Answer and explanation:
To find : List the sample space and tell whether you think the events are equally likely ?
Solution :
a) Toss 2 coins; record the order of heads and tails.
Let H is getting head and t is getting tail.
When two coins are tossed the sample space is {HH,HT,TH,TT}.
Total number of outcome = 4
As the outcome HT is different from TH. Each outcome is unique.
Events are equally likely since their probabilities [tex]\frac{1}{4}[/tex] are same.
b) A family has 3 children; record the number of boys.
Let B denote boy and G denote girl.
If there are 3 children then the sample space is
{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}
The possible number of boys are 0,1,2 and 3.
Number of boys Favorable outcome Probability
0 GGG [tex]\frac{1}{8}[/tex]
1 GGB,GBG,BGG [tex]\frac{3}{8}[/tex]
2 GBB,BGB,BBG [tex]\frac{3}{8}[/tex]
3 BBB [tex]\frac{1}{8}[/tex]
Since the probabilities are not equal the events are not equally likely.
c) Flip a coin until you get a head or 3 consecutive tails; record each flip.
Getting a head in a trial is dependent on the previous toss.
Similarly getting 3 consecutive tails also dependent on previous toss.
Hence, the probabilities cannot be equal and events cannot be equally likely.
d) Roll two dice; record the larger number
The sample space of rolling two dice is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Now we form a table that the number of time each number occurs as maximum number then we find probability,
Highest number Number of times Probability
1 1 [tex]\frac{1}{36}[/tex]
2 3 [tex]\frac{3}{36}[/tex]
3 5 [tex]\frac{5}{36}[/tex]
4 7 [tex]\frac{7}{36}[/tex]
5 9 [tex]\frac{9}{36}[/tex]
6 11 [tex]\frac{11}{36}[/tex]
Since the probabilities are not the same the events are not equally likely.
The sample spaces for different scenarios include (a) {HH, HT, TH, TT}, (b){0, 1, 2, 3}, (c){H, TH, TTH, TTTH}, and (d){1, 2, 3, 4, 5, 6}.
Let's explore the sample spaces and determine if the events are equally likely for each scenario:
(a) Toss 2 Coins; Record the Order of Heads and Tails
The sample space is: {HH, HT, TH, TT}. Each of these outcomes has an equal probability of occurring, since each coin flip is independent and there are two coins, making each combination equally likely.
(b) A Family Has 3 Children; Record the Number of Boys
The sample space can be represented by the number of boys: {0, 1, 2, 3}. Assuming that each child is equally likely to be a boy or a girl, the probabilities of each outcome differ. For example, having 1 boy (and thus 2 girls) has more combinations than having 3 boys or no boys at all.
(c) Flip a Coin Until You Get a Head or 3 Consecutive Tails; Record Each Flip
The sample space includes sequences that stop as soon as we get a head or reach 3 tails: {H, TH, TTH, TTTH}. These outcomes are not equally likely because the sequence lengths vary, affecting their probabilities.
(d) Roll Two Dice; Record the Larger Number
The sample space is: {1, 2, 3, 4, 5, 6}. However, the events are not equally likely. For example, getting a 6 as the larger number is more probable since it can occur in more pairs (e.g., (6,5), (6,4), etc.) compared to lower numbers.
The amounts due on a mobile phone bill in Ireland are normally distributed with a mean of € 53 and a standard deviation of € 15. If a monthly phone bill is chosen at random, find the probability that we get a phone bill between 47 and 74 euros.
Answer:
0.5746
Step-by-step explanation:
Lets call X monthly bill chosen. X is a random variable with distribution N(μ=53,σ =15). We first standarize X, lets call W = (X-53)/15, random variable obtained from X by substracting μ and dividing by σ. W is a random variable with distribution N(0,1) and the cummulative function of W, Φ, is tabulated. You can find the values of Φ on the attached file. Using this information we obtain
[tex] P(47 < X < 74) = P(\frac{47-53}{15} < \frac{X-53}{15} < \frac{74-53}{15}) = P(-0.4 < W < 1.4) = \phi(1.4) - \phi(-0.4) [/tex]
A standard normal random variable has a symmetric density function, as a result Φ(-0.4) = 1- Φ(0.4) = 1- 0.6554 = 0.3446. Also, Φ(1.4) = 0.9192. We conclude that Φ(1.4)-Φ(-0.4) = 0.9192-0.3446 = 0.5746, that is the probability that the phone bill is between 47 and 74 euros.
The local orchestra has been invited to play at a festival. There are 111 members of the orchestra and 6 are licensed to drive large multi-passenger vehicles. Busses hold 25 people but are much more expensive to rent. Passenger vans hold 12 people. make a system of equations to find the smallest number of busses the orchestra can rent. Let x= the number of busses and y= the number of vans make an equation representing the number of vehicles needed. __________ make an equation representing the total number of seats in vehicles for the orchestra members. __________ Solve the system of equations. How many busses does the orchestra need to rent? ____________ How many 12-passenger vans does the orchestra need to rent? ____________
- Make an equation representing the number of vehicles needed.
We have six drivers so
x + y ≤ 6
That's not really an equation; it's an inequality. We want to use all our drivers so we can use the small vans, so
x + y = 6
- Make an equation representing the total number of seats in vehicles for the orchestra members.
s = 25x + 12y
That's how many seats total; it has to be at least 111 so again an inequality,
25x + 12y ≥ 111
We solve it like a system of equations.
x + y = 6
y = 6 - x
111 = 25x + 12y = 25x + 12(6-x)
111 = 25x + 72 - 12x
111 - 72 = 13 x
39 = 13 x
x = 3
Look at that, it worked out exactly. It didn't have to.
y = 6 - x = 3
Answer: 3 buses, 3 vans
Using a System of equations, the number of buses and vans required are 3 respectively.
Using the system of equations :
Total number of buses :
b + v ≤ 6 - - - - (1)Total number of passengers :
25b + 12v ≥ 111 - - - - (2)From (1)
b = 6 - v - - - - - (3)Substitute (3) into (2)
25(6-v) + 12v = 111
150 - 25v + 12v = 111
-13v = 111 - 150
-13v = - 39
v = 39/13
v = 3
From (3)
b = 6 - 3
b = 3
Hence, the number of vans and buses required are 3 and 3 respectively.
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According to a survey of American households, the probability that the residents own two cars if annual household income is over $50,000 is 80%. Of the households surveyed, 60% had incomes over $50,000 and 70% had two cars. The probability that the residents of a household do not own two cars and have an income over $50,000 a year is __________?
Answer:
12%
Step-by-step explanation:
The probability that the residents of a household do not own two cars and have an income over $50,000 a year is 100% subtracted by the probability of owning two cars and having an income over $50,000(P(2∩50+) and the probability of not having an income over $50,000 (P(50-))
[tex]P = 1 - P(2\cap 50+) - P(50-)\\P = 1 - (0.6*0.7) - (1-0.6)\\P=1-0.48-0.40\\P=0.12[/tex]
Therefore, the probability that the residents of a household do not own two cars and have an income over $50,000 a year is 12%
Final answer:
To find the probability that a household has an income over $50,000 and does not own two cars, multiply the probability of a household not owning two cars (20%) by the probability of having an income over $50,000 (60%), which equals 12%.
Explanation:
The question is asking us to find the probability that the residents of a household do not own two cars and have an income over $50,000 a year. We know that the probability a household with income over $50,000 owns two cars is 80%, so the probability that such a household does not own two cars is 1 - 0.80 = 0.20 or 20%. Since 60% of households have an income over $50,000, we can calculate the probability that a household has an income over $50,000 and does not own two cars by multiplying these two probabilities: 0.60 * 0.20 = 0.12 or 12%.
In a Gallup telephone survey conducted on April 9-10, 2013, the person being interviewed was asked if they would vote for a law in their state that would increase the gas tax up to 20 cents a gallon, with the new gas tax money going to improve roads and bridges and build more mass transportation in their state. Possible responses were vote for, vote against, and no opinion. Two hundred ninety five respondents said they would vote for the law, 672 said they would vote against the law, and 51 said they had no opinion.
a. Do the responses for this question provide categorical or quantitative data?
b. What was the sample size for this Gallup poll?
c. What percentage of respondents would vote for a law increasing the gas tax?
d. Do the results indicate general support for or against increasing the gas tax to improve roads and bridges and build more mass transportation?
Answer:
a. categorical
b. 1018
c. 28.98%
d. Most voters interviewed are against the new tax
Step-by-step explanation:
Hello!
There was a poll made to know the voter opinion on a law that would increase the gas tax to 20 cents per gallon, this money is going to use for road and bridges improvement and build more mass transportation in the state.
Data:
295 support
672 against
51 no opinion
a. The study variable is "Opinion of the votes on the new gas tax" Categorized: "support; against; no opinion"
This is a categorical variable, that can take one number on a limited amount of possibles values, assigning each observation to a group of nominal categories based on a qualitative feature.
In this case, each voter is assigned to a category according to their opinion on the new tax.
b. To know the sample size you have to add the subtotals of each category:
n= 295 + 672 + 51 = 1018
c. To calculate the percentage of people supporting the new tax, you have to divide the subtotal by the total and multiply it by 100
[tex](\frac{295}{1018})*100[/tex] = 28.978% ≅ 28.98%
d. To answer this it's better to calculate the percentage of each category:
Support: 28.98%
Against: 66.01%
No opinion: 5.01%
Since the category "against" has the greater percentage, you can say that most voters interviewed are against the new gas tax
I hope it helps!
Chrissy walked 1/8 of a mile on Monday and 1/8 of a mile on Tuesday. How many miles did Chrissy walk on Monday and Tuesday combined? Express the answer in lowest terms.
1/16
1/8
1/4
1/2
Answer:
Chrissy walked 1/4 miles on Monday and Tuesday combined.
Step-by-step explanation:
1/8+1/8=2/8
2/8=1/4 you take half of both sides to get the lowest terms using simplification.
solve for −5(x+1)=−3(2x−2)
Answer:
x=11
Step-by-step explanation:
The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flash Drives randomly select a sample of 17 USB flash drives. If the mean weight of the USB flash drives is too heavy or too light, the machinery is shut down for adjustment; otherwise, the production process continues. The last sample showed a mean and standard deviation of 31.9 and 1.8 grams, respectively. Using α = 0.10, the critical t values are _______.
A. reject the null hypothesis and shut down the process.
B. reject the null hypothesis and do not shut down the process.
C. fail to reject the null hypothesis and do not shut down the process) do nothing.
D. fail to reject the null hypothesis and shut down the process.
The critical t values are reject the null hypothesis and shut down the process.
The correct option is (A).
To find the critical t values for a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex], we need to look up the t-distribution table or use statistical software.
Since we have a sample size of 17, the degrees of freedom ( df ) for the t-test will be [tex]\( n - 1 = 17 - 1 = 16 \).[/tex]
For a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom, we find the critical t value.
From the t-distribution table or using statistical software, the critical t value for [tex]\( \alpha = 0.10 \) and \( df = 16 \)[/tex] is approximately 1.337.
So, the critical t values are approximately [tex]\( \pm 1.337 \).[/tex]
Now, we compare the calculated t value for the sample mean to these critical values. If the calculated t value falls beyond these critical values, we reject the null hypothesis.
Since the problem does not provide the calculated t value, we cannot determine whether to reject the null hypothesis or not.
However, based on the information given, we can see that the mean weight of the USB flash drives in the last sample is 31.9 grams, which is heavier than the expected mean weight of 30 grams. If the calculated t value corresponds to a significantly larger value than the critical t value, we may reject the null hypothesis and conclude that the mean weight is significantly different from 30 grams, potentially leading to shutting down the process for adjustment.
Therefore, the correct answer would be:
A. reject the null hypothesis and shut down the process.
Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is: A. reject the null hypothesis and shut down the process.
To find the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and a sample size of n = 17, we can use a t-distribution table or a statistical software.
For a two-tailed test at a significance level of [tex]\( \alpha = 0.10 \)[/tex] and degrees of freedom df = n - 1 = 17 - 1 = 16, we need to find the critical t-values that correspond to the upper and lower tails of the t-distribution.
Using a t-distribution table or a statistical software, the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom are approximately [tex]\( t_{\alpha/2} = \pm 1.746 \).[/tex]
So, the critical t-values are approximately[tex]\( \pm 1.746 \).[/tex]
Now, let's compare the calculated t-value with the critical t-values:
The calculated t-value is given by:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean (31.9 grams)
- [tex]\( \mu \)[/tex] is the population mean (30 grams)
- s is the sample standard deviation (1.8 grams)
- n is the sample size (17)
Substitute the given values:
[tex]\[ t = \frac{31.9 - 30}{\frac{1.8}{\sqrt{17}}} \]\[ t \approx \frac{1.9}{\frac{1.8}{4.123}} \]\[ t \approx \frac{1.9}{0.437} \]\[ t \approx 4.35 \][/tex]
Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is:
A. reject the null hypothesis and shut down the process.
A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of Rin=80 L/minutes (min). The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y(t) denote the quantity of salt in the tank at time t. Assume that Rout=40L/min. (a) Set up and solve the differential equation for y(t). (b) What is the salt concentration when the tank overflows?
Final answer:
It would take roughly 22.22 hours to fill an 80,000 L swimming pool with a garden hose at a rate of 60 L/min. However, if diverting a moderate size river flowing at 5000 m³/s, it would fill the pool in approximately 0.016 seconds.
Explanation:
To estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min, we simply divide the total volume of the pool by the flow rate of the hose. For this calculation:
Time (min) = Total Volume (L) / Flow Rate (L/min)
Time = 80,000 L / 60 L/min = 1333.33 min
To convert minutes to hours, we divide by 60:
Time = 1333.33 min / 60 = 22.22 hours
For part (b), if you could divert a moderate size river, flowing at 5000 m³/s, into the pool, we need to convert this flow rate into liters per second (L/s) for consistency:
Flow Rate (L/s) = 5000 m³/s * 1000 L/m³
Flow Rate = 5000000 L/s
Now we can calculate the time it would take to fill the pool with this flow rate:
Time = 80,000 L / 5000000 L/s
Time = 0.016 seconds
Let A, B, C be events such that P(A) = 0.2 , P(B) = 0.3, P(C) = 0.4
Find the probability that at least one of the events A and B occurs if
(a) A and B are mutually exclusive;
(b) A and B are independent.
Find the probability that all of the events A, B, C occur if
(a) A, B, C are independent;
(b) A, B, C are mutually exclusive.
Answer:
At least one of the events A and B occurs:
(a) 0.5
(b) 0.44
All of the events A, B, C occur:
(a) 0.024
(b) 0
Step-by-step explanation:
Given:
P(A) = 0.2, P(B) = 0.3, P(C) = 0.4
At least one of the events A and B occurs:
(a) If A and B are mutually exclusive events, then their intersection is 0.
Probability of at least one of them occurring means either of the two occurs or the union of the two events. Therefore,
[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=0.2+0.3-0=0.5[/tex]
(b) If A and B are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,
[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=P(A)+P(B)+P(A)P(B)\\P(A\ or\ B)=0.2+0.3-(0.2)(0.3)=0.5-0.06=0.44[/tex]
All of the events A, B, C occur together:
(a) All of the events occurring together means the intersection of all the events.
If A, B, and C are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,
[tex]P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)\\P(A\cap B\cap C)=0.2\times 0.3\times 0.4=0.024[/tex]
(b) If A, B, and C are mutually exclusive events means events A, B, and C can't happen at the same time. Therefore, their intersection is 0.
[tex]P(A\cap B\cap\ C)=0[/tex]
Probability that at least one of the events A and B occurs:
(a) When A and B are mutually exclusive: 0.5 (b) When A and B are independent: 0.44
Probability that all of the events A, B, and C occur:
(a) When A, B, and C are independent: 0.024 (b) When A, B, and C are mutually exclusive: 0
To solve these probability questions, let's understand the given information and the rules of probability related to mutually exclusive and independent events.
Let P(A) = 0.2, P(B) = 0.3, and P(C) = 0.4.
1. Finding the probability that at least one of the events A and B occurs:
(a) When A and B are mutually exclusive:
Mutually exclusive events mean that A and B cannot occur at the same time. Therefore, the probability that at least one of them occurs is given by:
P(A OR B) = P(A) + P(B)
So, P(A OR B) = 0.2 + 0.3 = 0.5
(b) When A and B are independent:
Independent events mean that the occurrence of one event does not affect the occurrence of the other. The probability that at least one of them occurs is given by:
P(A OR B) = P(A) + P(B) - P(A AND B)
For independent events, P(A AND B) is calculated as: P(A AND B) = P(A) × P(B)
So, P(A AND B) = 0.2 × 0.3 = 0.06
Therefore, P(A OR B) = 0.2 + 0.3 - 0.06 = 0.44
2. Finding the probability that all of the events A, B, and C occur:
(a) When A, B, and C are independent:
For independent events, the probability that all of them occur is given by:
P(A AND B AND C) = P(A) × P(B) × P(C)
So, P(A AND B AND C) = 0.2 × 0.3 × 0.4 = 0.024
(b) When A, B, and C are mutually exclusive:
If events are mutually exclusive, it means they cannot all occur together. Therefore, the probability that all of them occur is:
P(A AND B AND C) = 0
What ratio is equivalent to 2 to 17
Answer:
4 to 34
Step-by-step explanation:
Answer:
There are a few correct equivalents to this. Please mark brainliest!!!
Step-by-step explanation:
4:34, 6:51, 8:68, and 10;85
A survey was conducted that asked 997 people how many books they had read in the past year. Results indicated that x overbar equals 13.2 books and sequels 18.9 books. Construct a 95 % confidence interval for the mean number of books people read. Interpret the interval.
Answer:
The 95% confidence interval would be given by (12.03;14.37)
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]\bar X=13.2[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=18.9 represent the sample standard deviation
n=997 represent the sample size
2) Calculate the confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=997-1=996[/tex]
Since the confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,996)".And we see that [tex]t_{\alpha/2}=1.96[/tex] and this value is exactly the same for the normal standard distribution and makes sense since the sample size is large enough to approximate the t distribution with the normal standard distribution.
Now we have everything in order to replace into formula (1):
[tex]13.2-1.96\frac{18.9}{\sqrt{997}}=12.03[/tex]
[tex]13.2+1.96\frac{18.9}{\sqrt{997}}=14.37[/tex]
So on this case the 95% confidence interval would be given by (12.03;14.37)
A survey of 61 comma 646 people included several questions about office relationships. Of the respondents, 26.5% reported that bosses scream at employees. Use a 0.05 significance level to test the claim that more than 1 divided by 4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to respond? Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
Answer:
H ₀ : p ≤ 0.25
H ₁: p > 0.25
Test statics value z = 12.91
Decision Rule:
Reject the value of null hypothesis
Conclusion:
If survey is online then there is sufficient evidence to claim . If the sample is voluntary response sample then conclusion is not valid.
Step-by-step explanation:
step by step explanation is shown in attachment.
The brightness of a picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in ¯x = 317.2 and s = 15.7 measured in microamps. (a) Find a 99% CI on mean current required.
Answer:
The 99% confidence interval for the mean would be (301.064;333.336) mA
Step-by-step explanation:
1) Notation and some definitions
n=10 sample selected
[tex]\bar x=317.2mA[/tex] sample mean for the sample tubes selected
[tex]s=15.7mA[/tex] sample deviation for the sample selected
Confidence = 99% or 0.99
[tex]\alpha=1-0.99=0.01[/tex] significance level
A confidence interval for the mean is used to "places boundaries around an estimated [tex]\bar X[/tex] so that the true population mean [tex]\mu[/tex] would be expected to lie within those boundaries with a confidence specified. If the uncertainty is large, then the interval between the boundaries must be wide; if the uncertainty is small, then the interval can be narrow"
2) Formula to use
For this case the sample size is <30 and the population standard deviation [tex]\sigma[/tex] is not known, so for this case we can use the t distributon to calculate the critical value. The first step would be calculate alpha
[tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], then we can calculate the degrees of freedom given by:
[tex]df=n-1=10-1=9[/tex]
Now we can calculate the critical value [tex]t_{\alpha/2}=3.25[/tex]
And then we can calculate the confidence interval with the following formula
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
3) Calculate the interval
Using the formula (1) and replacing the values that we got we have:
[tex]317.2 - 3.25\frac{15.7}{\sqrt{10}}=301.064[/tex]
[tex]317.2 + 3.25\frac{15.7}{\sqrt{10}}=333.336[/tex]
So then the 99% confidence interval for the mean would be (301.064;333.336)mA
Interpretation: A point estimate for the true mean of brightness level for the tubes in the population is 317.2mA, and we are 99% confident that the true mean is between 301.064 mA and 333.336 mA.
One hundred draws are made at random with replacement from a box with ninety-nine tickets marked "0" and one ticket marked "1." True or false, and explain:
(a) The sum will be around 1, give or take 1
(b) There is about a 68% chance that the sum will be in the range 0 to 2.
(a) False. The sum will not be around 1, give or take 1. (b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.
(a).
Each draw has a 99/100 probability of resulting in a sum of 0 and a 1/100 probability of resulting in a sum of 1. Since there are 100 draws made with replacement, the expected value of the sum is given by the average of the individual probabilities multiplied by the number of draws:Expected sum = (99/100 x 0) + (1/100 x 1)
= 1/100
= 0.01
Therefore, the expected value of the sum is 0.01, which is much closer to 0 than 1. While it is possible to observe a sum of 1 or close to 1 in a particular set of draws, it is not likely to be the average or typical outcome.
(b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.
To determine the probability, to consider the possible outcomes that result in a sum between 0 and 2.
The possible outcomes are:
Sum = 0: This can only happen if all 100 draws result in 0. The probability of this occurring is [tex](99/100)^{100} \approx 0.366[/tex].Sum = 1: This can only happen if exactly one draw results in 1, and the remaining 99 draws result in 0. The probability of this occurring is [tex]100 \times (1/100) \times (99/100)^{99} \approx 0.369[/tex].Sum = 2: This can only happen if exactly two draws result in 1, and the remaining 98 draws result in 0. The probability of this occurring is [tex](100 choose 2) \times (1/100)^2 \times (99/100)^{98} \approx 0.185.[/tex]Therefore, the total probability of the sum being in the range 0 to 2 is approximately:
[tex]0.366 + 0.369 + 0.185 = 0.92,[/tex]
which is far greater than 68%.
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Juan only paid $10 for his new letter. He got 25% discount what was the original price of the sweater?
Answer:
$12.50
Step-by-step explanation:
so Juan got his sweater at a 25% discount. So we woukd find 25% of 10 and then add it to $10
Answer:
Step-by-step explanation:
Juan only paid $10 for his new Sweater. He got a 25% discount
Let the original price of the sweater be $x. If he gets s 35% discount, it means that the price which she paid was lower than the original price.
Discount of 25% on the original price = 25/100 × x = 0.25x
The amount of money that Juan paid is the original price minus the discount of 25%. It becomes
x - 0.25x = 0.75x
Amount paid = $10. Therefore
0.75x = 10
x = 10/0.75 = $13.33
A Regional College uses the SAT to admit students to the school. The university notices that a lot of students apply even though they are not eligible. Given SAT scores are normally distributed, have a population mean of 500 and standard deviation of 100, what is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584?
Answer:
0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 100
n = 12
We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(greater than 525 but 584)
Standard error due to sampling =
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{100}{\sqrt{12}}[/tex]
[tex]P(525 < x < 584) = P(\displaystyle\frac{525 - 500}{\frac{100}{\sqrt{12}}} < z < \displaystyle\frac{584-500}{\frac{100}{\sqrt{12}}}) = P(0.866 < z < 2.909)\\\\= P(z \leq 2.909) - P(z < 0.866)\\= 0.998 - 0.807 = 0.191 = 19.1\%[/tex]
[tex]P(525 < x < 584) = 19.1\%[/tex]
0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.
Which shows a fractional and decimal equivalent of 9/20 ?
4/10=0.4
5/10=0.5
36/100=0.45
45/100=0.45
Answer:
45/100 = 0.45
Step-by-step explanation:
Divide 9 by 20.
9/20 = 0.45
Now write 36/100 as a decimal.
36/100 = 0.36
Write 45/100 as a decimal: 0.45
Answer: 45/100 = 0.45
For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 505 and a standard deviation of 170. The grading process of this year's exam has just begun. The average score of the 35 exams graded so far is 530.What is the probability that a sample of 35 exams will have a mean score of 530 or more if the exam scores follow the same distribution as in the past?
Answer:
[tex]P(\bar X>530)=1-0.808=0.192[/tex]
Step-by-step explanation:
1) 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".
Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=505,\sigma=170)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(505,\frac{170}{\sqrt{35}})[/tex]
2) Calculate the probability
We want this probability:
[tex]P(\bar X>530)=1-P(\bar X<530)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >530)=1-P(Z<\frac{530-505}{\frac{170}{\sqrt{35}}})=1-P(Z<0.87)[/tex]
[tex]P(\bar X>530)=1-0.808=0.192[/tex]
Using the central limit theorem, the probability of a sample mean of 530 or more is calculated considering the distribution of the sample mean, which is approximately normal due to the sample size.
Explanation:To calculate the probability that a sample of 35 exams will have a mean score of 530 or more, when the population mean is 505 with a standard deviation of 170, we use the concept of the sampling distribution of the sample mean. Since the sample size is 35, which is less than 10% of all psychology majors (assuming they number in the thousands), and the population distribution is normal or the sample size is large enough (which it is in this case), we can apply the Central Limit Theorem.
This theorem states that the sampling distribution of the sample mean will be approximately normally distributed with a mean equal to the population mean (μ = 505) and a standard deviation equal to σ/√n, where σ is the population standard deviation and n is the sample size (which is √35 in this case).
An ANOVA procedure is used for data obtained from five populations. Five samples, each comprised of 20 observations, were taken from the five populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are Select one:
A. 3 and 30
B. 4 and 30
C. 3 and 119
D. 3 and 116
E. None of the above answers is correct
Answer:
E. None of the above answers is correct
Step-by-step explanation:
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have [tex]5[/tex] groups and on each group from [tex]j=1,\dots,20[/tex] we have [tex]20[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
The degrees of freedom for the numerator on this case is given by [tex]df_{num}=df_{within}=k-1=5-1=4[/tex] where k =5 represent the number of groups.
The degrees of freedom for the denominator on this case is given by [tex]df_{den}=df_{between}=N-K=5*20-5=95[/tex].
And the total degrees of freedom would be [tex]df=N-1=5*20 -1 =99[/tex]
On this case the correct answer would be 4 for the numerator and 95 for the denominator.
E. None of the above answers is correct
A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of x¯=1.19 fluid ounces and the sample standard deviation is s=0.015 fluid ounces. Find a 95% two-sided confidence interval on the mean volume of syrup dispensed. Assume population is approximately normally distributed.
Answer: (1.1838, 1.1962)
Step-by-step explanation:
The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-
[tex]\overline{x}\pm t*\dfrac{s}{\sqrt{n}}[/tex],
where n= sample size
s= sample standard deviation.
[tex]\overline{x}[/tex]= sample mean
t* = Two-tailed critical t-value.
Given : n= 25
Degree of freedom : df = n-1 =24
Significance level [tex]=\alpha=1-0.95=0.95[/tex]
Now from students' t-distribution table , check the t-value for significance level [tex]\alpha/2=0.025[/tex] and df=24:
t*=2.0639
[tex]\overline{x}=1.19[/tex] fluid ounces
s= 0.015 fluid ounces
We assume that the population is approximately normally distributed
Now, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-
[tex]1.19\pm (2.0639)\dfrac{0.015}{\sqrt{25}}\\\\=1.19\pm (2.0639)(\dfrac{0.015}{5})\\\\=1.19\pm0.007728=(1.19-0.0061917,\ 1.19+0.0061917)\\\\=(1.1838083,\ 1.1961917)\approx(1.1838,\ 1.1962)[/tex]
∴ The required confidence interval = (1.1838, 1.1962)
To find the 95% two-sided confidence interval for the mean volume of syrup dispensed, use the t-distribution with the given sample mean, standard deviation, and size. The resulting interval is (1.183808, 1.196192) fluid ounces.
To calculate the 95% two-sided confidence interval for the mean volume of syrup dispensed, we can use the t-distribution, as the population standard deviation is unknown, and the sample size is less than 30.
Step-by-Step Explanation:
First, identify the sample mean (ar{x}), which is 1.19 fluid ounces, the sample standard deviation (s), which is 0.015 fluid ounces, and the sample size (n), which is 25.
Next, find the t-score that corresponds to a 95% confidence level and 24 degrees of freedom (n-1). You can use a t-distribution table or calculator for this, which typically gives you a t-score around 2.064.
Now calculate the margin of error (ME) using the formula ME = t*(s/sqrt{n}).
Finally, construct the confidence interval by adding and subtracting the margin of error from the sample mean: (1.19 - ME, 1.19 + ME).
Plugging in the values, we get ME = 2.064*(0.015/sqrt{25}) = 2.064*0.003 = 0.006192. So, the 95% confidence interval is (1.19 - 0.006192, 1.19 + 0.006192) = (1.183808, 1.196192).
Find the quadratic function f (x )equals ax squared plus bx plus c for which f (1 )equals negative 2, f (negative 3 )equals 26, and f (3 )equals 32.
Answer:
f(x) =4x² + x - 7
Step-by-step explanation:
f(x) = ax² + bx + c
f(1) = a + b + c
f(-3) = 9a - 3b + c
f(3) = 9a + 3b + c
a + b + c = -2 -----(1)
c = -2-a-b -----(2)
9a - 3b + c = 26 ------(3)
9a + 3b + c = 32 -------(4)
(2)->(3)
9a-3b-2-a-b = 26
8a-4b = 28
4(2a-b) = 28
2a-b = 7 ----(5)
(2)->(4)
9a+3b-2-a-b = 32
8a+2b = 34
2(4a+b) = 34
4a+b = 17 ----(6)
(5)+(6)
6a = 24
a = 4
sub a=4 into(6)
4(4)+b = 17
b = 1
sub a=4,b=1 into (2)
c = -2-4-1
c = -7
A firm that stocks lightbulbs gathers the following information: Demand = 19,500 units per year, Ordering cost = $25 per order, Holding cost = $4 per unit per year. The firm wants to calculate the: a) EOQ for the lightbulbs. b) Annual holding costs for the lightbulbs. c) Annual ordering costs for the lightbulbs.
Answer:
Step-by-step explanation:
Demand = 19,500 units per year (D)
Ordering cost = $25 per order (O)
Holding cost = $4 per unit per year (C)
a) [tex]EOQ=\sqrt{ \frac{2\times D\times O}{C}}[/tex]
[tex]EOQ=\sqrt{ \frac{2\times 19,500\times 25}{4}}[/tex]
= [tex]\sqrt{\frac{975000}{4}}[/tex]
= [tex]\sqrt{243,750}[/tex]
= 493.71044 ≈ 494
b) Annual holding cost = [tex]4\times(\frac{Q}{2})[/tex]
= [tex]4\times(\frac{494}{2})[/tex]
= 4 × 247
= 988
c) Annual ordering cost = [tex]O\times(\frac{D}{Q} )[/tex]
= [tex]25\times(\frac{19,500}{494} )[/tex]
= 25 × 39.47
= 986.75
AOQ = 494
Annual holding cost = 988
Annual ordering cost = 986.75
You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 30 apartments advertised in the local newspaper is $1,450. The standard deviation of the population is $220. Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.
Answer: 99% confidence interval would be [tex](1346.37,1553.63)[/tex]
Step-by-step explanation:
Since we have given that
Sample size = 30
Sample mean = $1450
Standard deviation = $220
We need to find the 99% confidence interval.
So, z = 2.58
So, confidence interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1450\pm 2.58\times \dfrac{220}{\sqrt{30}}\\\\=1450\pm 103.63\\\\=(1450-103.63,1450+103.63)\\\\=(1346.37,1553.63)[/tex]
Hence, 99% confidence interval would be [tex](1346.37,1553.63)[/tex]
A group of thirty people is selected at random. What is the probability that at least two of them will have the same birthday? (Round your answer to four decimal places.)
Answer:
0.706
Step-by-step explanation:
For uniformity, let's assuming none of the people selected is born on 29th of February. Therefore, the total possible birthday (sample space) for anybody selected is 365 days.
Probability of having at least two people having or not having the same birthday sum up to 1. It is easier to calculate the probability of people not having the same birthday.
The probability of at least two people having the same birthday = 1 - The probability of at least two people not having the same birthday
Pr (First Person selected is born in a day in a year) = [tex]\frac{365}{365}[/tex]
The next person selected is limited to 364 possible days
Pr (Second Person selected is born in a day in year) = [tex]\frac{364}{365}[/tex]
The next person selected is limited to 363 possible days
Pr (Third Person selected is born in a day in year) = [tex]\frac{363}{365}[/tex]
The next person selected is limited to 362 possible days
Pr (Fourth Person selected is born in a day in year) = [tex]\frac{362}{365}[/tex]
The next person selected is limited to 361 possible days
Pr (Fifth Person selected is born in a day in year) = [tex]\frac{361}{365}[/tex]
The next person selected is limited to 360 possible days
Pr (Sixth Person selected is born in a day in year) = [tex]\frac{360}{365}[/tex]
The next person selected is limited to 359 possible days
Pr (Seventh Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]
Looking at the pattern, the pattern continues by descending by 1 day from 365. The last person selected will have 336 possible days (365 – 29 days ) the other people before have used up 29 potential days.
Pr (Last Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]
The probability of selecting at least two people that will not have the same birthday, is the product of their 30 individual birthday probability.
[tex]\frac{365}{365} * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} * \frac{361}{365} * \frac{360}{365} * \frac{359}{365} ... \frac{337}{365} * \frac{336}{365} = [/tex]
This is a huge a hug e calculation to simplify. For simplicity the numerator and the denominator will be calculated separately.
The denominator is product of 365 in 30 times.
The denominator = 365³⁰
To simplify the numerator, the factorials method is used. Using factorials method, 365! = 365 × 364 × 363 × 362 × 361 × 360 × 359 × 358 ... × 3 × 2 × 1. But we only need the product of the integers from 365 to 336, so we’ll divide the unwanted numbers by dividing 365! by 335!
The numerator = [tex]\frac{365!}{335!}[/tex]
Therefore,
[tex]\frac{365!}{335! * 365^{30}} = 0.294[/tex]
The probability that no one selected in the group has the same birthday is 0.294. The probability that at least two people will have the same birthday is the complement of the probability that no one in the group has the same birthday.
1 - 0.294 = 0.706
The probability that at least two of them will have the same birthday = 0.706
The correct answer is approximately 0.7064.
To solve this problem, it is easier to calculate the probability that no one shares a birthday and then subtract this from 1 to find the probability that at least two people share a birthday.
Let's denote the probability that no two people share a birthday as P(no shared birthday). There are 365 days in a year, so the first person can have any birthday.
The second person then has 364 out of 365 days to have a different birthday, the third person has 363 out of 365 days, and so on, until the thirtieth person has 336 out of 365 days to avoid sharing a birthday with anyone else.
Thus, the probability that no one shares a birthday is:
[tex]\[ P(\text{no shared birthday}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \ldots \times \frac{336}{365} \][/tex]
We can calculate this product to find P(no shared birthday). Then, to find the probability that at least two people share a birthday, we subtract this from 1:
[tex]\[ P(\text{at least one shared birthday}) = 1 - P(\text{no shared birthday}) \][/tex]
Calculating the product for P(no shared birthday) and subtracting from 1 gives us the probability that at least two people in the group of thirty will have the same birthday.
When rounded to four decimal places, this probability is approximately 0.7064.
A cube has edges that are 3 inches in length. How many of these cubes will it take to completely fill a larger cube that has an edge of 1 foot?
64
Step-by-step explanation:
We will divide the volume of the larger cube with that if the smaller cube. However, we’ll first have to convert them to the same SI units;
12 inches = 1 foot
Therefore the volume of the larger cube;
12 * 12 * 12 = 1728 inches cubed
The volume of the smaller cube;
3 * 3 * 3 = 27 inches cubed
Divide the two;
1728/27
= 64
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To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were In the hypothesis test about cannabis use by conservatives and liberals, the P-value is extremely small. Which of the following errors is possible in this situation?
A. Type I only
B. Type II only
C. Type I and Type II
D. Neither Type I nor Type II
Answer:
A. TYPE 1 ERROR only
Step-by-step explanation:
In general terms:
‘a hypothesis has been rejected when it should have been accepted’. When this occurs, it is called a type I error, and,
‘a hypothesis has been accepted when it should have been rejected’.
When this occurs, it is called a type II error,
When testing a hypothesis, the largest value of probability which is acceptable for a type I error is called the level of significance of the test. The level of significance is indicated by the symbol α (alpha) and the levels commonly adopted are 0.1,0.05,0.01, 0.005 and 0.002.
A level of significance of 1%,say,0.01 means that 1 times in 100 the hypothesis has been rejected when it should have been accepted.
In significance tests, the following terminology is frequently adopted:
(i) if the level of significance is 0.01 or less, i.e. the confidence level is 9 9% or more, the results are considered to be highly significant, i.e. the results are considered likely to be correct,
(ii) if the level of significance is 0.05 or between 0.05and0.01,i.e.theconfidencelevelis95%or between 95% and 99%, the results are considered to be probably significant, i.e. the results are probably correct,
(iii) if the level of significance is greater than 0.05, i.e. the confidence level is less than 95%, the results are considered to be not significant, that is, there are doubts about the correctness of the results obtained.
You measure 44 textbooks' weights, and find they have a mean weight of 51 ounces. Assume the population standard deviation is 11.8 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places
Answer: (47.51, 54.49)
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= sample size .
[tex]\sigma[/tex] = population standard deviation.
[tex]\overline{x}[/tex] = sample mean
[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)
As per given , we have
[tex]\sigma=11.8\text{ ounces}[/tex]
[tex]\overline{x}=51 \text{ ounces}[/tex]
n= 44
Significance level for 95% confidence = [tex]\alpha=1-0.95=0.05[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]
Now, the 95% confidence interval for the true population mean textbook weight will be :-
[tex]51\pm (1.96)\dfrac{11.8}{\sqrt{44}}\\\\=51\pm(1.96)(1.7789)\\\\=51\pm3.486644\approx51\pm3.49\\\\=(51-3.49,\ 51+3.49)\\\\=(47.51,\ 54.49) [/tex]
Hence, the 95% confidence interval for the true population mean textbook weight. : (47.51, 54.49)
________ and currency risks are to key country success factors as land costs and ________ are to key region success factors. Cultural issues; zoning restrictions Labor cost; proximity to customers Land costs; air and rail systems Exchange rates; environmental impact All of the above are accurate relationships.
Country: Cultural issues, currency risks. Region: Land costs, air and rail systems. (Total: 10 words)
In assessing country and region success factors, certain variables hold significant weight. Country success factors encompass aspects such as cultural issues and currency risks, which directly influence a nation's economic stability and investment attractiveness.
Cultural nuances affect business practices and consumer behaviors, while currency risks impact trade competitiveness and profitability.
Conversely, region success factors pivot around considerations like land costs and transportation infrastructure, particularly air and rail systems. Land costs dictate the feasibility of establishing operations or acquiring property within a region, shaping investment decisions.
Meanwhile, efficient air and rail systems facilitate logistical operations and market access, enhancing the competitiveness of businesses located within the region.
These factors collectively contribute to a region's appeal for businesses seeking to establish or expand operations.
Thus, understanding and appropriately navigating these variables are crucial for organizations aiming to optimize their strategic positioning at both the national and regional levels, ensuring long-term success and sustainability in an increasingly complex global landscape.
The mean annual income for people in a certain city (in thousands of dollars) is 43, with a standard deviation of 29. A pollster draws a sample of 41 people to interview. What is the probability that the sample mean income is less than 42 (thousands of dollars)?
A) 0.4721 B) 0.4129 C) 0.3483 D) 0.5279
The probability that the average income of a 41-person sample would be less than $42,000, given a population with mean annual income of $43,000 and a standard deviation of $29,000, is 0.4129.
Explanation:This is a problem of probability concerning the Mean and Standard Deviation of the sampling distribution of the sample mean. We first have to find the Standard Error (SE) for the sampling distribution of the sample mean, which is given by SE = σ/√n , where σ is the standard deviation (29) and n is the sample size (41). Thus, SE = 29/√41 = 4.52.
Next, we find the Z score corresponding to the sample mean of 42. Z = (X - μ)/SE, where X is the sample mean (42), μ is the population mean (43), and SE is the standard error. So Z = (42 - 43)/4.52 = -0.22.
Looking up a Z score of -0.22 in a standard normal distribution table, we find a probability of 0.4129. Therefore, the answer is B) 0.4129.
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Final answer:
Using the Central Limit Theorem and provided statistics, the probability that the sample mean income is less than 42 thousand dollars is found by calculating a z-score and consulting standard normal distribution tables, resulting in a probability of 0.4129.
Explanation:
The task at hand is to calculate the probability that the sample mean income is less than 42 thousand dollars. Given the population mean (μ) of 43 thousand dollars and the standard deviation (σ) of 29 thousand dollars, with a sample size (n) of 41, we can use the Central Limit Theorem to determine that the sampling distribution of the sample mean will be normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
To find the probability, we first calculate the standard error (SE), which is:
SE = σ / √n = 29 / √41 ≈ 4.53
We then calculate the z-score for the sample mean of 42 thousand dollars as:
Z = (X - μ) / SE = (42 - 43) / 4.53 ≈ -0.22
Using standard normal distribution tables or a calculator, the probability of a z-score being less than -0.22 is approximately 0.4129. Thus, the answer is B) 0.4129.
Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0=2x2+3x-8?
Quadratic formula: x =StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction
A. 1.39
B. 2.00
C. 2.89
D. 3.50
Answer:
Step-by-step explanation:
The given quadratic equation is
2x^2+3x-8 = 0
To find the roots of the equation. We will apply the general formula for quadratic equations
x = -b ± √b^2 - 4ac]/2a
from the equation,
a = 2
b = 3
c = -8
It becomes
x = [- 3 ± √3^2 - 4(2 × -8)]/2×2
x = - 3 ± √9 - 4(- 16)]/2×2
x = [- 3 ± √9 + 64]/2×2
x = [- 3 ± √73]/4
x = [- 3 ± 8.544]/4
x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4
x = 5.544/4 or - 11.544/4
x = 1.386 or x = - 2.886
The positive solution is 1.39 rounded up to the nearest hundredth
Answer:
its a. 1.39
Step-by-step explanation: