Answer:
NO.
Step-by-step explanation:
From the question, we are given a random sample of 20 automotive batteries, a confidence interval =(0.22,0.33). Also, the population standard deviation is said to be less than 0.26 hour.
The confidence interval does NOT suggest that the variation in the batteries' reserve capacities is at an acceptable level.
REASON: If you look at the confidence interval, we can see that it contains the value for the standard deviation that is 0.26 and even MORE value than 0.26.
This does not give us any suggestion that the standard deviation is less than 0.26.
Final answer:
The confidence interval suggests that the variation in the batteries' reserve capacities is not at an acceptable level because the interval (0.22, 0.33) lies above the acceptable standard deviation of less than 0.26 hour.
Explanation:
In the context of the question, a constructed confidence interval for the population standard deviation of automotive batteries' reserve capacity is (0.22, 0.33 hours). To evaluate whether the variation in the batteries' reserve capacities is at an acceptable level, we would compare the interval to the stated acceptable level of a population standard deviation which should be less than 0.26 hour. Since the lower bound of the interval is above 0.22 and the entire confidence interval lies above 0.26, this suggests that the variability may be higher than the acceptable level, hence the variation is not at an acceptable level according to the specified criteria.
As for the example related to the NeverReady batteries, it demonstrates the use of sample data to question a company's claim about product performance by calculating the probability of obtaining a sample with a mean as low as or lower than observed when assuming that the company's claim is true. If the probability is very low, it casts doubt on the claim.
Smart phone users between 18 and 24 years of age send on average 64 text messages per day with a standard deviation of 10 messages about what percentage of smartphone users in this age groups in less than 51 text messages per day
Answer:
9.68%
Step-by-step explanation:
The revenue, in dollars, of a company that produces video game systems can be modeled by the expression 5x2 + 2x – 80. The cost, in dollars, of producing the video game systems can be modeled by 5x2 – x + 100, where x is the number of video game systems sold. If profit is the difference between the revenue and the cost, what expression represents the profit?
Profit can be modeled by the polynomial expression .
If 1,000 video game systems are sold, the company’s profit is $ .
Answer: 2,820 (use the profit equation from the first question )
Answer:
for edg its 3x - 180 and the other one is 2820
Step-by-step explanation:
bc i just took it
PLS HELPPPPPPP :((
Of 60 randomly chosen students surveyed, 16 chose the aquarium as their favorite field trip. There are 720 students in the school. Predict the number of students in the school who would choose the aquarium as their favorite field trip.
Answer:
Step-by-step explanation:
Final answer:
Using proportional reasoning from the sample survey where 16 out of 60 students favored the aquarium, it's predicted that approximately 192 out of 720 students in the entire school would prefer the aquarium as their favorite field trip.
Explanation:
We can predict the number of students who would choose the aquarium as their favorite field trip by using proportional reasoning based on the sample provided. In the survey, 16 out of 60 students chose the aquarium which gives us the ratio of 16/60 or approximately 0.267 (rounded to three decimal places).
To find out how many out of the entire school population of 720 students would likely choose the aquarium, we multiply the total number of students by this ratio:
(Total number of students) times (Ratio favoring aquarium)
720 times 0.267 = 192
Therefore, we would predict that approximately 192 students in the school would choose the aquarium as their favorite field trip.
An opinion poll asks a simple random sample of 100 college seniors how they view their job prospects. In all, 53 say "good." Does the poll give convincing evidence to conclude that more than half of all seniors think their job prospects are good? If p = the proportion of all college seniors who say their job prospects are good, what are the the hypotheses for a test to answer this question?
Answer:
We are tryng to proof if more than half of all seniors think their job prospects are good so that would be the alternative hypothesis and the complement rule would be the null hypothesis.:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
Step-by-step explanation:
Information provided
n=100 represent the random sample ofcollege senior selected
X=53 represent the college seniors who say good
[tex]\hat p=\frac{53}{100}=0.53[/tex] estimated proportion of seniors who think their job prospects are good
[tex]p_o=0.5[/tex] is the value that we want to test
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We are tryng to proof if more than half of all seniors think their job prospects are good so that would be the alternative hypothesis and the complement rule would be the null hypothesis.:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we have:
[tex]z=\frac{0.53 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=0.6[/tex]
The p value for this case would be given by
[tex]p_v =P(z>0.6)=0.274[/tex]
Mr. Acosta, a sociologist, is doing a study to see if there is a relationship between the age of a young adult (18 to 35 years old) and the type of movie preferred. A random sample of 93 adults revealed the following data. Test whether age and type of movie preferred are independent at the 0.05 level. Person's Age Movie 18-23 yr 24-29 yr 30-35 yr Row Total Drama 10 14 10 34 Science Fiction 11 9 10 30 Comedy 7 9 13 29 Column Total 28 32 33 93 (a) What is the level of significance
Answer:
There is a relationship between the age of a young adult and the type of movie preferred.
Step-by-step explanation:
In this case we need to test whether there is a significance relationship between the age of a young adult and the type of movie preferred.
The hypothesis can be defined as:
H₀: Age of a young adult and movie preference are independent.
Hₐ: Age of a young adult and movie preference are not independent.
The test statistic is given as follows:
[tex]\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}[/tex]
Consider the table below.
The formula to compute the expected frequencies is:
[tex]E_{i}=\frac{\text{Row Total}\times \text{Column Total}}{\text{Total}}[/tex]
The Chi-square test statistic value is:
[tex]\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}=2.6013[/tex]
The significance level of the test is, α = 0.05.
The degrees of freedom of the test is,
df = (r - 1)(c - 1)
= (3 - 1)(3 - 1)
= 2 × 2
= 4
Compute the p-value as follows:
p-value = 0.6266
*Use a Chi square table.
As the p-value is more than the significance level the null hypothesis was failed to be rejected.
Thus, concluding that there is a relationship between the age of a young adult (18 to 35 years old) and the type of movie preferred.
If rice costs $1.50 per kilogram, determine the cost of a pound of rice.
Given that $1.50 is the cost of one kilogram of rice and that one kilogram equals about 2.20462 pounds, the cost of one pound of rice will be about $0.68.
Explanation:To convert the costs from kilogram to pound, you need to know that one kilogram equals about 2.20462 pounds. Hence, if rice costs $1.50 per kilogram, we can calculate the cost per pound by dividing $1.50 by 2.20462.
This results in a cost of approximately $0.68 per pound of rice.
So, if you're looking to purchase a pound of rice under this cost scheme, it would set you back about $0.68.
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Find the value of x
Answer:
x =89
Step-by-step explanation:
The sum of the angles of a quadrilateral are 360 degrees
88+154+x+29 = 360
Combine like terms
271 +x = 360
271+x-271 = 360-271
x =89
The volume of Box A
of Box A is the volume of Box B. What is the height of Box A
has a base area of 32 square centimeters?
32 cm?
10 cm
Box B
Box A
8 cm
16 cm
Answer:
4cm; 1.79 cm.
Step-by-step explanation:
Okay, this question is about the Calculation involving the volume of a box with a square base and a height.
So, the volume of the box = length × width × height ---------------------------(1).
Therefore, we are going to make use of Equation (1) to determine the solution to this question, so that we have;
For Box A; We are given our volume to be equal to 32cm^3, height = 2cm, length and width = x cm.
For Box B = 32 = 2x^2.
x^2 = 16..
x = 4.
Note that Length = width.
For box B=> 32 = 10x^2.
x = √3.2 cm. = 1.79 cm.
Final answer:
The height of Box A with a base area of 32 square centimeters is 1 cm.
Explanation:
To find the height of Box A, we need to use the formula for the volume of a rectangular box, which is V = length * width * height. Given that the base area of Box A is 32 cm2 and assuming that the volume of Box A is equal to the volume of Box B, we can express their volumes as proportional to each other based on their dimensions.
Box A has a volume of 32 cm³ with a base area of 32 square centimeters.
Volume of Box A = Base Area x Height = 32 cm² x Height = 32 cm³
Height of Box A is 1 cm.
The sum of a brother's and sister's ages is 57 years.If the brother is 5 years older than his sister,find their ages.
Answer: Brother = 31, sister = 26
Step-by-step explanation:
first, take 57, and minus the difference of the ages (5) = 52. divide that between both of them, getting 26, and 26. Since the brother is 5 years older, add 5 to one of the 26's and you get 26, and 31.
Check if its correct: 26+31=57
We can make formulas for this.
Assume x=male age and y=female age.
From this problem, we know that x+y=57 and x-5=y since the brother is older. We can do substitution and plug in the second formula into the first.
x+(x-5)=57
2x-5=57 Add like terms (x)
2x=62 Isolate
x=31
Then plug this into the original to find out y.
x+y=57
31+y=57
y=26
Consider a drug testing company that provides a test for marijuana usage. Among 317 tested subjects, results from 25 subjects were wrong (either a false positive or a falsenegative). Use a 0.05 significance level to test the claim that less than 10 percent of the test results are wrong. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0H0: pequals=0.10.1 Upper H 1H1: pless than<0.10.1 B. Upper H 0H0: pless than<0.10.1 Upper H 1H1: pequals=0.10.1 C. Upper H 0H0: pequals=0.10.1 Upper H 1H1: pgreater than>0.10.1 D. Upper H 0H0: pequals=0.10.1 Upper H 1H1: pnot equals≠0.10.1 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is nothing. (Round to two decimal places as needed.) Identify theP-value for this hypothesis test. The P-value for this hypothesis test is nothing. (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test. A. Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. B. RejectReject Upper H 0H0. There is notis not sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. C. RejectReject Upper H 0H0. There isis sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong. D. Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to warrant support of the claim that less than 1010 percent of the test results are wrong.
Answer:
We conclude that we fail to reject [tex]H_0[/tex] as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.
Step-by-step explanation:
We are given that among 317 tested subjects, results from 25 subjects were wrong.
We have to test the claim that less than 10 percent of the test results are wrong.
Let p = proportion of subjects that were wrong.
So, Null Hypothesis, [tex]H_0[/tex] : p = 10% {means that 10 percent of the test results are wrong}
Alternate Hypothesis, [tex]H_A[/tex] : p < 10% {means that less than 10 percent of the test results are wrong}
The test statistics that would be used here One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of test results that were wrong = [tex]\frac{25}{317}[/tex] = 0.08
n = sample of tested subjects = 317
So, test statistics = [tex]\frac{0.08-0.10}{\sqrt{\frac{0.08 (1-0.08)}{317} } }[/tex]
= -1.31
The value of z test statistics is -1.31.
Now, the P-value of the test statistics is given by the following formula;
P-value = P(Z < -1.31) = 1 - P(Z [tex]\leq[/tex] 1.31)
= 1 - 0.9049 = 0.095
Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test. Since our test statistics is more than the critical value of z as -1.31 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that we fail to reject [tex]H_0[/tex] as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.
The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean student-loan debt is $27,524 and the standard deviation is $6,000. Is there sufficient evidence to support the student's claim at a 5% significance level? Preliminary: Is it safe to assume that n ≤ 5 % of all college students in the local area? No Yes Is n ≥ 30 ? Yes No Test the claim: Determine the null and alternative hypotheses. Enter correct symbol and value. H 0 : μ = H a : μ Determine the test statistic. Round to two decimals. t = Find the p -value. Round to 4 decimals. p -value = Make a decision. Fail to reject the null hypothesis. Reject the null hypothesis. Write the conclusion. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area. There is not sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 25235
For the alternative hypothesis,
µ > 25235
This is a right tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 27524
µ = population mean = 25235
s = samples standard deviation = 6000
t = (27524 - 25235)/(6000/√100) = 3.815
We would determine the p value using the t test calculator. It becomes
p = 0.000119
Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
The graph at the right
shows how a much a
company charges to ship
an item priority based on
the weight of the package.
According to the graph
how much would it cost to
ship a package that
weighs 3 pounds?
Answer:
$5
Step-by-step explanation:
Read the labels and find the relationship. Go the the bottom line labeled 'weight' and find the three pound mark. Then move up to find where the line is, showing the price for that weight. Once there, move to the left, labeled 'cost' and find the price for that weight.
Answer:
a.) $5
Step-by-step explanation:
On the x-axis (weight), find the 3-pound mark. Move up to find the price. As you can see, there are two possible prices for 3 pounds: $5 and $10.
Look at the dots. If a circle is filled in, it includes that value. If it is empty, it does not include the indicated value. Therefore, the dark blue circle is the correct price.
At this spot, move over to the left to find the price:
$5
:Done
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A survey by the Pew Research Center found that social networking is popular in many nations around the world. The file here contains the level of social media networking (measured as the percent of individuals polled who use social networking sites) and the GDP per capita based on purchasing power parity (PPP) for each of 25 selected countries. At the 0.001 level of significance, is there a significant linear relationship between GDP and social media usage?
Answer:
Check the explanation
Step-by-step explanation:
Social Median GDP (PPP), Y X^2 Y^2 XY
usage (%), X
18085 66 327067225 4356 1193610
15833 56 250683889 3136 886648
6117 47 37417689 2209 287499
6559 42 43020481 1764 275478
9948 40 98962704 1600 397920
15264 41 232989696 1681 625824
17522 50 307020484 2500 876100
4661 34 21724921 1156 158474
9833 54 96687889 2916 530982
5186 22 26894596 484 114092
18579 69 345179241 4761 1281951
19105 67 365001025 4489 1280035
13580 59 184416400 3481 801220
12115 49 146773225 2401 593635
15614 44 243796996 1936 687016
5289 45 27973521 2025 238005
7508 29 56370064 841 217732
11521 43 132733441 1849 495403
2824 32 7974976 1024 90368
3542 27 12545764 729 95634
1855 36 3441025 1296 66780
2056 35 4227136 1225 71960
3142 7 9872164 49 21994
1473 13 2169729 169 19149
Total 227211 1007 2984944281 48077 11307509
Sample size: n=24
Now,
Syy= 5824.95833
833909342.6
ry 1774114.125 ry
The coefficient of correlation is :
0.805
The coefficient of correlation indicates a strong and positive correlation between the variables.
(b)
For hypotheses correct option is (a).
The test statistics is
= 6.36
Degree of freedom: df=n-2= 22
The p-value is: 0.000
The critical value of t is to/2 = 2.07
Since p-value is less than 0.05 so we reject the null hypothesis. There is significant evidence ....
Jeff want to build a model to predict the relationship between graduate GPA and first job salary. He collected data below. Jeff want to use the regression model GPA (x) Annual Salary y xy X*x 2.8 19,815 55,482.00 7.84 3.4 23,037 78,325.80 11.56 3.2 21,963 70,281.60 10.24 3.8 25,185 95,703.00 14.44 3.2 22,396 71,667.20 10.24 3.4 24,388 82,919.20 11.56 Sum 19.8 136,784.00 454,378.80 65.88 Q27. What is the approximate intercept b0? Find the closest
Answer:
Approximate intercept is 4,515.33
Step-by-step explanation:
The intercept of the GPA grades and the first job salary can be determined by using the intercept function in excel spreadsheet.
In order to ascertain the intercept,we apply the intercept formula:
=intercept(known values of y,known values of x)
known values of y are the dependent variables.The first annual salary in this scenario is dependent upon the GPA of the graduate in question
Known values of x are the independent variables upon which the dependent ones rely for their values i.e the GPA itself
As found in the attached,the intercept is 4,515.33
A random survey of enrollment at 35 community colleges across the United States yielded the following figures:
6,414; 1,549; 2,108; 9,348; 21,829; 4,300; 5,942; 5,721; 2,825; 2,044; 5,481; 5,200; 5,855; 2,749; 10,012; 6,355; 26,999; 9,412; 7,681; 3,199; 17,498; 9,199; 7,381; 18,312; 6,557; 13,711; 17,769; 7,494; 2,769; 2,862; 1,264; 7,286; 28,166; 5,082; 11,621.
Assume the underlying population is normal.
Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. State the confidence interval
Final answer:
To construct a 95% confidence interval for the population mean enrollment at community colleges in the United States, we can use the formula: Confidence Interval = sample mean ± (critical value) × (sample standard deviation / √n). Using the provided sample data, the confidence interval is approximately (5,814.61, 9,173.27).
Explanation:
To construct a 95% confidence interval for the population mean enrollment at community colleges in the United States, we can use the formula:
Confidence Interval = sample mean ± (critical value) × (sample standard deviation / √n)
Given the sample enrollment figures, we can calculate the sample mean as 7,493.94 and the sample standard deviation as 6,650.43. The critical value for a 95% confidence level for a normal distribution is approximately 1.96. With n = 35, the sample size, we can substitute these values into the formula to find the confidence interval:
Confidence Interval = 7,493.94 ± (1.96) × (6,650.43 / √35)
Calculating the values, the confidence interval for the population mean enrollment at community colleges in the United States is approximately (5,814.61, 9,173.27).
B) the perimeter of the shape is 25.71cm
calculate the value of the radius x.
take pi to be 3.142
We have been given that the perimeter of a semicircle is 25.71 cm. We are asked to find the value of the radius.
We know that perimeter of a semicircle is [tex]\pi r+2r[/tex], where r is radius of perimeter.
Upon equating perimeter of semicircle formula with 25.71, we will get:
[tex]\pi r+2r=25.71[/tex]
[tex]3.142r+2r=25.71[/tex]
[tex]5.142r=25.71[/tex]
[tex]\frac{5.142r}{5.142}=\frac{25.71}{5.142}[/tex]
[tex]r=5[/tex]
Therefore, the radius of the semicircle is 5 cm.
One 50 pound of grass seed is enough to seed 10,000 square feet of ground. If your lawn is 400 feet by 285 feet how many packages of seed do you need to seed your entire lawn?
To seed the entire lawn, you will need 11.4 packages of grass seed.
Explanation:To determine the number of packages of grass seed needed, we need to find the area of the lawn. The area of a rectangle is given by the formula length * width. In this case, the length is 400 feet and the width is 285 feet. Multiplying these values together, we get an area of 114,000 square feet. Now, we can determine the number of packages needed by dividing the area of the lawn by the area covered by one package of seed. Since one package covers 10,000 square feet, we divide 114,000 by 10,000 to get 11.4 packages.
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A total of 12 packages needed.
To determine how many packages of grass seed are needed to seed a lawn of 400 feet by 285 feet, follow these steps:
Calculate the lawn's total area by multiplying its length by its width: 400 feet * 285 feet = 114,000 square feet.
Since one 50-pound package of grass seed is enough for 10,000 square feet, divide the lawn's total area by 10,000 square feet to find the number of packages needed: 114,000 square feet / 10,000 square feet per package = 11.4 packages.
Because you can't purchase a fraction of a package, you'll need to round up to the nearest whole number, which means you require 12 packages of grass seed to cover the entire lawn.
It's also important to consider the seeding rate when determining the amount of seed required. However, in this case, the seeding rate is not needed since the question specifies one package covers a set amount of square footage.
Mrs. Jones wants to cover half of her 5 x 7 rectangular bulletin board with two different colors (white and black) as show in the picture below. How much black paper does she need?
A bucket contains exactly 3 marbles, one red, one blue and one green. A person arbitrarily pulls out each marble one at a time. Given the following: Event 1: The first marble removed is non-red Event 2: The last marble removed is non-red Which of the following statements is true? Group of answer choices The probability of either of these events happening is 33%. The probability of Event 1 occurring is the same as the probability of Event 2 occurring. Event 1 is independent of Event 2. The probability of both events happening is 50%.
Answer:
Check the explanation
Step-by-step explanation:
here are 6 possible sequences of removing 3 marbles. Let R be the red marble, B be the blue marble and G be the green marble. The possible sequences are: RBG, RGB, BGR, BRG, GRB and GBR.
Let A denotes Event 1 and B denotes Event 2.
The favorable cases for A are: BGR, BRG, GRB and GBR.
So, P(A) = 4/6 = 2/3.
The favorable cases for B are: RBG, RGB, BRG and GRB.
So, P(B) = 4/6 = 2/3.
The probability of both events happening is P(A \cap B).
Favorable cases for (A \cap B) are BRG and GRB.
Thus, P(A \cap B) = 2/6 = 1/3.
The probability of either of these events happening is P(A \cup B).
Favorable cases for (A \cup B) are RBG, RGB, BRG, BGR, GRB and GBR.
Thus, P(A \cup B) = 6/6 = 1.
Also, P(A \cap B) \neq P(A) P(B).
So, A and B are not independent.
Hence, Option (D) is the only true statement. (Ans).
Foster makes sandwiches at a deli. There are 10 1/2 pounds of turkey at the deli. How many 1/4 -pound turkey sandwiches can he make?
Answer:
foster can make 42 1/4-pound turkey sandwiches!!!
explanation: 10.5 divided by 1/4 is 42
Answer:
convert pounds to ounces. theres 16 oz. in a pound.
16x10lb=160 oz + 8 additional ounces = 168 oz of meat. theres 4 oz in 1/4 lb so 168 ÷ 4 = 42 sandwiches
A Broadway theater has 700 seats, divided into orchestra, main, and balcony seating. Orchestra seats sell for $ 50 comma main seats for $ 40 comma and balcony seats for $ 25. If all the seats are sold, the gross revenue to the theater is $ 26 comma 250. If all the main and balcony seats are sold, but only half the orchestra seats are sold, the gross revenue is $ 22 comma 750. How many are there of each kind of seat?There are orchestra seats?
main seats?
balcony seats?
Answer:
orchestra seats = 140 seats
main seats = 350 seats
balcony seats = 210 seats
Step-by-step explanation:
Let x = number of orchestra seats
Let y = number of main seats
Let z = number of balcony seats
Theater has 700 seats. Thus;
Total seat equation is;
x + y + z = 700 - - - - (1)
We are told Orchestra seats sell for $ 50, main seats for $ 40, and balcony seats for $ 25 and a gross revenue of $26,250
Thus ;
all seats revenue equation is;
50x + 40y + 25z = 26250 - - - (eq2)
Now, we are told that, when all the main and balcony seats are sold, but only half the orchestra seats are sold, the gross revenue is $ 22,750
Thus, we have;
½(50x) + 40y + 25z = 22,750
Which gives ;
25x + 40y + 25z = 22,750 ---(eq3)
Solving the 3 equations simultaneously, we have;
x = 140
y = 350
z = 210
Renee has a triangular garden with an area of 24 square feet. Which drawing shows Renee's garden?
3 ft.
6 ft.
6
ft.
Answer:
3 ft
Step-by-step explanation:
What is the area of 1” by 3”
Answer:
2 sq. ft. ?
Step-by-step explanation:
hope we can be friends
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Roni recently bought a car for $18,000. Her research shows the car will depreciate by an average of $1,500 per year. If x represents the number of years Roni owns the car, which of the following equations represents the value of the car after x years?
A.
y = 18,000x - 1,500
B.
y = 1,500x - 18,000
C.
y = 18,000 + 1,500x
D.
y = 18,000 - 1,500x
Answer:
D is the answer
Step-by-step explanation:
If every year the value decreases by $1,500 that variable will change, but the price he bought the car at will not. That is why the x variable(standing for years) would go into the 1,500. And if each year the value decreases the answer would be D
Answer: D. y=18,000 - 1,500x
Step-by-step explanation: the answer is d.) because the cars value will go down as time pas, so 1,500 represent the money and x could represent one year and you subtract it from 18,000 which is the original value of the car.
Find the degree of the monomial.
8m2n4
Answer:
6
Step-by-step explanation:
The degree of 8m^2n^4 is the sum of the exponents of the variables:
2 + 4 = 6
The monomial has degree 6.
Maria drew a number line divided into 8 equal parts what fraction names point b on the number line ?
Answer:
The fraction that named point b is one-quarter
Step-by-step explanation:
Given
Nunber line divided into 8 parts (see attachment)
Required
Name the fraction
Your question is incomplete. However, I'll make use of my attachment to answer the question.
If you follow the steps I'll highlight, you'll arrive at your answer.
To name a fraction using a number line, first you need to count the total number of partitions on the number line.
In this case, it is 8.
Then we count the number of partitions or points from the index point till the point that needs to be named.
Between index 0 and point b, there are 2 points.
The next step is to represent this using a fraction.
The fraction is 2/8
I.e 2 points out of a total of 8 points
Fraction = 2/8 (Simplify fraction)
Fraction = ¼
The fraction can no longer be simplified;
The next step is to name it.
¼ in words in 0ne quarter.
Hence, the fraction that named point b on the number line is one-quarter
The management of ABC Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using a 0.1 level of significance, can we conclude that the assembly time using the new method is faster? Use only the P-Value approach. State H0 and Ha (20 pts)
Answer:
Null hypothesis:[tex]\mu \geq 42.3[/tex]
Alternative hypothesis:[tex]\mu < 42.3[/tex]
[tex]t=\frac{40.6-42.3}{\frac{2.7}{\sqrt{24}}}=-3.085[/tex]
The degrees of freedom are given by:
[tex]df=n-1=24-1=23[/tex]
Then the p value can be calculated with this probability:
[tex]p_v =P(t_{(23)}<-3.085)=0.0026[/tex]
If we compare the p value and the significance lvel of 0.1 we see that the p value is lower than the signficance level we can reject the null hypothesis and we can conclude that the the assembly time using the new method is faster
Step-by-step explanation:
Information given
[tex]\bar X=40.6[/tex] represent the sample mean for the new method
[tex]s=2.7[/tex] represent the sample standard deviation
[tex]n=24[/tex] sample size
[tex]\mu_o =42.3[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the assembly time using the new method is faster, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 42.3[/tex]
Alternative hypothesis:[tex]\mu < 42.3[/tex]
The statistic for this case is given:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
The statistic would be given by:
[tex]t=\frac{40.6-42.3}{\frac{2.7}{\sqrt{24}}}=-3.085[/tex]
The degrees of freedom are given by:
[tex]df=n-1=24-1=23[/tex]
Then the p value can be calculated with this probability:
[tex]p_v =P(t_{(23)}<-3.085)=0.0026[/tex]
If we compare the p value and the significance lvel of 0.1 we see that the p value is lower than the signficance level we can reject the null hypothesis and we can conclude that the the assembly time using the new method is faster
A 2-column table with 10 rows. Column 1 is labeled Classes with entries Hall, Benny, Leggo, Talle, Flower, Gomez, Range, Book, Toledo, Rich. Column 2 is labeled Watch News with entries 35, 45, 26, 32, 46, 38, 39, 40, 26, 72. Number 72 is crossed out.
Carol wanted to see how the data would change if she did not include the outlier in her study. The results are shown in the table with the outlier crossed out.
Find each listed value.
Range =
Median =
Lower quartile and upper quartile =
Interquartile range =
Which value was affected the most?
Answer:
range: 20
median: 38
Lower quartile and upper quartile: 29 and 42.5
Interquartile range: 13.5
Which value was affected the most: range
The value of the range was mostly affected after the outlier crossed out.
This is determined by calculating all the values before and after the outlier 72.
How the range of data is calculated?The range of data is calculated by taking the difference between the highest value and the lowest value of the data set.
Range=maximum value - minimum value
Given data:Given a table of data points as 35, 45, 26, 32, 46, 38, 39, 40, 26, 72.
Finding the values given in the list before crossing out the outlier 72:
Range: 72-26=46
Median: 38.5
Arranging in the ascending order
26, 26, 32, 35, 38, 39, 40, 45, 46, 72
Then, median=[tex]\frac{38+39}{2}[/tex]=38.5
Lower quartile: Q1=32
Upper quartile: Q3=45
IQR: Q3-Q1=45-32=13
Finding the values given in the list after crossing out the outlier 72:
The data set now has 9 elements with data points as 35, 45, 26, 32, 46, 38, 39, 40, 26
Range: 46-26=20
Median: 38
Arranging in the ascending order
26, 26, 32, 35, 38, 39, 40, 45, 46
Since odd number of data points, the median is at the middle point which is 38
Lower quartile: Q1=29
[tex]Q_1=\frac{1}{4} (n+1)[/tex] th term (n=9)
Q1=10/4=2.5
So, the median is between 26 and 32 which is 29
Upper quartile: Q3=42.5
[tex]Q_3=\frac{3}{4} (n+1)[/tex] th term (n=9)
Q3=7.5
So, the median is between 40 and 45 which is 42.5
IQR: Q3-Q1=42,5-29=13.5
From these results, it is observed that only the range is affected mostly when outlier 72 is removed.
Learn more about the range here:
https://brainly.com/question/16399146
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DEL
Simplify this algebraic expression. 8b -
2b - 4*
ELLER
Step-by-step explanation:
Given expression is [tex]8b -2b -4[/tex]
To simplify any expression we combine like terms
Any expression contains constant and variables.
the terms that contains variables are 8b and -2b .the constant term is -4
To simplify combine the like terms 8b and -2b
[tex]8b-2b= 6b[/tex]
[tex]8b -2b -4[/tex]
[tex]6b -4[/tex]
So the simplified expression is [tex]6b -4[/tex]
in simplest form 3n-13-5n+6n
Answer:
THE ANSWER IS 4n - 13