Answer:
a)
[tex]H_0: \pi\geq0.78\\\\H_a: \pi<0.78[/tex]
b) The Type I error occurs when we reject a null hypothesis that is actually true. In this case, it means we conclude that the arrival time have improved, when it didn't.
The Type II error occurs when we accept a null hypothesis that is actually false. In this case, although the arrival times have really improved, the evidence from the sample was not enough to show that improvement.
c) In this case, the Type I error is more serious, because it gives the wrong impression of improvement and no further actions will be taken to reduce the times.
Step-by-step explanation:
a) If you want to determine if the responders are arriving within 8 minutes of the call more often, you have to evaluate the proportion of accidents in which the arrival time is less than 8 minutes and compare it with the known proportion of π=0.78.
The sample parameter "p: proportion of accidents with arrival time of 8 minutes or less" will be used to test the hypothesis.
The null and alternative hypothesis will be:
[tex]H_0: \pi\geq0.78\\\\H_a: \pi<0.78[/tex]
Final answer:
The hypotheses for a significance test to determine whether first responders are arriving more often within 8 minutes are H0: p = 0.78 and HA: p > 0.78, with p representing the proportion of calls responded to within this timeframe. A Type I error involves mistakenly concluding an improvement, while a Type II error occurs by overlooking an actual improvement. The potential demotivating effect of a Type II error may render it more serious in this context.
Explanation:
The question seems to revolve around the concept of hypothesis testing in statistics and how it applies to the emergency response times in a city. The parameter of interest would be the proportion of emergency calls responded to within 8 minutes. So for part (a), the hypotheses could be stated as follows:
H0: p = 0.78 (The proportion of calls responded to within 8 minutes is 78% as it was last year.)
HA: p > 0.78 (The proportion of calls responded to within 8 minutes has increased from last year.)
In part (b), a Type I error would occur if the city concludes that the proportion of calls responded to within 8 minutes has increased when in reality, it has not. The consequence of a Type I error would be misallocating resources based on false success. A Type II error would occur if the city fails to recognize an actual improvement in response times. The consequence of this could lead to a lack of recognition and continued encouragement for first responders who have actually improved.
Part (c) asks which error is more serious. A Type II error may be considered more serious in this setting, as failing to acknowledge and react to an actual improvement could demotivate emergency personnel and affect future performances, possibly leading to life-threatening delays for accident victims.
I needd helpppppppp pleaseeeee
Answer:
hi guada!
Step-by-step explanation:
well the answer i think is 35.
180-155= 25
180-120-25=35
Use a formula to solve the problem.
An airplane flew from Chicago to San Francisco in 3.25 hours. If the cities are 1,950 miles apart, what was the average speed of the plane?
Answer: The average speed of the plane is 600 miles per hour.
Step-by-step explanation:
The formula for determining average speed is expressed as
Average speed = total distance travelled/total time taken
The airplane flew from Chicago to San Francisco in 3.25 hours. Therefore, time = 3.25 hours
The cities are 1,950 miles apart. This means that the distance travelled is 1950 miles. Therefore,
Average speed = 1950/3.25 = 600 miles per hour.
Answer:
Speed = 600 mile per hour
Step-by-step explanation:
Given values:
Time = 3.25hours
Distance = 1950miles
Formula: speed = distance ÷ time
Therefore:
Speed = 1950 ÷ 3.25
Speed = 1950 ÷ 325/100
Speed = 1950 × 100/325
Speed= 195000/325
Speed = 600ml^h
How is the graph of the parent quadratic function transformed to produce the graph of y = negative (2 x + 6) squared + 3?
We start with the parent function
[tex]f(x)=x^2[/tex]
The first child function would be
[tex]g(x)=(2x)^2[/tex]
We have multiplied the input of the function by a constant: we have
[tex]g(x)=f(2x)[/tex]
This kind of transformation result in a horizontal stretch/compression. If the multiplier is greater than 1, we have a compression. So, this first child causes a horizontal compression with compression rate 2.
The second child function would be
[tex]h(x)=(2x+6)^2[/tex]
We added 6 to the input of the function: we have
[tex]h(x)=g(x+6)[/tex]
This kind of transformation result in a horizontal translation. If the constant added is positive, we translate to the left. So, this second child causes a translation 6 units to the left.
The third child function would be
[tex]l(x)=-(2x+6)^2[/tex]
We changed the sign of the previous function (i.e. we multiplied it by -1): we have
[tex]l(x)=-h(x)[/tex]
This kind of transformation result in a vertical stretch/compression. If the multiplier is greater than 1 we have a stretch, if it's between 0 and 1 we have compression. If it's negative, we reflect across the x axis, and then apply the stretch/compression. In this case, the multiplier is -1, so we only reflect across the x axis.
The fourth child function would be
[tex]m(x)=-(2x+6)^2+3[/tex]
We added 3 to previous function: we have
[tex]m(x)=l(x)+3[/tex]
This kind of transformation result in a vertical translation. If the constant added is positive, we translate upwards. So, this last child causes a translation 3 units up.
Recap
Starting from the parent function [tex]y=x^2[/tex], we have to:
Compress the graph horizontall, with scale factor 2;Translate the graph 6 units to the left;Reflect the graph across the x axis;Translate the graph 3 units upNote that the order is important!
Answer:
B
Step-by-step explanation:
i know the other answer was a little confusing, but they did more, so feel free to give them brainliest, just wanted to help clarify :)
edgenuity 2020
Consider the following sample data for two variables. x y 7 7 8 5 5 9 3 7 9 7 Calculate the sample covariance. b. Calculate the sample correlation coefficient. c. Describe the relationship between x and y.
Answer:
a) Sample co-variance = -1.5
b) Sample correlation = -0.4404152
c) Weak negative relationship between X and Y.
Step-by-step explanation:
a) by sample co-variance = sum ((x - xbar)* (y-ybar)/n-1
#.... use the following program in R
x = c(7,8, 5, 3, 9)
y = c(7,5,9,7,7)
sv = sum((x-mean(x))*(y-mean(y)))/4
cor(x,y)
####################################
The CFO of the company believes that an appropriate annual interest rate on this investment is 4%. What is the present value of this uneven cash flow stream, rounded to the nearest whole dollar
Find attached the missing cash flow stream
Answer:
Present value = $1,685,334 (rounded to the nearest whole)Explanation:
Since the cash flow stream is uneven, you must discount each stream individually and after you have discounted every stream you can add each preset value to find the net present value.
The formula for present value is:
[tex]\text{Present Value} = \dfrac{\text{Future Value}}{(1+r)^n}\\\\\ Where:\\ \\ r = \text{Rate of return=interest rate}\\ \\ n = \text{Number of periods}[/tex]
1. Year 1:
Future value = $250,000r = 4%n = 1[tex]Present\text{ }value=\$250,000/(1+0.04)^1=\$240,384.62[/tex]
2. Year 2:
Future value = $20,000r = 4%n = 2[tex]Present\text{ }value=\$20,000/(1+0.04)^2=\$18,491.12[/tex]
3. Year 3:
Future value = $330,000r = 4%n = 3[tex]Present\text{ }value=\$330,000/(1+0.04)^3=\$293,368.80[/tex]
4. Year 4:
Future value = $450,000r = 4%n = 4[tex]Present\text{ }value=\$450,000/(1+0.04)^4=\$384,661.89[/tex]
5. Year 5:
Future value = $550,000r = 4%n = 5[tex]Present\text{ }value=\$550,000/(1+0.04)^5=\$452,059.91[/tex]
6. Year 6:
Future value = $375,000r = 4%n = 6[tex]Present\text{ }value=\$375,000/(1+0.04)^6=\$296,367.95[/tex]
Total present value = $240,384.62 + $18,491.12 + $293,368.80 + $384,661.89 + $452,059.91 + $296,367.95 + $ 296,367.95
Total present value = $1,685,334.29 = $ 1,685,334
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 170 engines and the mean pressure was 7.5 pounds/square inch (psi). Assume the population variance is 0.36. The engineer designed the valve such that it would produce a mean pressure of 7.4 psi. It is believed that the valve does not perform to the specifications. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to two decimal places.
Answer:
2.17
Step-by-step explanation:
As the population variance is known we can calculate population standard deviation and so, z-test statistic will be used.
[tex]z=\frac{xbar-pop mean}{\frac{pop SD}{\sqrt{n} } }[/tex]
population variance=σ²=0.36
population standard deviation=σ=√0.36=0.6
The value of population mean can be achieved through null hypothesis.
Null hypothesis : μ=7.4
Alternative hypothesis : μ≠7.4
population mean=μ=7.4
xbar=7.5
n=170
z=(7.5-7.4)/(0.6/√170)
z=0.1/(0.6/13.038)
z=0.1/0.046
z=2.17
Thus, the calculated z-test statistic is 2.17.
Let $a=-1$, let $b=3$, and let $c=-5$. Calculate $b(abc+5ab)+b(c+a)$.
Answer: $b(abc+5ab)+b(c+a)$ = - 18
Step-by-step explanation:
If $a = $ - 1, $b = $3 and $c = $- 5, then
abc = - 1 × 3 × - 5 = 15
5ab = 5 × - 1 × 3 = - 15
Then,
$b(abc+5ab) = 3(15 - 15)
Opening the brackets, it becomes
3 × 0 = 0
(c + a)$ = - 5 - 1 = - 6
Then,
b(c+a)$ = 3 × - 6 = - 18
Therefore,
$b(abc+5ab)+b(c+a)$ = 0 + (- 18)
$b(abc+5ab)+b(c+a)$ = 0 - 18 = - 18
Final answer:
After substituting the given values into the expression and simplifying, the result of the calculation is -18.
Explanation:
To calculate b(abc+5ab)+b(c+a) with given values of a=-1, b=3, and c=-5, first, substitute these values into the expression:
b(abc+5ab)+b(c+a) = 3((-1)(3)(-5)+5(-1)(3))+3((-5)+(-1))
Next, simplify inside the parentheses:
3((15)+(-15))+3(-6)
Now, simplify further:
3(0)+3(-6)
This reduces to:
0 - 18
And the final result is:
-18
203 red, 117 white, and 28 blue. She asked the students to draw marbles from the box, one at a time and without looking. What is the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same colour?
Answer:
7
Step-by-step explanation:
Since the goal is to draw three marbles of the same colour, regardless of which colour that is, the worst possible scenario would be drawing two marbles of each color in the first six picks (2 red, 2 white and 2 blue). At this point, with the 7th pick, no matter what colour marble the student picks will form three of the same kind.
Therefore, the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same colour is 7.
Final answer:
To ensure at least three marbles of the same color, a minimum of 349 marbles must be drawn from the box.
Explanation:
Question: What is the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same color?
When considering the worst-case scenario, you would need to take marbles until you have at least 3 of the same color.
For this situation, where you have 203 red, 117 white, and 28 blue marbles:
It's possible that you could pick 202 red, 117 white, 27 blue, and still not have 3 of the same color. The next pick would guarantee 3 of the same color, giving a minimum of 349 marbles in total.
The probability distribution of customers that walk into a coffee shop on any given day of the week is described by a Normal distribution with mean equal to 100 and standard deviation equal to 20. What is the probability that no more than 80 customers walk into the coffee shop next Monday
Answer:
15.87% probability that no more than 80 customers walk into the coffee shop next Monday
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 100, \sigma = 20[/tex]
What is the probability that no more than 80 customers walk into the coffee shop next Monday?
This is the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 100}{20}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
So there is a 15.87% probability that no more than 80 customers walk into the coffee shop next Monday
Find the area between y = 8 sin ( x ) y=8sin(x) and y = 8 cos ( x ) y=8cos(x) over the interval [ 0 , π ] . [0,π]. (Use decimal notation. Give your answer to three decimal places.)
Answer:
0.416 au
Step-by-step explanation:
Let y1=8sin(x) and y2=8cos(x), we must find the area between y1 and y2
[tex]\int\limits^\pi _0{(8cos(x)-8sin(x))} \, dx = 8\int\limits^\pi _0{(cos(x)-sin(x))} \, dx =\\8(sin(x)+cos(x)) evaluated(0-\pi )=\\8(sin(\pi )-sin(0))+8(cos(\pi )-cos(0))=\\8(0.054-0)+8(0.998-1)=8(0.054)+8(-0.002)=0.432-0.016=0.416[/tex]
Type a proposition involving p, q, r and s that is true just when at least two of the propositional variables are true. For example, your proposition would be true in any case for which p and r are both true but false when, say, p is true while q, r and s are all false.
Answer:
(X and Y are real numbers)
p) X > 0
q) Y > 0
r) XY > 0
s) (X > 0 ∧ Y > 0) ∨ (X = Y = 0)
Step-by-step explanation:
Lets assume that a pair of the propositions are true, and we will show that the other 2 are also true. There are 6 possible cases:
1) If p and q are true, then XY has to be positive because it is the product of positive numbers. And the first option of proposition s is true
2) If p and r are true, then Y = XY/X > 0 because it is the division of tow positive numbers, thus q is true. Since p and q are true, then so it is s.
3) If p and s are true, then option 2 of S is impossible, thus X > 0 and Y>0, hence q is true. SInce p and q are true, so is r.
4) If q and r are true, we can obtain X by dividing the positive numbers XY and Y, thus X>0, and p and s are true.
5) is q and s are true, an argument similar at the one made in 3) shows that both p and r are true.
6) if r and s are true, then the second part of s cant be true, therefore X>0 and Y>0, as a consecuence, p and q are true.
Now, lets show that, individually, no proposition implies the others:
p clearly doesnt imply q. X can be positive while Y is negative. Similarly Y can be positive while X is negative, thus q doesnt imply p either.
If XY > 0, then it can be that both X and Y are negative, thus r might not imply neither p nor q.
and if only s is true, then it may be that the second part is the one that is true, thereofre X=Y=0 and p, q and r are all false in this case.
As a consecuence, you need 2 of the propositions to be true so that all 4 are true, but as we show above, by having any pair of the propositions true, then all 4 are automaticallyy true.
A study was interested in determining if walking 2 miles a day lowered someone's blood pressure.Twenty people's blood pressure was measured. Then, ten of these individuals were randomly selected from the initial 20 people. These ten were told to walk 2 miles a day for 6 weeks and to eat as they normally did. The other ten were told to eat as they normally would. After six weeks, their blood pressure levels were measured again. What type of study is this? Group of answer choices observational study experiment anecdotal evidence
Answer:
This is an "Experiment" type of study
Step-by-step explanation:
Researchers use various methods or techniques to conduct research and draw conclusions from it. Some of these methods are given below
1. Experiment
2. Observational study
3. Anecdotal evidence
1. Experiment: In this type of study, researchers have control over their setup and they can give directions or apply any treatment on their subjects.
2. Observational study: In this type of study, researchers cannot give directions or apply any treatment on their subjects but rather they can only observe what is going on.
3. Anecdotal evidence: Also refers to experience and can be defined as personal experience based on something.
The study that we are given is clearly an example of "experiment" since half of the participants were to told walk 2 miles a day for 6 weeks. The researchers have control over their subjects in this case and they gave directions to their subjects to do this and that.
This could have been an observational study if researchers had selected 10 people who walked 2 miles a day for 6 weeks and 10 people who did not walk 2 miles a day for 6 weeks and simply measured their blood pressure and come to a conclusion. This way, it would have been a case of observational study.
Answer:
Experiment
Step-by-step explanation:
Experimental study design is a type of study design in which the investigator or researcher is in complete control of the research environment. For example, a researcher may want to assess the effects of certain treatments on some experimental units. The researcher or investigator decides the type of treatment, the type of experimental unit to use, the time, the allocation and assessment procedures of the treatment and effects respectively. In Experimental study design, the investigator or researcher is in complete control of the exposure and the result should therefore provide a stronger evidence of an association or lack of association between an exposure and a health problem than would an observational study where the investigator or researcher is usually a passive observer as he only observes and analyses facts and events as they occur naturally or anecdotal evidence which is just someones personal experience or testimony without scientific proof that can be based on some measurement or experimentation.
A batch of 484 containers for frozen orange juice contains 7 that are defective. Two are selected, at random, without replacement from the batch. a) 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).
Answer:
1.242% probability that the second one selected is defective given that the first one was defective
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
We have
484 containers
7 are defective
a) What is the probability that the second one selected is defective given that the first one was defective?
After the first one was selected(defective), we have
483 containers
6 defective
6/483 = 0.01242
0.01242 = 1.242% probability that the second one selected is defective given that the first one was defective
A company makes pens. They sell each pen for $99.
Their revenue is represented by R = 9x.
The cost to make the pens is $11 each with a one time start up cost of $4000.
Their cost is represented by C = 1x + 4000.
a) Find the profit, P, when the company sells 1000 pens.
b) Find the number of pens they need to sell to break even.
Answer:
a) Profit = 84000
b) at break even number of pens sold = 46
Step-by-step explanation:
Profit = Selling price - Cost price
Total revenue generated = 99 * 1000
Total revenue generated = 99000
Total cost on making the pen = 11 * 1000
Total cost on making the pen = 11000
Total cost including the initial cost = 11000 + 4000
Total cost including the initial cost = 15000
Profit = 99000 - 15000
Profit = 84000
Break even is when the cost are equal to Revenue thus no profit or loss
Revenue = total cost (break even)
9x = 1x + 4000
9x - x = 4000
8x = 4000
x = 500
At breakeven Revenue = 9 * 500
At breakeven Revenue = 4500
since one pen is sold at 99 therefore at break even number of pens sold = 4500/99 = 45.45( to 2 decimal place)
at break even number of pens sold = 46
To find the profit, substitute the selling quantity into the revenue and cost equations and subtract the cost from the revenue. To find the break-even point, set the profit to zero and solve for the selling quantity.
Explanation:To find the profit, P, when the company sells 1000 pens, we first calculate the revenue by substituting x = 1000 into the revenue equation: R = 9x. Therefore, R = 9(1000) = 9000. Next, we calculate the cost by substituting x = 1000 into the cost equation: C = 1x + 4000. Therefore, C = 1(1000) + 4000 = 5000. Finally, we find the profit by subtracting the cost from the revenue: P = R - C. Therefore, P = 9000 - 5000 = 4000. The company's profit when selling 1000 pens is $4000.
To find the number of pens needed to break even, we set the profit to zero and solve for x. Therefore, P = R - C = 0. Substituting the revenue and cost equations, we have 9x - (1x + 4000) = 0. Simplifying this equation gives us 8x - 4000 = 0. Solving for x, we get x = 500. The company needs to sell 500 pens to break even.
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A company that manufactures video cameras produces a basic model and a deluxe model. Over the past year, 30% of the cameras sold have been of the basic model. Of those buying the basic model, 44% purchase an extended warranty, whereas 40% of all deluxe purchasers do so. If you learn that a randomly selected purchaser has an extended warranty, how likely is it that he or she has a basic model?
Answer:
the probability is 0.32 (32%)
Step-by-step explanation:
defining the event W= has extended warranty , then
P(W)= probability of purchasing the basic model * probability of purchasing extended warranty given that has purchased the basic model + probability of purchasing the deluxe model * probability of purchasing extended warranty given that has purchased the deluxe model = 0.3 * 0.44 + 0.7 * 0.40 = 0.412
then using the theorem of Bayes for conditional probability and defining the event B= has the basic model , then
P(B/W)= P(B∩W)/P(W)= 0.3 * 0.44/0.412 =0.32 (32%)
where
P(B∩W)= probability of purchasing the basic model and purchasing the extended warranty
P(B/W) = probability of purchasing the basic model given that has purchased the extended warranty
Final answer:
Using Bayes' theorem, there is approximately a 32% chance that a customer who purchased an extended warranty has a basic model camera.
Explanation:
The question asks us to calculate the probability of a randomly selected purchaser having a basic model given that they have an extended warranty. We can use Bayes' theorem to solve this. Let's denote B as the event of buying a basic model, D as the event of buying a deluxe model, and W as the event of purchasing an extended warranty. From the information provided, we can infer :
P(B) = 0.30 (30% of cameras sold are basic models)
P(D) = 1 - P(B) = 0.70 (since if it's not a basic model, it's a deluxe model)
P(W|B) = 0.44 (44% of basic model buyers purchase an extended warranty)
P(W|D) = 0.40 (40% of deluxe model buyers purchase an extended warranty)
We are interested in P(B|W), the probability that a randomly selected purchaser who bought an extended warranty has a basic model. Using Bayes' theorem:
P(B|W) = (P(W|B) * P(B)) / ((P(W|B) * P(B)) + (P(W|D) * P(D)))
Plugging in the values:
P(B|W) = (0.44 * 0.30) / ((0.44 * 0.30) + (0.40 * 0.70))
P(B|W) = (0.132) / (0.132 + 0.28)
P(B|W) = 0.132 / 0.412 = approximately 0.320
So, there is approximately a 32% chance that a customer with an extended warranty has purchased a basic model camera.
Suppose that, of all the customers at a coffee shop,70% purchase a cup of coffee;40% purchase a piece of cake;20% purchase both a cup of coffee and a piece of cake.Given that a randomly chosen customer has purchased a piece of cake, what is the probability that he/she has also purchased a cup of coffee
Answer:
0.50
Step-by-step explanation:
The probability that a customer has purchased a cup of coffee given that they have also purchased a piece of cake is determined by the percentage of customers who purchase both coffee and cake (20%) divided by the percentage of customers who purchase cake (40%):
[tex]P(Coffee|Cake) = \frac{P(Coffee\cap Cake)}{P(Cake)}\\P(Coffee|Cake) =\frac{0.20}{0.40}=0.50[/tex]
50% of the customers also purchased coffee given that they have purchased a piece of cake.
Final answer:
To find the probability that a customer purchased a cup of coffee given they purchased a cake, we use conditional probability, resulting in a 50% chance.
Explanation:
The question requires us to use conditional probability to find the probability that a customer who purchased a piece of cake also purchased a cup of coffee. The known probabilities are that 70% of customers purchase coffee, 40% purchase cake, and 20% purchase both.
To find the probability that a customer purchased coffee given they purchased cake, we use the formula for conditional probability: P(A|B) = P(A ∩ B) / P(B), where A is the event 'customer buys coffee' and B is the event 'customer buys cake'. Given P(A ∩ B) = 20% or 0.2, and P(B) = 40% or 0.4, the calculation is P(A|B) = 0.2 / 0.4.
So, the conditional probability is 0.5 or 50%.
You prepare decorations for a party. How many ways are there to arrange 12 blue balloons, 9 green lanterns and 6 red ribbons in a row, such that no two ribbons are next to each other?
Answer: 27! - [22! * 6!]
Step-by-step explanation:
Decorations available = 12 blue ballons, 9 Green Lanterns and 6 red ribbons
To determine the number if ways of arrangement for if the ribbons must not be together, we must first determine the number of possible ways to arrange these decorations items if there are no restrictions.
Total number of ways to arrange them = [12+9+6]! = 27! = 1.089 * 10^28
If this ribbons are to be arranged distinctively by making sure all six of them are together, then we arrange the 6 of them in different ways and consider the 6ribbons as one entity.
Number of ways to arrange 6 ribbons = 6!.
To arrange the total number of entity now that we have arranged this 6ribbons together and taking them as one entity become: = (12 + 9 + 1)!. = 22!
Number of ways to arrange if all of the ribbons are taken as 1 and we have 22 entities in total becomes: 22! * 6!.
Hence, to arrange these decoration items making sure no two ribbons are together becomes:
= 27! - (22! * 6!)
= 1.08880602 * 10^28
To arrange the decorations, first calculate the number of ways to arrange the balloons and lanterns, then use the concept of stars and bars to calculate the number of ways to place the ribbons. The product of these two calculations gives the total number of ways to arrange the decorations.
Explanation:To arrange the balloons, lanterns, and ribbons in a row without any two ribbons being next to each other, we need to consider their positions separately. We can arrange the balloons and lanterns first, making sure no two ribbons are adjacent. The number of ways to arrange the balloons and lanterns is (12+1)C12 x (9+1)C9 = 13C12 x 10C9 = 13 x10 = 130.
Now, let's place the ribbons using the concept of stars and bars. We can consider the spaces between the balloons and lanterns as 'bars' and the ribbons as 'stars.' Since there are 13 spaces between the 12 balloons and lanterns, we have 13 bars. We need to place 6 ribbons (stars) in these 13 spaces. The number of ways to do this is (13-6+1)C6 = 8C6 = 28.
The total number of ways to arrange the decorations is the product of the number of ways to arrange the balloons and lanterns and the number of ways to place the ribbons. Therefore, the answer is 130 x 28 = 3640.
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(a) The mean age at death is 15 years and the standard deviation is 7 years. What percentage of the dinosaurs' ages were within 1 standard deviation of the mean? (Answer as a whole number.)
Answer:
68% of the dinosaurs' ages were within 1 standard deviation of the mean.
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.
What percentage of the dinosaurs' ages were within 1 standard deviation of the mean?
By the Empirical Rule, 68% of the dinosaurs' ages were within 1 standard deviation of the mean.
Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 33% of healthy people (those who have not had heart attacks) have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 15% probability.
A. If someone has periodontal disease, what is the probability that he or she will have a heart attack?
B. If 38% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack?
Final answer:
To find the probability of someone with periodontal disease having a heart attack, we can use conditional probability and Bayes' theorem. The probability of having a heart attack given that someone has periodontal disease is 79%. Using Bayes' theorem, we can calculate the probability of a person with periodontal disease having a heart attack.
Explanation:
To answer part A, we can use conditional probability. The probability of having a heart attack given that someone has periodontal disease is represented by P(heart attack | periodontal disease). According to the given information, 79% of people who have suffered a heart attack had periodontal disease. Therefore, P(heart attack | periodontal disease) = 0.79.
To answer part B, we can use Bayes' theorem. The probability of a person with periodontal disease having a heart attack is represented by P(heart attack | periodontal disease). According to the given information, the probability of having a heart attack in the community is 38%.
Therefore, P(heart attack) = 0.38. Let's use Bayes' theorem: P(heart attack | periodontal disease) = (P(periodontal disease | heart attack) × P(heart attack)) / P(periodontal disease). We know that P(periodontal disease | heart attack) = 0.79 from part A.
The probability of having periodontal disease in the community is represented by P(periodontal disease). In the given information, it says that only 33% of healthy people have periodontal disease. Therefore, P(periodontal disease) = 0.33. Substituting these values into Bayes' theorem, we can calculate P(heart attack | periodontal disease).
In France gasoline is 2.096 per liter. There are 3.78541178 liters per gallon. If the Euro is trading at 1.762 then the equivalent price per gallon is
Answer:
equivalent price/ gallon = 13.98 US$ / gallon
Step-by-step explanation:
Assuming that the exchange rate is 1.762 US$ /€ , then
thus
equivalent price/ gallon = liters/gallon * euros / liter * dollars/euro
equivalent price/ gallon = 3.78541178 liters/ gallon * 2.096 € /liter * 1.762 US$ /€ = 13.98 US$ / gallon
equivalent price/ gallon = 13.98 US$ / gallon
if the actual price would be different from the equivalent , then there would be an arbitrage opportunity ( profit with no risk)
To find the price per gallon in France converted to dollars, multiply the price per liter by liters per gallon to get the Euro amount, then convert to dollars using the exchange rate. The result is approximately $13.97 per gallon.
Explanation:The question asks to determine the price of gasoline per gallon in France when converted to Euros, considering the given petrol price per liter and the exchange rate. To calculate this, we will first convert the price from liters to gallons and then convert Euros to the equivalent value in using the exchange rate.
Calculate the price per gallon in Euros by multiplying the price per liter by the number of liters in a gallon:Therefore, the equivalent price of gasoline per gallon in France, when converted to dollars, is approximately $13.97.
In a multicriteria decision problem____________.a. it is impossible to select a single decision alternative.b. the decision maker must evaluate each alternative with respect to each criterion.c. successive decisions must be made over time.d. each of these choices are true.
Answer:
Option b.
Step-by-step explanation:
In a situation of a multicriteria decision problem, the decision-maker can tackle a given situation in at least two alternatives but due to some conflicting objectives, he is unable to choose the best option among the alternatives.
Therefore, in a multicriteria decision problem, the decision-maker must evaluate each alternative with respect to each criterion.
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight should the citation designation be established
Answer:
The weight of the citation designation should be at 4.8432 pounds.
Explanation:
Given
Mean [tex]= 3.2 pounds.[/tex]
Standard deviation[tex]= 0.8 pound.[/tex]
Step 1:
Consider 'y' as one of the top weight, that is, [tex]y = 2 \% = 2.054 pounds.[/tex]
Let 'x' be the weight of the citation designation.
[tex]y = \frac{x-mean}{standard\ deviation}[/tex]
[tex]=2.054 = \frac{x-3.2}{0.8}[/tex]
[tex]=2.054\times 0.8 = x-3.2[/tex]
[tex]=1.6432 = x-3.2[/tex]
[tex]x = 1.6432+3.2[/tex]
[tex]x = 4.8432[/tex]
Thus, at 4.8432 pounds citation designation be established.
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https://brainly.com/question/18675815Final answer:
The citation catfish designation should be established at a weight of at least 4.84 pounds, which corresponds to the top 2% of a normal distribution given the mean is 3.2 pounds and the standard deviation is 0.8 pounds.
Explanation:
The weight at which a citation catfish designation should be established is calculated by identifying the z-score that corresponds to the top 2% of a normal distribution. Given that the average weight of a catfish is 3.2 pounds with a standard deviation of 0.8 pounds, we can use a z-table to find the z-score for the 98th percentile, which typically is around 2.05.
Using the z-score formula: z = (X - mean) / standard deviation, we rearrange to solve for the unknown X (catfish weight): X = z * standard deviation + mean. Plugging in the numbers: X = 2.05 * 0.8 + 3.2, we find that the citation catfish should weigh at least 4.84 pounds.
Please help! Will Mark Brainliest! I'm struggling on how to get the answer so please clarify.
Question Choose the correct simplification of the expression: (3m/n)^4
Options:
81m^4/n^4
12m^4/n
3m^4/n^4
81m^4/n
Picture is posted with the problem for clarification. The subject is multiplying and dividing monomials.
Answer:
Option 1
Step-by-step explanation:
(3m/n)⁴ = (3⁴×m⁴)/n⁴
= 81m⁴/n⁴
Answer the following
Answer:
Step-by-step explanation:
f(x) = 3x-2
g(x) = 1/2(x²)
f(0) = 3x-2 = 3(0)-2 = 0-2 = -2
f(-1) = 3x-2 = 3(-1)-2 = -3-2 = -5
g(4) = 1/2(4²) = 1/2(16) = 8
g(-1) = 1/2(-1²) = 1/2(1) = 1/2
A boy has color blindness and has trouble distinguishing blue and green. There are 75 blue pens and 25 green pens mixed together in a box. Given that he picks up a blue pen, there is a 80% chance that he thinks it is a blue pen and a 20% chance that he thinks it is a green pen. Given that he picks a green pen, there is an 90% chance that he thinks it is a green pen and a 10% chance that he thinks it is a blue pen. Assume that the boy randomly selects one of the pens from the box.
a) What is the probability that he picks up a blue pen and recognizes it as blue?
b) What is the probability that he chooses a pen and thinks it is blue?
c) (Given that he thinks he chose a blue pen, what is the probability that he actually chose a blue pen?
Answer:
a) There is a 60% probability that he picks up a blue pen and recognizes it as blue.
b) There is a 62.5% probability that he chooses a pen and thinks it is blue.
c) Given that he thinks he chose a blue pen, there is a 96% probability that he actually chose a blue pen.
Step-by-step explanation:
We have these following probabilities
A 75% probability that a pen is blue
A 25% probability that a pen is green
If a pen is blue, an 80% probability that the boy thinks it is a blue pen
If a pen is blue, a 20% probability that the boy thinks it is a green pen.
If a pen is green, a 90% probability that the boy thinks it is a green pen.
If a pen is green, a 10% probability that that the boy thinks it is a blue pen.
a) What is the probability that he picks up a blue pen and recognizes it as blue?
There is a 75% probability that he picks up a blue pen.
There is a 80% that he recognizes a blue pen as blue.
So
[tex]P = 0.75*0.8 = 0.6[/tex]
There is a 60% probability that he picks up a blue pen and recognizes it as blue.
b) What is the probability that he chooses a pen and thinks it is blue?
There is a 75% probability that he picks up a blue pen.
There is a 80% that he recognizes a blue pen as blue.
There is a 25% probability that he picks up a green pen
There is a 10% probability that he thinks a green pen is blue.
So
[tex]P = 0.75*0.80 + 0.25*0.10 = 0.625[/tex]
There is a 62.5% probability that he chooses a pen and thinks it is blue.
c) (Given that he thinks he chose a blue pen, what is the probability that he actually chose a blue pen?
There is a 62.5% probability that he thinks that he choose a blue pen.
There is a 60% probability that he chooses a blue pen and think that it is blue.
So
[tex]P = \frac{0.6}{0.625} = 0.96[/tex]
Given that he thinks he chose a blue pen, there is a 96% probability that he actually chose a blue pen.
Final answer:
The answer provides the probabilities related to a color-blind boy picking and identifying blue and green pens accurately.So,a)The overall probability is 0.75 * 0.80 = 0.60,b)0.62,c)0.968.
Explanation:
a) What is the probability that he picks up a blue pen and recognizes it as blue?
Let's calculate this by considering the probabilities given:
Probability of picking a blue pen: 75/100 = 0.75
Probability of recognizing a blue pen as blue given it is blue: 0.80
The overall probability is 0.75 * 0.80 = 0.60.
b) What is the probability that he chooses a pen and thinks it is blue?
This includes both him picking a blue pen and thinking it is blue or picking a green pen and mistakenly thinking it is blue. It can be calculated as 0.75 * 0.80 + 0.25 * 0.10 = 0.62
c) Given he thinks he chose a blue pen, what is the probability he actually chose a blue pen?
This involves computing the conditional probability using Bayes' theorem:
P(chose blue pen | thinks it's blue) = P(chose blue pen and thinks it's blue) / P(thinks it's blue) = (0.75 * 0.80) / 0.62 = 0.968.
The operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, each of which has a nominal value of 575 millimeters. A sample of five tires of each grade is selected, and the results representing the inner diameters of the tires, ranked from smallest to largest, are as follows:
Grade X
568 570 575 578 584
Grade Y
573 574 575 577 578
a. For each of the two grades of tires, compute the mean, median, and mode.
b. Compute and Interpret S
c. Which grade of tire is providing better quality? Explain.
d. What would be the effect on your answers in (a) if the last value for grade Y were
888 instead of 578?
Answer:
Step-by-step explanation:
given that the operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, each of which has a nominal value of 575 millimeters. A sample of five tires of each grade is selected, and the results representing the inner diameters of the tires, ranked from smallest to largest, are as follows:
Grade X
568 570 575 578 584
Mean = Total/5 = 575
Median = middle entry = 575
Mode = nil
Grade Y
573 574 575 577 578
Mean = sum/5 = 575.4
Median = middle entry =575
Mode = nil
PLEASE HELP 50 COINS!!!!
Answer:
Therefore,
[tex]AB=16.25\ units[/tex]
The Measurement of AB is 16.25 units.
Step-by-step explanation:
In Right Angle Triangle ABC
m∠C=90°
AC = 10.01 .....(Adjacent Side to angle A)
m∠A=52°
cos 52 ≈ 0.616
To Find:
AB = ? (Hypotenuse)
Solution:
In Right Angle Triangle ABC, Cosine Identity,
[tex]\cos A= \dfrac{\textrm{side adjacent to angle A}}{Hypotenuse}\\[/tex]
Substituting the values we get
[tex]\cos 52= \dfrac{AC}{AB}=\dfrac{10.01}{AB}[/tex]
But cos 52 ≈ 0.616 ....Given
[tex]AB=\dfrac{10.01}{0.616}=16.25\ units[/tex]
Therefore,
[tex]AB=16.25\ units[/tex]
The Measurement of AB is 16.25 units.
Consider the plane which passes through the three points: (−6,−4,−6) , (−2,0,−1), and (−2,1,1). Find the vector normal to this plane which has the form:
Answer:
The normal vector is [tex]{\bf n} \ =\ 3{\bf i} -8{\bf j} +4{\bf k}[/tex]
Step-by-step explanation:
Let
[tex]{\bf b}=\langle\,x_1,\,y_1,\,z_1\, \rangle\,, \ \ {\bf r} = \langle\,x_2,\,y_2,\,z_2\, \rangle\,, \ \ {\bf s} = \langle\,x_3,\,y_3,\,z_3\, \rangle.[/tex]
The vectors
[tex]\overrightarrow{QR} \ = \ {\bf r} - {\bf b} \,, \qquad \overrightarrow{QS} \ = \ {\bf s} - {\bf b} \,,[/tex]
then lie in the plane. The normal to the plane is given by the cross product
[tex]{\bf n} = ({\bf r} - {\bf b})\times ( {\bf s} - {\bf b})[/tex]
We have the following points:
[tex]Q(-6,\,-4,\,-6)\,, \ \ R(-2,\,0,\,-1)\,, \ \ S(-2,\,1,\,1)\,.[/tex]
when the plane passes through [tex]Q,\, R[/tex], and [tex]S[/tex], then the vectors
[tex]\overrightarrow{QR} = \langle\, -2-(-6),\, 0-(-4),\, -1-(-6)\, \rangle\,, \overrightarrow{QS}= \langle\, -2-(-6),\, 1-(-4),\, 1-(-6)\, \rangle\,,\\\\\overrightarrow{QR} = \langle\, 4,\, 4,\, 5\, \rangle\,,\qquad \overrightarrow{QS}= \langle\, 4,\, 5,\, 7\, \rangle\,[/tex]
lie in the plane. Thus the cross-product
[tex]{\bf n} \ = \ \left|\begin{array}{ccc}{\bf i} & {\bf j} & {\bf k} \\ 4 & 4 & 5 \\ 4 & 5 & 7 \end{array}\right| \ = \begin{pmatrix}4\cdot \:7-5\cdot \:5&5\cdot \:4-4\cdot \:7&4\cdot \:5-4\cdot \:4\end{pmatrix} \ = \ 3{\bf i} -8{\bf j} +4{\bf k}[/tex]
is normal to the plane.
The normal vector to the plane passing through the points (-6,-4,-6), (-2,0,-1), and (-2,1,1) is (-6, -8, 4). This is found by obtaining two vectors from the given points and then taking the cross product of these vectors.
Explanation:To find the vector normal to the plane passing through the points (-6, -4, -6), (-2, 0, -1), and (-2, 1, 1), we first need to find two vectors lying in the plane that originate from the same point. Let's take the first point as common and obtain the vectors AB and AC:
AB = B - A = (-2 - (-6), 0 - (-4), -1 - (-6)) = (4, 4, 5)
AC = C - A = (-2 - (-6), 1 - (-4), 1 - (-6)) = (4, 5, 7)
Now, we can find the normal vector to the plane by taking the cross-product of these two vectors. The cross product of two vectors provides a new vector which is perpendicular to the original vectors:
N = AB x AC = (4*7 - 5*5, 5*4 - 4*7, 4*5 - 4*4) = (-6, -8, 4)
Therefore, the normal vector to the plane is (-6, -8, 4).
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A bag contains 5 red marbles, 8 blue marbles, and 3 green marbles. If a green marble is drawn you win $5, and if a red marble is drawn you win $1. Drawing a blue marble causes you to lose $1. Suppose you make 81 draws out of the bag Your chances remain the same, the number of red, blue and green marbles doesn't change from draw to draw. This corresponds to drawing how many tickets at random?
Final answer:
The number of tickets drawn at random can be calculated using the formula for the number of combinations.
Explanation:
To determine the number of tickets drawn at random in this scenario, we need to calculate the total number of possible outcomes. In this case, Maria is drawing marbles without replacement, meaning that the number of marbles in the bag decreases with each draw. We can use the formula for the number of combinations to calculate the total number of outcomes. The formula is nCr = n! / (r!(n-r)!), where n is the total number of marbles in the bag and r is the number of marbles drawn. In this case, n = 5+8+3 = 16 and r = 2. Therefore, the number of tickets drawn at random is 16C2 = 120.
"Suppose you draw a single card from a standard deck of 52 cards. How many ways arethere to draw either queen or a heart?"
Answer:
16 ways
Step-by-step explanation:
We are given that
Total cards=52
Total queen in deck of cards=4
Total cards of heart=13
We know 1 card of queen is heart
Number of ways drawing one card of heart out of 13=[tex]13C_1[/tex]
Using combination formula ;[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
Number of ways drawing one card of heart out of 13=[tex]\frac{13!}{1!12!}=\frac{13\times 12!}{12!}=13[/tex]
Number of ways of drawing one card of queen out of 4=[tex]4C_1=\frac{4!}{3!}[/tex]
Number of ways of drawing one card of queen out of 4=[tex]\frac{4\times 3!}{3!}=4[/tex]
Total number of ways of drawing one card out of 52 cards=[tex]13+4-1=16[/tex]