Answer:
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 50.3 minutes
Standard deviation = 3.6 minutes
Within what range would you expect 95% of your customers to stay in your store?
Within 2 standard deviations of the mean.
So from
50.3 - 2*3.6 = 43.1 minutes
To
50.3 + 2*3.6 = 57.5 minutes
You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.
You roll the same die three times. Consider the possible outcomes if the order of the resultsis not recorded (meaning, e.g., that 1-2-1 and 2-1-1 are considered the same outcome).(a) Order the possible results in lexicographical order and show by direct counting that thenumber of possible outcomes is
Answer: 56 outcomes
Step-by-step explanation:
Tossing a die 3 times and not having the order recorded, we have the following outcome.
[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]
[1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],
[1,3,3] [1,3,4] [1,3,5] [1,3,6]
[1,4,4] [1,4,5] [1,4,6]
[1,5,5] [1,5,6]
[1,6,6]
[2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],
[2,3,3] [2,3,4] [2,3,5] [2,3,6]
[2,4,4] [2,4,5] [2,4,6]
[2,5,5] [2,5,6]
[2,6,6]
[3,3,3] [3,3,4] [3,3,5] [3,3,6]
[3,4,4] [3,4,5] [3,4,6]
[3,5,5] [3,5,6]
[3,6,6]
[4,4,4] [4,4,5] [4,4,6]
[4,5,5] [4,5,6]
[4,6,6]
[5,5,5] [5,5,6]
[5,6,6]
[6,6,6]
Hence, By direct counting, the number of possible outcome is 56 outcomes.
Find the sample space for the experiment.
A taste tester ranks three varieties of yogurt, A, B, and C, according to preference.
Answer:
S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}
And we have 6 possible outcomes for this case.
Step-by-step explanation:
By definition the sample space of an experiment "is the set of all possible outcomes or results of that experiment".
For the case described here: "A taste tester ranks three varieties of yogurt, A, B, and C, according to preference.".
Assuming that we have three varieties of yogurt {A,B,C}
We denote the event (A,B,C) like this: A is the first, B the second and C the third in the rank. And for example the event (C,B,A) means that C is on the 1th position of the rank, B on the 2nd position and A on the 3th position.
The sampling space denoted by S and is given by:
S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}
And we have 6 possible outcomes for this case.
The sample space for ranking three yogurt varieties (A, B, C) consists of 6 permutations: (A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), and (C, B, A).
The problem involves finding all possible ways a taste tester can rank three varieties of yogurt: A, B, and C. In this context, a sample space represents all possible outcomes of the experiment.
Since the taste tester is ranking the yogurt varieties, we are looking for all permutations of the set {A, B, C}.
A permutation is an arrangement of all the members of a set in a specific order. For three items, the total number of permutations is given by 3! (3 factorial), which equals 6.
Therefore, the sample space S, containing all possible rankings, is:
(A, B, C)(A, C, B)(B, A, C)(B, C, A)(C, A, B)(C, B, A)Three people have been nominated for president of a class. From a poll, it is estimated that the first candidate has a 37% chance of winning and the second candidate has a 44% chance of winning. What is the probability that the third candidate will win?
Answer:
19%
Step-by-step explanation:
We assume that the class has to have a president. So the chances of none of the candidates winning is 0%. In turn we expect that one of the candidates will win the elections, therefore the chances of someone winning is 100%. Therefore the chances of the third candidate winning can be calculated by removing the chances of the other two candidates winning
P(third candidate winning) = 100% - (37%+44%) = 19%
You receive a fax with six bids (in millions of dollars):2.2,1.3,1.9,1.2 2.4 and x is some number that is too blurry to read. Without knowing what x is, the median a. Is 1.9 b. Must be between 1.3 and 2.2 c. Could be any number between 1.2 and 2.4
The median of a set of bids can be found by arranging them in numerical order and selecting the middle value.
Explanation:The median is the middle value of a set of data arranged in numerical order. In this case, we have a set of six bids: 2.2, 1.3, 1.9, 1.2, 2.4, and x (blurred number). To find the median, we first need to arrange the bids in numerical order:
1.21.31.92.22.4xSince there are six bids, the middle value will be the fourth number in the ordered list. Therefore, the median is 2.2.
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In a large sample of customer accounts, a utility company determined that the average number of days between when a bill was sent out and when the payment was made is 32 with a standard deviation of 7 days. Assume the data to be approximately bell-shaped.. Between what two values will approximately 68% of the numbers of days be?. Estimate the percentage of customer accounts for which the number of days is between 18 and 46.. Estimate the percentage of customer accounts for which the number of days is between 11 and 53.
Answer:
Question 1: between 25 and 39 daysQuestion 2: about 95%Question 3: about 99.7%
Explanation:
Question 1. Between what two values will approximately 68% of the numbers of days be?.
You can answer based on the 68-95-99.7% rule. As per this rule, about 68% of the data of a normal distribution (bell shaped) are within one standard deviation of the mean.
Here the mean is 32 day and the standard deviation is 7 day. Then 68% are in the interval 32 days ± 7 days.
That is:
32 days + 7 days = 39 days32 days - 7 days = 25 daysConsequently, approximately 68% of the numbers of days will be between 25 and 39 days.
Question 2. Estimate the percentage of customer accounts for which the number of days is between 18 and 46.
First must determine the Z-scores both both values X = 18 and X = 46
The formula is:
[tex]Z-score = (X-mean)/(standard\text{ }deviation)[/tex]
For X = 18[tex]Z-score=(18-32)/7=-2[/tex]
For X = 46[tex]Z-score=(46-32)/7=2[/tex]
Hence, you want to estimate the percentage of customers accounts for which the the number of days is within 2 standard deviations of the mean.
As per the 68-95-99.7 rule about 95% of the data are within 2 standard deviations of the mean. You can calculate it also from a standard normal distribution table, finding the area to the left of Z-score = - 2 and subtracting the area to the right of Z-score equal to 2: That is: 0.9772 - 0.0228 = 0.9484 = 95.44% ≈ 95%.
Question 3. Estimate the percentage of customer accounts for which the number of days is between 11 and 53.
Again, determine the Z-scores for the two values, X = 11 and X = 53.
For X = 11:[tex]Z-score=(11-32)/7=-3[/tex]
For X = 53:[tex]Z-score=(53-32)/7=3[/tex]
Hence, you want to estimate the probability of the number of days s between - 3 and 3 standard deviations.
Such probability is about 99.7%, according to the 68-95-99.7 rule.
If you use a standard distritution table you will find that the area to the right of the Z-score of -3 is 0.99865, thus the probability of the Z-score be to the right of 3 is 1 - 0.99865 = 0.00135.
And the probability in between -3 and 3 standard deviations is 0.99865 - 0.00135 = 0.9973 = 99.73% ≈ 99.7%.
Estimate the percentage of bills for number 39 when a bill was sent and when payment was made Answer:
Step-by-step explanation: yes
A single tree produces about 2.6x102 pounds of oxygen each year. the amazon rainforest has about 3.9 X 10 [ ] pounds per day enter your answer by filling in the box
Answer:
The answer to your question is 2.8 x 10¹¹
Step-by-step explanation:
Data
1 tree produces 2.6 x 10² pounds of oxygen/year
number of trees = 3.9 x 10¹¹
pounds of oxygen per day = ?
Process
1.- Divide the pounds of oxygen by 365, to get the pounds of oxygen per day.
2.6 x 10² / 365 = 0.712
2.- Multiply the number of trees by the pounds of oxygen per day
3.9 x 10¹¹ x 0.712 = 2.8 x 10¹¹ pounds of oxygen
Roll one fair, six-sided die, numbered 1 through 6. Let A be the event you will roll an even number. Let B be the event you will roll a prime number. Enter 1 for Yes, and enter 0 for No. = Yes = No 1 (a) Are A and B independent events? (b) Are A and B mutually exclusive events?
Answer:
a) not independent
b) not mutually exclusive
Step-by-step explanation:
Given:
- A 6 sided die is rolled
- Event A is rolling an even number
- Event B is rolling a prime number
Find
- (a) Are A and B independent events?
- (b) Are A and B mutually exclusive events?
Solution:
- We will find the probability of each event:
set(Even number: A) = {2, 4, 6} = 3 outcomes
set(Prime number: B) = {2 , 3, 5} = 3 outcomes
- The probabilities are:
P(A) = 3/6 = 0.5
P(B) = 3/6 = 0.5
- For Event A and B to be independent then the following condition must match:
P ( A & B ) = P(A)*P(B)
set (A&B) = {2} = 1 outcome
P(A&B) = 1/6
1 / 6 = 0.5*0.5
1/6 = 0.25 ...... NOT INDEPENDENT
- For Event A and B to be mutually exclusive then the following condition must match:
P(A&B) = 0
P(A&B) = 1/6
Hence, we can say the events are NOT MUTUALLY EXCLUSIVE
No, events A and B are not independent events because the probability of both events occurring is not equal to the product of their individual probabilities. Furthermore, events A and B are not mutually exclusive events because they can both occur simultaneously if the number rolled is 2.
No, events A and B are not independent events. In order for two events to be independent, the probability of both events occurring should be equal to the product of the probabilities of each event occurring individually. However, in this case, the probability of rolling an even number (event A) is 3/6 (since there are three even numbers out of six total outcomes), and the probability of rolling a prime number (event B) is 2/6 (since there are two prime numbers out of six total outcomes). Therefore, P(A and B) is not equal to P(A) * P(B), indicating that events A and B are dependent.
Moreover, events A and B are not mutually exclusive events either. Mutually exclusive events are events that cannot occur at the same time, meaning they have no outcomes in common. In this case, event A (rolling an even number) and event B (rolling a prime number) can both occur simultaneously if the number rolled is 2, which is both even and prime. Therefore, P(A and B) is not equal to 0, indicating that events A and B are not mutually exclusive.
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What symbol is used for the arithmetic mean when it is a sample statistic? What symbol is used when the arithmetic mean is a population parameter?
Answer:
for sample = xbar
population = μ
Step-by-step explanation:
The arithmetic mean for sample can be represented by xbar and it can be calculated as
xbar=∑xi/n
Where xi represents data values and n represents number of data values in a sample.
The arithmetic mean for population can be represented by μ and it can be calculated as
μ=∑xi/N
Where xi represents data values and N represents number of data values in a population.
The symbol for the arithmetic mean when it's a sample statistic is 'x', and 'μ' when it's a population parameter. Sample mean (x) refers to the mean of a sample group, while population mean (μ) refers to the mean of an entire population.
Explanation:The symbol that is used for the arithmetic mean when it is a sample statistic is 'x' (pronounced as x-bar). This is used to denote the mean of a sample. On the other side, the symbol that is used when the arithmetic mean is a population parameter is 'μ' (pronounced as mu). This symbol is representative of the mean of an entire population.
Take for example a group of 50 students in a class who took a test. If you wanted to find the sample mean (x) of their scores, you'd add all their scores and divide by the total number of students (which is 50 in this case). However, if you were to calculate the population mean (μ) of the scores of all students in all high school classes in the nation, you'd add their scores and divide by the total number.
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Jacinta babysat for 2 1/2 hours each Saturday for 6 weeks she was paid $8.50 each hour what is the total amount that Jacinta was paid for babysitting
Answer:
Jacinta was paid $127.5 for babysitting.
Step-by-step explanation:
This problem can be solved by consecutive rules of three.
Jacinta babysat for 2 1/2 hours each Saturday for 6 weeks
How many hours did she work?
She worked for 6 Saturdays, 2.5 hours in each. So
1 Saturday - 2.5 hours
6 Saturdays - x hours
[tex]x = 6*2.5[/tex]
[tex]x = 15[/tex]
She worked 15 hours.
Paid $8.50 each hour
15*$8.50 = 127.5
Jacinta was paid $127.5 for babysitting.
A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacement from the batch. What is the probability that the second one selected is defective given that the first one was defective? Round your answer to five decimal places (e.g. 98.76543). What is the probability that both are defective? Round your answer to seven decimal places (e.g. 98.7654321). What is the probability that both are acceptable? (e.g. 98.765). Three containers are selected, at random, without replacement, from the batch. What is the probability that the third one selected is defective given that the first and second one selected were defective? (e.g. 98.765). What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? Round your answer to five decimal places (e.g. 98.76543). What is the probability that all three are defective? (e.g. 98.765).
Answer:
When Two containers are selected(a) Probability that the second one selected is defective given that the first one was defective = 0.00450
(b) Probability that both are defective = 0.0112461
(c) Probability that both are acceptable = 0.986
2. When Three containers are selected
(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.
(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451
(c) Probability that all three are defective = 6.855 x [tex]10^{-8}[/tex] .
Step-by-step explanation:
We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.
Total containers = 445
Defective ones = 3
Non - Defective ones = 442 { Acceptable ones}
Two containers are selected, at random, without replacement from the batch.(a) Probability that the second one selected is defective given that the first one was defective is given by;
Since we had selected one defective so for selecting second the available
containers are 444 and available defective ones are 2 because once
chosen they are not replaced.
Hence, Probability that the second one selected is defective given that the first one was defective = [tex]\frac{2}{444}[/tex] = 0.00450
(b) Probability that both are defective = P(first being defective) +
P(Second being defective)
= [tex]\frac{3}{445} + \frac{2}{444}[/tex] = 0.0112461
(c) Probability that both are acceptable = P(First acceptable) + P(Second acceptable)
Since, total number of acceptable containers are 442 and total containers are 445.
So, Required Probability = [tex]\frac{442}{445}*\frac{441}{444}[/tex] = 0.986
Three containers are selected, at random, without replacement from the batch.(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;
Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .
Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = [tex]\frac{1}{443}[/tex] = 0.002.
(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;
Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.
Hence, Required probability = [tex]\frac{2}{443}[/tex] = 0.00451 .
(c) Probability that all three are defective = P(First being defective) +
P(Second being defective) + P(Third being defective)
= [tex]\frac{3}{445}* \frac{2}{444} * \frac{1}{443}[/tex] = 6.855 x [tex]10^{-8}[/tex] .
Final answer:
The probability values are found by considering the number of defective and non-defective containers remaining at each step of selection.
Explanation:
The probability that the second one selected is defective given that the first one was defective can be found by looking at the remaining pool of containers. Initially, there are 3 defective containers out of 445.
After picking one, there are 2 defective containers left out of 444. The probability is therefore calculated as 2/444 = 0.00450 (rounded to five decimal places).
The probability that both containers are defective is calculated by multiplying the probability of selecting a defective container on the first draw with the probability of selecting a defective container on the second draw: (3/445) * (2/444) = 0.00003 (rounded to seven decimal places).
The probability that both containers are acceptable is the probability of not selecting a defective container twice: (442/445) * (441/444) = 0.98876.
If the first and second containers selected are defective, the probability that the third one selected is defective comes from the remaining one defective container out of 443: 1/443 = 0.00226.
If the first container is defective and the second one is not, the probability that the third one selected is defective is 2/443 = 0.00451 (rounded to five decimal places).
The probability that all three containers selected are defective is the product of the probabilities of selecting a defective container each time without replacement: (3/445) * (2/444) * (1/443) = 0.00002.
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into two pieces. Steve takes the first piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. What should the lengths of the wires be so that the total area of the circle and square combined is as small as possible? (Round your answers to two decimal places.)
Answer:
a) the length of the wire for the circle = [tex](\frac{60\pi }{\pi+4}) in[/tex]
b)the length of the wire for the square = [tex](\frac{240}{\pi+4}) in[/tex]
c) the smallest possible area = 126.02 in² into two decimal places
Step-by-step explanation:
If one piece of wire for the square is y; and another piece of wire for circle is (60-y).
Then; we can say; let the side of the square be b
so 4(b)=y
b=[tex]\frac{y}{4}[/tex]
Area of the square which is L² can now be said to be;
[tex]A_S=(\frac{y}{4})^2 = \frac{y^2}{16}[/tex]
On the otherhand; let the radius (r) of the circle be;
2πr = 60-y
[tex]r = \frac{60-y}{2\pi }[/tex]
Area of the circle which is πr² can now be;
[tex]A_C= \pi (\frac{60-y}{2\pi } )^2[/tex]
[tex]=( \frac{60-y}{4\pi } )^2[/tex]
Total Area (A);
A = [tex]A_S+A_C[/tex]
= [tex]\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2[/tex]
For the smallest possible area; [tex]\frac{dA}{dy}=0[/tex]
∴ [tex]\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0[/tex]
If we divide through with (2) and each entity move to the opposite side; we have:
[tex]\frac{y}{18}=\frac{(60-y)}{2\pi}[/tex]
By cross multiplying; we have:
2πy = 480 - 8y
collect like terms
(2π + 8) y = 480
which can be reduced to (π + 4)y = 240 by dividing through with 2
[tex]y= \frac{240}{\pi+4}[/tex]
∴ since [tex]y= \frac{240}{\pi+4}[/tex], we can determine for the length of the circle ;
60-y can now be;
= [tex]60-\frac{240}{\pi+4}[/tex]
= [tex]\frac{(\pi+4)*60-240}{\pi+40}[/tex]
= [tex]\frac{60\pi+240-240}{\pi+4}[/tex]
= [tex](\frac{60\pi}{\pi+4})in[/tex]
also, the length of wire for the square (y) ; [tex]y= (\frac{240}{\pi+4})in[/tex]
The smallest possible area (A) = [tex]\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})[/tex]
= 126.0223095 in²
≅ 126.02 in² ( to two decimal places)
What is the area of this composite shape?
27
35
60
40
Answer:
Answer: 40 sq. in.
Step-by-step explanation:
First we gotta find the area of triangle part of the shape, we can see the left side of the shape is 5 in. , so we subtract the 3 from 5 which gives 2 the height of the triangle.
Now, we find the length for the base of the triangle, the top part of the shape is 7 in. , so we subtract 7 from 12 which gives us 5 in. as the base
Now, we find the area of the triangle:
A = [tex]\frac{1}{2}[/tex]b × h
A = ([tex]\frac{1}{2}[/tex] × 5 in,) × 2 in
A = 5 sq. in.
Now we find the area for the rectangle:
A = b × h
A = 5 x 7
A = 35 sq. in
Finally, we add the areas together
35 sq. in. + 5 sq. in. = 40 sq. in.
We get our answer 40 sq. in,
An electronics company has two production facilities, denoted A and B. During an average week, facility A produces 2000 computer monitors and 10,000 flat panel televisions, and facility B produces 3000 computer monitors and 13,000 flat panel televisions.
Determine the number of weeks of production from A and B required to produce 28,000 monitors and 132,000 televisions.
Answer:
8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.
Step-by-step explanation:
Let the number of weeks of production for team A be x and for team B be y.
In a week, facility A produces 2,000 computer monitors.
In a week ,facility A produces 3,000 computer monitors.
Total monitors to be produced = 28,000
[tex]2000x+3000y=28,000[/tex]
[tex]2x+3y=28[/tex]..[1]
In a week, facility A produces 10,000 flat panel televisions.
In a week ,facility A produces 13,000 flat panel televisions.
Total flat panel televisions to be produced = 132,000
[tex]10,000x+13,000y=132,000[/tex]
[tex]10x+13y=132[/tex]..[2]
Solving equation [1] and [2] by eliminationg method.
5 × [1] + (-1) × [2]
[tex]10x+15y=140[/tex]
[tex]-10x-13y=-132[/tex]
y = 4
[tex]x = \frac{28-3\times 4}{2}=\frac{16}{2}=8[/tex]
8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.
To produce 28,000 monitors and 132,000 televisions, it would take approximately 3.5 weeks for monitors and 3.67 weeks for televisions.
Explanation:To determine the number of weeks of production required, we need to divide the total number of monitors and televisions needed by the production rate of each facility.
For monitors: Facility A produces a total of 2000 + 3000 = 5000 monitors per week, and Facility B produces 3000 monitors per week. Therefore, the total production rate for monitors is 5000 + 3000 = 8000 monitors per week.
For televisions: Facility A produces a total of 10,000 + 13,000 = 23,000 televisions per week, and Facility B produces 13,000 televisions per week. Therefore, the total production rate for televisions is 23,000 + 13,000 = 36,000 televisions per week.
To produce 28,000 monitors and 132,000 televisions, it would take:
Monitors: 28,000 / 8000 = 3.5 weeks
Televisions: 132,000 / 36,000 = 3.67 weeks
Therefore, the number of weeks required to produce the desired quantities is approximately 3.5 weeks for monitors and 3.67 weeks for televisions.
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Swinging Sammy Skor's batting prowess was simulated to get an estimate of the probability that Sammy will get a hit. Let 1- HIT and 0 OUT. The output from the simulation was as follows 10001001001110000111 1000011001111 000001111 Estimate the probability that he gets a hit. A) 0.301 B) 0.286 C) 0.452 D) 0.476
Answer:
The answer is D) 0.476
Step-by-step explanation:
If 1 represents HIT, and 0 represents OUT, the probability that Sammy will get a hit = the number of HITS (1s) ÷ the output from the simulation (i.e., total number of HITs and OUTs in the simulation)
Where:
the number of HITS (1s) = 20
the output from the simulation (i.e., total number of HITs and OUTs in the simulation) = 42
therefore, the probability that Sammy will get a hit = 20 ÷ 42 = 0.476
This implies that the correct answer is D) 0.476
Find all x ∈ Z satisfying each of the following equations.
(a) 3x ≡ 2 (mod 7)
(b) 5x + 1 ≡ 13 (mod 23)
(c) 5x + 1 ≡ 13 (mod 26)
(d) 9x ≡ 3 (mod 5)
(e) 5x ≡ 1 (mod 6)
(f) 3x ≡ 1 (mod 6)
Answer:
answers are shown in the file attached
Step-by-step explanation:
The detailed step are as shown in the attached file.
The solution of the all the equations are:
a) x = {10 + 7k}, where k is an integer.
b) x = {2 + 23k}, where k is an integer.
c) x = {18 + 26k}, where k is an integer.
d) x = {2 + 5k}, where k is an integer.
e) x = {5 + 6k}, where k is an integer.
f) x = {1 + 6k}, where k is an integer.
To find all integer solutions (x ∈ Z) for each equation, we need to solve them using modular arithmetic.
For each equation, we will use different approaches depending on the specific form of the equation.
Let's solve them one by one:
(a) 3x ≡ 2 (mod 7):
To find x, we'll multiply both sides of the equation by the modular inverse of 3 modulo 7, which is 5 (since 3 × 5 ≡ 1 (mod 7)):
3x × 5 ≡ 2 × 5 (mod 7)
15x ≡ 10 (mod 7)
Now, we find the smallest non-negative integer solution by dividing both sides by the greatest common divisor (GCD) of 15 and 7 (which is 1 since 15 and 7 are coprime):
x ≡ 10 (mod 7)
The solution set is x = {10, 17, 24, 31, ...}.
Since we are looking for integer solutions, we can simplify it to x = {10 + 7k}, where k is an integer.
(b) 5x + 1 ≡ 13 (mod 23):
Subtract 1 from both sides:
5x ≡ 12 (mod 23)
Next, we'll find the modular inverse of 5 modulo 23, which is 14 (since 5 * 14 ≡ 1 (mod 23)):
5x × 14 ≡ 12 × 14 (mod 23)
70x ≡ 168 (mod 23)
Now, reduce the coefficients to the smallest positive residue modulo 23:
x ≡ 2 (mod 23)
The solution set is x = {2, 25, 48, ...}, which can be simplified to x = {2 + 23k}, where k is an integer.
(c) 5x + 1 ≡ 13 (mod 26):
This equation is the same as the previous one. Follow the same steps:
5x ≡ 12 (mod 26)
Find the modular inverse of 5 modulo 26, which is 21 (since 5 × 21 ≡ 1 (mod 26)):
5x × 21 ≡ 12 × 21 (mod 26)
105x ≡ 252 (mod 26)
Reduce the coefficients to the smallest positive residue modulo 26:
x ≡ 18 (mod 26)
The solution set is x = {18, 44, 70, ...}, which can be simplified to x = {18 + 26k}, where k is an integer.
(d) 9x ≡ 3 (mod 5):
To find x, we'll multiply both sides by the modular inverse of 9 modulo 5, which is 4 (since 9 × 4 ≡ 1 (mod 5)):
9x × 4 ≡ 3 × 4 (mod 5)
36x ≡ 12 (mod 5)
Reduce the coefficients to the smallest positive residue modulo 5:
x ≡ 2 (mod 5)
The solution set is x = {2, 7, 12, ...}, which can be simplified to x = {2 + 5k}, where k is an integer.
(e) 5x ≡ 1 (mod 6):
To find x, we'll multiply both sides by the modular inverse of 5 modulo 6, which is 5 (since 5 × 5 ≡ 1 (mod 6)):
5x × 5 ≡ 1 × 5 (mod 6)
25x ≡ 5 (mod 6)
Reduce the coefficients to the smallest positive residue modulo 6:
x ≡ 5 (mod 6)
The solution set is x = {5, 11, 17, ...}, which can be simplified to x = {5 + 6k}, where k is an integer.
(f) 3x ≡ 1 (mod 6):
To find x, we'll multiply both sides by the modular inverse of 3 modulo 6, which is 3 (since 3 × 3 ≡ 1 (mod 6)):
3x × 3 ≡ 1 × 3 (mod 6)
9x ≡ 3 (mod 6)
Reduce the coefficients to the smallest positive residue modulo 6:
x ≡ 1 (mod 6)
The solution set is x = {1, 7, 13, ...}, which can be simplified to x = {1 + 6k}, where k is an integer.
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The United States Coast Guard assumes the mean weight of passengers in commercial boats is 185 pounds. The previous value was lower, but was raised after a tragic boating accident. The standard deviation of passenger weights is 26.7 pounds. The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample. (Round your answer to 3 decimal places. Example: If the answer is 0.8976 then you would enter 0.898 in the answer box.)
Answer:
There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.
Step-by-step explanation:
To solve this problem, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85[/tex]
The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.
The probability of an extreme value below the mean.
This is the pvalue of Z when X = 177.6.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{177.6 - 185}{3.85}[/tex]
[tex]Z = -1.92[/tex]
[tex]Z = -1.92[/tex] has a pvalue of 0.0274.
So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.
The probability of an extrema value above the mean.
Measures above the mean have a positive z score.
So this probability is 1 subtracted by the pvalue of [tex]Z = 1.92[/tex]
[tex]Z = 1.92[/tex] has a pvalue of 0.9726.
So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.
Total:
2*0.0274 = 0.0548 = 0.055
There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.
Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedurex1+4x2 =112x1+7x2=18Find the solution to the system of equations.(Simplify your answer. Type an ordered pair.)
Answer:
-5; 4
Step-by-step explanation:
The given linear system is:
[tex]x_1+4x_2=11\\2x_1+7x_2=18[/tex]
Multiplying the first equation by -2 and adding to the second gives:
[tex]x_1+4x_2=11\\2x_1-2x_1_+7x_2-8x_2=18 -22\\\\x_1+4x_2=11\\-x_2=-4[/tex]
Multiply the second equation by 4 and add to the first to find x1:
[tex]x_1+4x_2-4x_2=11-16\\-x_2=-4\\\\x_1=-5\\x_2=4[/tex]
The order pair for the solution of the system is -5; 4.
Fill in the blanks to rewrite the following statement with variables: Given any positive real number, there is a positive real number that is smaller. (a) Given any positive real number r, there is s such that __________ s is __________ . (b) For any __________ , __________ such that s < r.
Answer:
a) there is s such that r>s and s is positive
b) For any r>0 , there exists s>0 such that s<r
Step-by-step explanation:
a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.
b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.
The statement with variables filled in is: (a) Given any positive real number r, there is s such that s is a positive real number and s is less than r. (b) For any positive real number r, there exists s such that s < r.
The statement can be rewritten with variables as follows:
(a) Given any positive real number r, there is s such that s is a positive real number and s is less than r.
(b) For any positive real number r, there exists s such that s < r.
The concept here is that for any positive real number, you can always find another positive real number that is less than the given number. This is fundamental in understanding the density of real numbers on the number line, where between any two distinct real numbers, there are infinitely many other real numbers.
If a ball is thrown into the air with a velocity of 36 ft/s, its height in feet t seconds later is given by y = 36t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds 28 Incorrect: Your answer is incorrect. ft/s (ii) 0.1 seconds ft/s (iii) 0.05 seconds ft/s(iv) 0.01 second.(v) t = 2.
a. The average velocity for the time period beginning at t = 2 and lasting 0.5 seconds is -36 ft/s.
b. The average velocity for the time period beginning at t = 2 and lasting 0.1 seconds is -29.6 ft/s.
c. The average velocity for the time period beginning at t = 2 and lasting 0.05 seconds is -40.8 ft/s.
d. The average velocity for the time period beginning at t = 2 and lasting 0.01 seconds is 3.84 ft/s.
e. The estimated instantaneous velocity at t = 2 is -32 ft/s.
(a) To find the average velocity over a time period, we use the formula for average velocity:
[tex]\[ \text{Average Velocity} = \frac{\text{Change in height}}{\text{Change in time}} \][/tex]
Given [tex]\( y = 36t - 16t^2 \)[/tex], we'll find the heights at [tex]\( t = 2 \)[/tex] and [tex]\( t = 2.5 \)[/tex], then calculate the difference.
At [tex]\( t = 2 \), \( y = 36(2) - 16(2)^2 = 72 - 64 = 8 \)[/tex] ft.
At [tex]\( t = 2.5 \), \( y = 36(2.5) - 16(2.5)^2 = 90 - 100 = -10 \)[/tex] ft.
Change in height = [tex]\( -10 - 8 = -18 \)[/tex] ft.
Change in time = [tex]\( 2.5 - 2 = 0.5 \)[/tex] sec.
Average velocity = [tex]\( \frac{-18}{0.5} = -36 \)[/tex] ft/s.
(b) Following the same procedure as (a), at [tex]\( t = 2.1 \), \( y = 36(2.1) - 16(2.1)^2 = 75.6 - 70.56 = 5.04 \)[/tex] ft.
Change in height = [tex]\( 5.04 - 8 = -2.96 \)[/tex] ft.
Change in time = [tex]\( 2.1 - 2 = 0.1 \)[/tex] sec.
Average velocity = [tex]\( \frac{-2.96}{0.1} = -29.6 \)[/tex] ft/s.
(c) Continuing the process, at [tex]\( t = 2.05 \), \( y = 36(2.05) - 16(2.05)^2 = 73.8 - 67.84 = 5.96 \)[/tex] ft.
Change in height = [tex]\( 5.96 - 8 = -2.04 \)[/tex] ft.
Change in time = [tex]\( 2.05 - 2 = 0.05 \)[/tex] sec.
Average velocity = [tex]\( \frac{-2.04}{0.05} = -40.8 \)[/tex] ft/s.
(d) Applying the method to [tex]\( t = 2.01 \), \( y = 36(2.01) - 16(2.01)^2 = 72.36 - 64.3216 = 8.0384 \)[/tex] ft.
Change in height = [tex]\( 8.0384 - 8 = 0.0384 \)[/tex] ft.
Change in time = [tex]\( 2.01 - 2 = 0.01 \)[/tex] sec.
Average velocity = [tex]\( \frac{0.0384}{0.01} = 3.84 \)[/tex] ft/s.
(e) To estimate the instantaneous velocity at t = 2, we find the derivative of [tex]\( y = 36t - 16t^2 \)[/tex] with respect to [tex]\( t \)[/tex], which is [tex]\( v(t) = 36 - 32t \)[/tex]. Plugging in [tex]\( t = 2 \)[/tex],
we get [tex]\( v(2) = 36 - 32(2) = -32 \)[/tex] ft/s.
An English class consists of 23 students, and three are to be chosen to give speeches in a school competition. In how many different ways can the teacher choose the team, given the following conditions?
Answer: Check the attached
Step-by-step explanation:
A retail store owner offers a discount on product A and predicts that, the customers would purchase products B and C in addition to product A. Identify the technique used to make such a prediction. a. Data query b. Simulation c. Data mining d. Data dashboards
Answer:
C. Data Mining
Step-by-step explanation:
As data mining is a technique which is used predict and to find patterns or relationships among elements of the data in a large database. It also facilitate the enterprise to predict the future trends.
26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+b2≤c2+d2. Show that ∼ ∼ is reflexive and transitive but not symmetric.
Answer:
~ is reflexive.
~ is asymmetric.
~ is transitive.
Step-by-step explanation:
~ is reflexive:
i.e., to prove [tex]$ \forall (a, b) \in \mathbb{R}^2 $[/tex], [tex]$ (a, b) R(a, b) $[/tex].
That is, every element in the domain is related to itself.
The given relation is [tex]$\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$[/tex]
Reflexive:
[tex]$ (a, b) \sim (a, b) $[/tex] since [tex]$ a^2 + b^2 = a^2 + b^2 $[/tex]
This is true for any pair of numbers in [tex]$ \mathbb{R}^2 $[/tex]. So, [tex]$ \sim $[/tex] is reflexive.
Symmetry:
[tex]$ \sim $[/tex] is symmetry iff whenever [tex]$ (a, b) \sim (c, d) $[/tex] then [tex]$ (c, d) \sim (a, b) $[/tex].
Consider the following counter - example.
Let (a, b) = (2, 3) and (c, d) = (6, 3)
[tex]$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $[/tex]
[tex]$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $[/tex]
Hence, [tex]$ (a, b) \sim (c, d) $[/tex] since [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]
Note that [tex]$ c^2 + d^2 \nleq a^2 + b^2 $[/tex]
Hence, the given relation is not symmetric.
Transitive:
[tex]$ \sim $[/tex] is transitive iff whenever [tex]$ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $[/tex] then [tex]$ (a, b) \sim (e, f) $[/tex]
To prove transitivity let us assume [tex]$ (a, b) \sim (c, d) $[/tex] and [tex]$ (c, d) \sim (e, f) $[/tex].
We have to show [tex]$ (a, b) \sim (e, f) $[/tex]
Since [tex]$ (a, b) \sim (c, d) $[/tex] we have: [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]
Since [tex]$ (c, d) \sim (e, f) $[/tex] we have: [tex]$ c^2 + d^2 \leq e^2 + f^2 $[/tex]
Combining both the inequalities we get:
[tex]$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $[/tex]
Therefore, we get: [tex]$ a^2 + b^2 \leq e^2 + f^2 $[/tex]
Therefore, [tex]$ \sim $[/tex] is transitive.
Hence, proved.
It is known that a cable with a cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the cable is proportional to its cross-sectional area, what size cable is needed to hold 70007000 lb?
Answer:
0.84 square in
Step-by-step explanation:
Since the capacity of the cable is proportional to its cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold
2500 / 0.3 = 8333.33 lb/in
To old 7000 lb it the cross-sectional area would need to be
7000 / 8333.33 = 0.84 square in
Answer:
0.84 square in.
Step-by-step explanation:
Since the capacity of the cable is proportional to its cross-sectional area.
Mathematically,
C = k * A
A1 = 0.3 sq
C1 = 2500 lb
C2 = 7000 lb
C1/A1 = C2/A2
2500/0.3 = 7000/C2
= 7000 / 8333.33
= 0.84 square in.
1. purchase a Toyota 4runner for 25,635. promised your daughter the SUV will be hers when the car value is worth 10,000. 2. the car dealer said the SUV will depreciate in value approximately 3,000 per year. 3. write a linear equation in which y represents the total value of the car and x represents the age of the car.
Answer:buy here both
Step-by-step explanation:
Answer:
y = -3000x + 25,635
Step-by-step explanation:
well if the inital value of the car is $25,635 this means that when
x = 0 y = 25,635
(0 , 25635)
this will be our first point
Now if you tell us that in 1 year depreciate in value $3,000
this means thaw when
x = 1 y = 25,635 - 3,000
x = 1 y = 22,635
(1, 22635)
Now that we have 2 points we can have the equation
First we take the slope as follows
m = (y2 - y1) / (x2 - x1)
m = (22,635 - 25,635) / (1 - 0)
m = -3000 / 1
m = -3000
after calculating the slope we have to replace it in the following formula
(y - y1) = m (x - x1)
y - 25,635 = -3000 ( x - 0)
y - 25,635 = -3000x
y = -3000x + 25,635
Finally we replace the value of y by 10000
10,000 = -3000x + 25,635
10,000 - 25,635 = -3000x
-15,635 = -3000x
-15,635/-3000 = x
5.21167 = x years
These are the years it would take for the value to be 10,000
to know the days we simply multiply by 365
5.21167 * 365 = 1902.26 days
A principal of $1900 was invested at 3.75% interest, compounded annually. Let be the number of years since the start of the investment. Let y be the value of the investment, in dollars.
Write an exponential function showing the relationship between y and t?
Answer:
[tex]y(t) = \$1900*1.0375^t[/tex]
Step-by-step explanation:
If y is the future value of the $1900 investment, in dollars, after t years at a rate of 3.75% per year, compounded annually. The exponential function that describes the relationship between the variables y and t is:
[tex]y(t) = \$1900*(1.0375)^t[/tex]
This relationship means that for every year t, the amount y will increase by a factor of 1.0375.
To represent the value of an investment of $1900 at a 3.75% annual interest rate compounded annually, the exponential function y = 1900(1 + 0.0375)^t is used, where t is the number of years.
Explanation:The question asks for an exponential function that shows the relationship between the value of an investment, y, and time, t, when the principal is $1900 and the interest rate is 3.75% compounded annually. The standard formula for an investment compounded annually is y = P(1 + r)t, where P is principal, r is the annual interest rate as a decimal, and t is the number of years.
For this specific case, the function would be y = 1900(1 + 0.0375)t. This represents the value of the investment after t years, given the initial investment and interest rate provided.
Ask Your Teacher Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 5 + sec(x), −π 3 ≤ x ≤ π 3 , y = 7; about y = 5
Answer:
V=15.44
Step-by-step explanation:
We have a formula
V=\int^{π/3}_{-π/3} A(x) dx ,
where A(x) calculate as cross sectional.
We have:
Inner radius: 5 + sec(x) - 5= sec(x)
Outer radius: 7 - 5=2, we get
A(x)=π 2²- π· sec²(x)
A(x)=π(4-sec²(x))
Therefore, we calculate the volume V, and we get
V=\int^{π/3}_{-π/3} A(x) dx
V=\int^{π/3}_{-π/3} π(4-sec²(x)) dx
V=[ π(4x-tan(x)]^{π/3}_{-π/3}
V=π·(8π/3-2√3)
V=15.44
We use a site geogebra.org to plot the graph.
The volume of the solid obtained by rotating the region bounded by y = 5 + sec(x), -π/3 ≤ x ≤ π/3, y = 7 about y = 5 is calculated using the disc method formula in calculus, resulting in a volume of 8π²/3 cubic units.
Explanation:The region that we need to rotate around the line y=5 is bounded by the curves y=7 and y=5+sec(x) over the interval -π/3 ≤ x ≤ π/3. The resulting solid is a disc shaped object. We can find the volume of this solid using the disc method formula in calculus:
V = π ∫ [R(x)]² dx
, where R(x) = radius function. The radius function is the absolute difference between y=7 (the upper curve) and y=5 (the line of rotation), which equals 2. Therefore, the integral becomes:
V = π ∫ from -π/3 to π/3 [2]² dx
, which simplifies to
V = 4π [x] from -π/3 to π/3
. Finally, evaluate this expression by subtracting the lower limit from the upper limit, giving
V = 4π(2π/3) = 8π²/3
cubic units.
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A manufacturer of personal computers tests competing brands and finds that the amount of energy they require is normally distributed with a mean of 285 kwh and a standard deviation of 9.1 kwh. If the lowest 25% and the highest 30% are not included in a second round of tests, what are the upper and lower limits for the energy amounts of the remaining computers?
Answer:
[tex]a=285 -0.674*9.1=278.87[/tex]
So the value of height that separates the bottom 25% of data from the top 75% is 278.87.
[tex]a=285 +0.524*9.1=289.77[/tex]
So the value of height that separates the bottom 70% of data from the top 30% is 289.77.
The answer would be 278.87 and 289.77
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the amount of energy of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(285,9.1)[/tex]
Where [tex]\mu=285[/tex] and [tex]\sigma=9.1[/tex]
Lowest 25%
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.75[/tex] (a)
[tex]P(X<a)=0.25[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.674<\frac{a-285}{9.1}[/tex]
And if we solve for a we got
[tex]a=285 -0.674*9.1=278.87[/tex]
So the value of height that separates the bottom 25% of data from the top 75% is 278.87.
Highest 30%
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.30[/tex] (a)
[tex]P(X<a)=0.70[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.524<\frac{a-285}{9.1}[/tex]
And if we solve for a we got
[tex]a=285 +0.524*9.1=289.77[/tex]
So the value of height that separates the bottom 70% of data from the top 30% is 289.77.
The lower limit for the energy amounts of the remaining computers is 279.03 kwh, and the upper limit is 290.74 kwh.
Explanation:To find the upper and lower limits for the energy amounts of the remaining computers, we need to find the z-scores corresponding to the lowest 25% and highest 30% of the distribution.
Using the invNorm function in a calculator, we find that the z-score for the lowest 25% is approximately -0.674 and the z-score for the highest 30% is approximately 0.524.
To find the corresponding energy amounts, we use the formula:
Lowest Energy Amount = Mean + (Z-score * Standard Deviation)
Highest Energy Amount = Mean + (Z-score * Standard Deviation)
Substituting the z-scores and the given values, we get:
Lowest Energy Amount = 285 + (-0.674 * 9.1) = 279.03 kwh
Highest Energy Amount = 285 + (0.524 * 9.1) = 290.74 kwh
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The Quick Change Oil Company has a number of outlets in the metropolitan Seattle area. The daily number of oil changes at the Oak Street outlet in the past 20 days are: 65 98 55 62 79 59 51 90 72 56 70 62 66 80 94 79 63 73 71 85 The data are to be organized into a frequency distribution. a. How many classes would you recommend
Answer:
5 classes.
Step-by-step explanation:
You can use the [tex]2^k[/tex] rule to determine the number of classes for a frequency distribution.
The [tex]2^k[/tex] rule says that [tex]2^k\geq n[/tex] where
[tex]k[/tex] is the number of classes
[tex]n[/tex] is the number of the data points
We know that the number of data points is [tex]n[/tex] = 20.
Next, we start searching for [tex]k[/tex] so that we can get a number 2 to the [tex]k[/tex] that is larger that the number of data points.
[tex]2^4=16\\2^5=32[/tex]
This suggests that you should use 5 classes.
Final answer:
For this data, it is recommended to use 5 classes.
Explanation:
To organize the data into a frequency distribution, you need to determine the number of classes.
The recommended number of classes can vary depending on the data.
One popular rule is to use the square root of the number of data points to determine the number of classes.
In this case, there are 20 data points, so the square root is approximately 4.47.
Since you can't have a fraction of a class, you can round up to the nearest whole number, which gives you a recommendation of 5 classes.
Find the balance on 1,800 deposited at 5% annually for 4 years
Answer:
The balance after 5 years will be $2,187.91.
Step-by-step explanation:
This problem can be solved by the following formula:
[tex]A = P(1 + r)^{t}[/tex]
In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.
In this problem, we have that:
[tex]P = 1800, r = 0.05, t = 4[/tex]
We want to find A
So
[tex]A = P(1 + r)^{t}[/tex]
[tex]A = 1800(1+0.05)^{5}[/tex]
[tex]A = 2,187.91[/tex]
The balance after 5 years will be $2,187.91.
Answer: 2188
Step-by-step explanation:
so the person deposited 1800 and expects an annual raise in the amount of 5 percent
so the equation for 1 year is
1800(1+(5/100))=the answer
but for four years u will have to power the bract by 4
180(1+(5/100))^4= 2187.9
aka 2188
Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Visa credit card and B be the analogous event for a MasterCard where P(A) = 0.45, P(B) = 0.35, and P(A ❩ B) = 0.30. Calculate and interpret each of the following probabilities (a Venn diagram might help). (Round your answers to four decimal places.)(a) P(B | A)(b) P(B' | A)(c) P(A | B)(d) P(A' | B)
Answer:
a) [tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]
Represent the probability that the event B occurs given that the event A occurs first
b) [tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]
Represent the probability that the event B no occurs given that the event A occurs first
c) [tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]
Represent the probability that the event A occurs given that the event B occurs first
d) [tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]
Represent the probability that the event A no occurs given that the event B occurs first
Step-by-step explanation:
For this case we have the following probabilities given for the events defined A and B
[tex] P(A) = 0.45, P(B) = 0.35, P(A \cap B) =0.30[/tex]
For this case we can begin finding the probability for the complements:
[tex] P(B') =1-P(B) = 1-0.35= 0.65[/tex]
[tex] P(A') =1-P(A) = 1-0.45= 0.55[/tex]
For this case we are interested on the following probabilities:
Part a
[tex] P(B|A)[/tex]
For this case we can use the Bayes theorem and we can find this probability like this:
[tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]
Represent the probability that the event B occurs given that the event A occurs first
Part b
[tex] P(B'|A) = \frac{P(B' \cap A)}{P(A}[/tex]
And for this case we can find [tex] P(B' \cap A) =P(A) -P(A\cap B)= 0.45-0.3=0.15[/tex]
And if we replace we got:
[tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]
Represent the probability that the event B no occurs given that the event A occurs first
Part c
[tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]
Represent the probability that the event A occurs given that the event B occurs first
Part d
[tex] P(A'|B) = \frac{P(A' \cap B)}{P(B}[/tex]
And for this case we can find [tex] P(A' \cap B) =P(B) -P(A\cap B)= 0.35-0.3=0.05[/tex]
And if we replace we got:
[tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]
Represent the probability that the event A no occurs given that the event B occurs first
Conditional probabilities are calculated as the ratio of the intersection of two events to the probability of the given event. P(B|A) is approximately 0.6667, P(B'|A) is approximately 0.3333, P(A|B) is approximately 0.8571, and P(A'|B) is approximately 0.1429.
When considering the probability of randomly selecting a student at a university who has a Visa (Event A) or a MasterCard (Event B), we can use the given probabilities to calculate the following:
P(B|A) is the probability that a selected individual has a MasterCard given they have a Visa. It is calculated as P(A [tex]\cap[/tex] B) / P(A). Given P([tex]A \cap B[/tex]) = 0.30 and P(A) = 0.45, P(B|A) = 0.30 / 0.45 which rounds to 0.6667.
P(B'|A) is the probability that a selected individual does not have a MasterCard given they have a Visa. It is 1 - P(B|A), which is 1 - 0.6667, rounding to 0.3333.
P(A|B) is the probability that a selected individual has a Visa given they have a MasterCard. It is calculated as P(A \\cap B) / P(B). Given P(A [tex]\cap[/tex] B) = 0.30 and P(B) = 0.35, P(A|B) = 0.30 / 0.35 which rounds to 0.8571.
P(A'|B) is the probability that a selected individual does not have a Visa given they have a MasterCard. It is 1 - P(A|B), which is 1 - 0.8571, rounding to 0.1429.