex]\[ z = \frac{5.7 - 6}{0.2} = \frac{-0.3}{0.2} = -1.5 \][/tex]2. Find [tex]\(P(Z > -1.5)\):[/tex]
Using the Z-table, [tex]\(P(Z < -1.5) = 0.0668\)[/tex]Thus, [tex]\(P(Z > -1.5) = 1 - 0.0668 = 0.9332\)[/tex]So, [tex]\(P(x > 5.7) = 0.9332\)[/tex](f) [tex]\(P(x > 5)\)[/tex]
1. Calculate the [tex]\(z\)[/tex]-score for [tex]\(x = 5\):[/tex]
[tex]\[ z = \frac{5 - 6}{0.2} = \frac{-1}{0.2} = -5 \][/tex]2. Find [tex]\(P(Z > -5)\):[/tex]
Since [tex]\(z = -5\)[/tex] is far in the tail of the standard normal distribution, [tex]\(P(Z < -5)\)[/tex] is almost 0.Thus, [tex]\(P(Z > -5) \approx 1\)[/tex]So, [tex]\(P(x > 5) \approx 1\)[/tex]Expand. Your answer should be a polynomial in standard form. (x + 5)(x +3)
Answer: x^2+8x+15
Step-by-step explanation:
(x+5)(x+3)
x^2+3x+5x+15
x^2+8x+15
To expand the polynomial (x + 5)(x + 3), multiply each term from the first binomial by each term from the second binomial and combine like terms.
Explanation:
To expand the polynomial (x + 5)(x + 3), we need to use the distributive property. We multiply each term from the first binomial by each term from the second binomial. So, we have:
x * x = x^2x * 3 = 3x5 * x = 5x5 * 3 = 15Combining like terms, we get x^2 + 3x + 5x + 15. Simplifying further, we have the expanded polynomial:
x^2 + 8x + 15.
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"A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.3 minutes with a standard deviation of 1.5 minutes. At the 0.01 significance level, test the claim that the mean waiting time is less than 10 minutes. Use the P-value method of testing hypotheses."
Answer:
We conclude that the mean waiting time is less than 10 minutes.
Step-by-step explanation:
We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.
Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.3 minutes with a standard deviation of 1.5 minutes.
Let [tex]\mu[/tex] = mean waiting time.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 10 minutes {means that the mean waiting time is less than 10 minutes}
The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean waiting time = 7.3 minutes
s = sample standard deviation = 1.5 minutes
n = sample of occasions = 18
So, test statistics = [tex]\frac{7.3-10}{\frac{1.5}{\sqrt{18}}}[/tex] ~ [tex]t_1_7[/tex]
= -7.637
The value of t test statistics is -7.637.
Now, the P-value of the test statistics is given by the following formula;
P-value = P( [tex]t_1_7[/tex] < -7.637) = Less than 0.05%
Since, our P-values of test statistics is less than the level of significance as 0.05% < 1%, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the mean waiting time is less than 10 minutes.
What is 2 3/8 ÷ 1 1/4?
Answer:
1 9/10 or 1.9
Answer:
1.9
Step-by-step explanation:
A sphere and a cylinder have the same radius and height. The volume of the cylinder is 18 cm
What is the volume of the sphere?
O 12 cm
O 24 cm
O 36 cm
Mark this and return
Save and Exit
Next
Sub
Answer: b
Step-by-step explanation: ik
25y = 15y +75
y= 7.5
infinite solutions
no solution
y= 750
Answer:
y= 7.5
Step-by-step explanation:
25y = 15y +75
-15y -15y
10y = 75
÷10 ÷10
y = 7.5
Matthew invests $500 into an account with a 2.5% interest that is compounded quarterly. How much money will he have in this account if he keeps it for 10 years? Round your answer to the nearest cent. Do NOT round until you have calculated the final answer.
Answer:
Step-by-step explanation:
CORRECT ANSWER IS : $641.51
Use the compound interest formula and substitute the given values: A=$500(1+0.0254)4(10). Simplify using the order of operations: A=$500(1.00625)40=$500(1.283026821)≈ $641.51.
After ten years, Matthew will have $641.51 in the account, rounded to the closest penny.
What is compound interest?The interest earned on savings that are computed using both the original principal and the interest accrued over time is known as compound interest. [tex]A = P(1 + r/n)^{nt[/tex], where P is the principal balance, r is the interest rate, n is the number of times interest is compounded annually, and t is the number of years, which is the formula for compound interest.
We can use the formula for compound interest to calculate how much money Matthew will have in the account after 10 years:
[tex]A = P(1 + r/n)^{nt[/tex]
Where:
A = final amount
P = principal (initial amount invested) = $500
r = annual interest rate (as a decimal) = 0.025
n = number of times interest is compounded per year = 4 (quarterly)
t = time in years = 10
Plugging in the values, we get:
[tex]A = 500(1 + 0.025/4)^(4*10)\\A = 500(1.00625)^{40}\\A = 500(1.305011)[/tex]
A = $641.51
Therefore, Matthew will have $641.51 in the account after 10 years, rounded to the nearest cent.
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Alex has five rolls of shelf paper that is 800 cm long.She wants to use the to line the 1-meter wide shelves in her pantry. How many 1-meter wide can she line with the paper?
Answer:
Alex can line eight 1-meter wide with the paper.
Step-by-step explanation:
- Alex has five rolls of shelf paper that is 800cm.
- She wants to use the paper to line the 1-meter wide shelves in her pantry.
- We want to determine how many 1-meter wide she can line with the paper.
- First, we know that
100cm = 1m
- we need to determine how many meters are in 800cm.
100cm = 1m
800cm = xm
100x = 800
x = 800/100
= 8
Therefore, 800cm is equivalent to 8m
Alex can line eight 1-meter wide with the paper.
To determine the number of 1-meter wide shelves Alex can line with the 800 cm long shelf paper, convert the total length to meters and divide by the shelf width. Alex can line 8 shelves with the paper.
To find out how many 1-meter wide shelves Alex can line with the 800 cm long shelf paper, we need to convert the total length of the paper to meters to match the shelf width.
Convert 800 cm to meters: 800 cm = 8 meters
Divide the total length of the paper by the width of each shelf: 8 meters / 1 meter = 8 shelves
Alex can line 8 shelves with the 1-meter wide shelf paper she has.
evaluate 6+xwhen x=3
Answer:
9
Step-by-step explanation:
Plug in 3 for x, and the equation is 6 + 3, which is equal to 9.
Answer:
6 + x = 9
Step-by-step explanation:
If x equals 3
6 plus x (when x equals 3)
Thus, 6 plus 3
= 9
9 will be your answer
The senior class at a very small high school has 25 students. Officers need to be elected for four positions: President, Vice-President, Secretary, and Treasurer. a. In how many ways can the four officers be chosen? b. If there are 13 girls and 12 boys in the class, in how many ways can the officers be chosen if the President and Treasurer are girls and the Vice-President and Secretary are boys?
Answer:
(a) The total number of ways to select 4 officers from from 25 students is 12,650.
(b) The total number of ways the four officers are selected such that the President and Treasurer are girls and the Vice-President and Secretary are boys is 5,148.
Step-by-step explanation:
(a)
It is provided that there are a total of n = 25 students.
Officers need to be elected for four positions:
President, Vice-President, Secretary, and Treasurer.
k = 4
In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.
The formula to compute the combinations of k items from n is given by the formula:
[tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]
Compute the number of ways to select 4 students from from 25 as follows:
[tex]{25\choose 4}=\frac{25!}{4!(25-4)!}[/tex]
[tex]=\frac{25!}{4!\times 21!}\\\\=\frac{25\times 24\times 23\times 22\times 21!}{4!\times 21!}\\\\=\frac{25\times 24\times 23\times 22}{4\times 3\times 2\times 1}\\\\=12650[/tex]
Thus, the total number of ways to select 4 officers from from 25 students is 12,650.
(b)
It is provided that of the 25 students, there are 13 girls and 12 boys in the class.
For the post of President and Treasurer only girls are selected.
For the post of Vice-President and Secretary only boys are selected.
Compute the number of ways to select 2 girls for the post of President and Treasurer as follows:
[tex]{13\choose 2}=\frac{13!}{2!(13-2)!}[/tex]
[tex]=\frac{13!}{2!\times 11!}\\\\=\frac{13\times 12\times 11!}{2!\times 11!}\\\\=\frac{13\times 12}{ 2\times 1}\\\\=78[/tex]
Compute the number of ways to select 2 boys for the post of Vice-President and Secretary as follows:
[tex]{12\choose 2}=\frac{12!}{2!(12-2)!}[/tex]
[tex]=\frac{12!}{2!\times 10!}\\\\=\frac{12\times 11\times 10!}{2!\times 10!}\\\\=\frac{12\times 11}{ 2\times 1}\\\\=66[/tex]
The number of ways the four officers are selected such that the President and Treasurer are girls and the Vice-President and Secretary are boys is:
[tex]{13\choose 2}\times {12\choose 2}=78\times 66=5148[/tex]
Thus, the total number of ways the four officers are selected such that the President and Treasurer are girls and the Vice-President and Secretary are boys is 5,148.
Find positive numbers x and y satisfying the equation xyequals15 such that the sum 3xplusy is as small as possible. Let S be the given sum. What is the objective function in terms of one number, x? Sequals nothing (Type an expression.) The interval of interest of the objective function is nothing. (Simplify your answer. Type your answer in interval notation.) The numbers are xequals nothing and yequals nothing. (Type exact answers, using radicals as needed.)
Answer:
[tex]x = \sqrt{5}\\\\y = \frac{15}{ \sqrt{5} }[/tex]
Step-by-step explanation:
According to the information of the problem
[tex]xy = 15[/tex]
And
[tex]S = 3x+y[/tex]
If you solve for [tex]y[/tex] on the first equation you get that
[tex]y = {\displaystyle \frac{15}{x}}[/tex]
then you have that
[tex]S = {\displaystyle 3x + \frac{15}{x} }[/tex]
If you find the derivative of the function you get that
[tex]S' = {\displaystyle 3 - \frac{15}{x^2}} = 0\\[/tex]
The equation has two possible solutions but you are looking for POSITIVE numbers that make [tex]S[/tex] as small as possible.
Then
[tex]x = \sqrt{5}\\\\y = \frac{15}{ \sqrt{5} }[/tex]
A dealer bought a snowboard for E45 and sold it for E54. What was her percentage profit?
Answer:
20%
Step-by-step explanation:
(54/45) - 1 = 0.2 or 20%
Express the confidence interval
24.4
%
<
p
<
32.6
%
in the form of
ˆ
p
±
M
E
.
Answer:
[tex]\hat p = \frac{0.244+0.326}{2}=0.285[/tex]
[tex] ME = \frac{0.326-0.244}{2}=0.041[/tex]
[tex] 0.285 \pm 0.041[/tex]
Step-by-step explanation:
For this case we have a confidence interval given as a percent:
[tex] 24.4\% \leq p \leq 32.6\%[/tex]
If we express this in terms of fraction we have this:
[tex] 0.244 \leq p \leq 0.326 [/tex]
We know that the confidence interval for the true proportion is given by:
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
And thats equivalent to:
[tex]\hat p \pm ME[/tex]
We can estimate the estimated proportion like this:
[tex]\hat p = \frac{0.244+0.326}{2}=0.285[/tex]
And the margin of error can be estimaed using the fact that the confidence interval is symmetrical
[tex] ME = \frac{0.326-0.244}{2}=0.041[/tex]
And then the confidence interval in the form desired is:
[tex] 0.285 \pm 0.041[/tex]
A shirt regularly priced at 36.00$ was on sale for 25% off. What was the sale price?
A.9.00$
B.24.00$
C.27.00$
D.48.00$
E. None correct
Answer:$27
Step-by-step explanation:
cost price(cp)=$36
Percentage off=25
sale price=sp
Percentage off=(cp-sp)/cp x 100
25=(36-sp)/36 x 100
Cross multiplying we get
25x36=100(36-sp)
900=100(36-sp)
Divide both sides by 100 we get
900/100=100(36-sp)/100
9=36-sp
Collect like terms
sp=36-9
sp=27
The editor of a textbook publishing company is deciding whether to publish a proposed textbook. Information on previous textbooks published show that 20 % are huge successes, 30 % are modest successes, 30 % break even, and 20 % are losers. Before a decision is made, the book will be reviewed. In the past, 99 % of the huge successes received favorable reviews, 70 % of the moderate successes received favorable reviews, 40 % of the break-even books received favorable reviews, and 20 % of the losers received favorable reviews. If the textbook receives a favorable review, what is the probability that it will be huge success?
Answer:
34.86% probability that it will be huge success
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Receiving a favorable review.
Event B: Being a huge success.
Information on previous textbooks published show that 20 % are huge successes
This means that [tex]P(B) = 0.2[/tex]
99 % of the huge successes received favorable reviews
This means that [tex]P(A|B) = 0.99[/tex]
Probability of receiving a favorable review:
20% are huge successes. Of those, 99% receive favorable reviews.
30% are modest successes. Of those, 70% receive favorable reviews.
30% break even. Of those, 40% receive favorable reviews.
20% are losers. Of those, 20% receive favorable reviews.
Then
[tex]P(A) = 0.2*0.99 + 0.3*0.7 + 0.3*0.4 + 0.2*0.2 = 0.568[/tex]
Finally
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.2*0.99}{0.568} = 0.3486[/tex]
34.86% probability that it will be huge success
A line passes through the point (4, −9) and has a slope of 5/2.
Write an equation in slope-intercept form for this line.
The equation of the line that has a slope of 5/2 and passes through the point (4, -9) is y = 5/2x -19.
Explanation:The equation of a line in slope-intercept form is given by the formula y = mx + b, where m is the slope and b is the y-intercept. We're given that this line has a slope of 5/2 and it passes through the point (4, -9).
Plugging the known values into the equation, we get: -9 = (5/2) * 4 + b. By solving for b, we find the y-intercept of the line.
-9 = 10 + b --> b = -9 - 10 = -19
So the equation of the line in slope-intercept form is y= 5/2x - 19
Learn more about Equation of a Line here:https://brainly.com/question/33578579
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A men’s softball league is experimenting with a yellow baseball that is easier to see during night games. One way to judge the effectiveness is to count the number of errors. In a preliminary experiment, the yellow baseball was used in 10 games and the tradi- tional white baseball was used in another 10 games. The number of errors in each game was recorded and is listed here. Can we infer that there are fewer errors on average when the yellow ball is used?
Answer:
Yes. There is enough evidence to support the claim that there are fewer errors on average when the yellow ball is used.
Step-by-step explanation:
The question is incomplete:
The sample data is:
Yellow 5 2 6 7 2 5 3 8 4 9
White 7 6 8 5 9 11 8 3 6 10
This is a hypothesis test for the difference between populations means.
The claim is that there are fewer errors on average when the yellow ball is used.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0[/tex]
The significance level is α=0.05.
The sample 1 (yellow ball errors), of size n1=10 has a mean of 5.1 and a standard deviation of 2.42.
The sample 2 (white balls errors), of size n2=10 has a mean of 7.3 and a standard deviation of 2.41.
The difference between sample means is Md=-2.2.
[tex]M_d=M_1-M_2=5.1-7.3=-2.2[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{2.42^2+2.41^2}{10}}\\\\\\s_{M_d}=\sqrt{\dfrac{11.665}{10}}=\sqrt{1.166}=1.08[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-2.2-0}{1.08}=\dfrac{-2.2}{1.08}=-2.037[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=10+10-2=18[/tex]
This test is a left-tailed test, with 18 degrees of freedom and t=-2.037, so the P-value for this test is calculated as (using a t-table):
[tex]P-value=P(t<-2.037)=0.028[/tex]
As the P-value (0.028) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that there are fewer errors on average when the yellow ball is used.
Yes, there are fewer errors on average when the yellow ball is used and this can be determined by using the given data.
The Hypothesis test is carried out in which null and alternate hypothesis is given below:
[tex]\rm H_0 : \mu_1-\mu_2=0[/tex]
[tex]\rm H_a : \mu_1-\mu_2<0[/tex]
Now, determine the sample mean difference.
[tex]\rm M_d = M_1-M_2 = 5.1-7.3 = -2.2[/tex]
Now, determine the estimated standard error using the below formula:
[tex]\rm s =\sqrt{\dfrac{\sigma^2_1+\sigma^2_2}{n}}[/tex]
[tex]\rm s =\sqrt{\dfrac{(2.42)^2+(2.41)^2}{10}}[/tex]
s = 1.08
So, the t-statistics can be calculated as:
[tex]\rm t = \dfrac{M_d-(\mu_1-\mu_2)}{s}[/tex]
[tex]\rm t = \dfrac{-2.2-0}{1.08}=-2.037[/tex]
Now, determine the degree of freedom.
[tex]\rm df = n_1+n_2-1[/tex]
df = 10 + 10 - 2
df = 18
Now, for this test, the p-value is 0.028 which is less than the significance level. Therefore, the null hypothesis is rejected.
For more information, refer to the link given below:
https://brainly.com/question/4454077
Q 3.19: In 2011 and 2015, the study was held to determine the proportion of people who read books. 948 people of 1200 said they read at least one book in the last 3 months in 2011. 1080 people of 1500 said they read at least one book in the last 3 months in 2015. Find the 95% confidence interval for the difference in proportions.
Answer:
[tex] (0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376[/tex]
[tex] (0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024[/tex]
And the 95% confidence interval for the difference of the two proportions is given by:
[tex] 0.0376 \leq p_1 -p_2 \leq 0.1024[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] X_1 = 948[/tex] number of people that they read at least one book in the last 3 months in 2011
[tex]n_1 = 1200[/tex] the sample size selected for 2011
[tex] X_2 = 1080[/tex] number of people that they read at least one book in the last 3 months in 2015
[tex]n_2 = 1500[/tex] the sample size selected for 2015
The estimated proportions people that they read at least one book in the last 3 months for each year are given by:
[tex]\hat p_1 = \frac{948}{1200}= 0.79[/tex]
[tex]\hat p_2 = \frac{1080}{1500}= 0.72[/tex]
And the confidence interval for the true difference of proportions is given by:
[tex](\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
The confidence level is 95% so then the significance is 0.05 or 5% and [tex]\alpha/2 =0.025[/tex] and the critical value for this case using the normal standard distribution is:
[tex]z_{\alpha/2}=\pm 1.96[/tex]
And replacing into the confidence interval formula we got:
[tex] (0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376[/tex]
[tex] (0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024[/tex]
And the 95% confidence interval for the difference of the two proportions is given by:
[tex] 0.0376 \leq p_1 -p_2 \leq 0.1024[/tex]
Consider rolling two 6-sided dice. One of them is a fair die. The other is unfair, where the numbers 1-4 are all equally likely to be rolled, but the number 5 is twice as likely as the number 1 to be rolled, and the number 6 is 3 times as likely as the number 1 to be rolled. What are these values from the probability distribution for the set of possible outcomes for the sum of the two dice
Answer:
is there a pic i need more info...
Step-by-step explanation:
A circle is centered on a point B. Points A, C and D lie on its circumference. if
Answer:
m\angle ABC=124^o
Step-by-step explanation:
A bag contains 20 marbles of which 4 are red what is the probability that a randomly selected marble will be red
Answer:
1/5
Step-by-step explanation:
4 of the 20 marbles are red, so the probability is 4/20 = 1/5.