Answer: 24 feet
Step-by-step explanation:
By using Pythagoras rule:
Let x be the high up the house does the ladder reached.
X^2 + 5^2= 25^2
X^2 = 25^2 - 5^2
x^2 = 625 - 25
x^2 = 600
Square both side
x = sqrt(600)
x= 24.495
x = 24 feet
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
1. The probability of getting exactly one tail
2. The probability of getting exactly two tails
3. The probability of getting a head on the first toss
4. The probability of getting a tail on the last toss
5. The probability of getting at least one head
6. The probability of getting at least two heads
Answer:
1) 0.375
2) 0.375
3) 0.5
4) 0.5
5) 0.875
6) 0.5
Step-by-step explanation:
We are given the following in the question:
Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]
1. The probability of getting exactly one tail
P(Exactly one tail)
Favorable outcomes ={HHT, HTH, THH}
[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]
2. The probability of getting exactly two tails
P(Exactly two tail)
Favorable outcomes ={ HTT,THT, TTH}
[tex]\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375[/tex]
3. The probability of getting a head on the first toss
P(head on the first toss)
Favorable outcomes ={HHH, HHT, HTH, HTT}
[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]
4. The probability of getting a tail on the last toss
P(tail on the last toss)
Favorable outcomes ={HHT,HTT,THT,TTT}
[tex]\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]
5. The probability of getting at least one head
P(at least one head)
Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}
[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]
6. The probability of getting at least two heads
P(Exactly one tail)
Favorable outcomes ={HHH, HHT, HTH,THH}
[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]
The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.35, the analogous probability for the second signal is 0.55, and the probability that he must stop at at least one of the two signals is 0.75.
What is the probability that he must stop:
A. at both signals?
B. at the first signal but not at the second one?
C. at exactly one signal?
Answer:
a) 0.15
b) 0.2
c) 0.6
Step-by-step explanation:
We are given the following in the question:
A: Stopping at first signal
B: Stopping at second signal
P(A) = 0.35
P(B) = 0.55
Probability that he must stop at at least one of the two signals is 0.75
[tex]P(A\cup B) = 0.75[/tex]
a) P(at both signals)
[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\0.75 = 0.35 + 0.55 - P(A\cap B)\\P(A\cap B) = 0.35 + 0.55 - 0.75 = 0.15[/tex]
0.15 is the probability that motorist stops at both signals.
b) P(at the first signal but not at the second one)
[tex]P(A\cap B') = P(A) - P(A\cap B)\\P(A\cap B') = 0.35 - 0.15 = 0.2[/tex]
0.2 is the probability that motorist stops at the first signal but not at the second one.
c) P(at exactly one signal)
[tex]P(A\cap B') + P(A\cap 'B) = P(A\cup B) - P(A\cap B) \\P(A\cap B') + P(A\cap 'B) = 0.75 - 0.15 = 0.6[/tex]
0.6 is the probability that the motorist stops at exactly one signal.
Final answer:
To calculate various probabilities related to stopping at traffic signals, we use given values to determine a 15% chance of stopping at both signals, a 20% chance of stopping at the first but not the second, and a 60% chance of stopping at exactly one signal.
Explanation:
The question involves calculating probabilities of stopping at traffic signals. Given are the probabilities of stopping at the first (0.35) and second (0.55) signals, and the probability of stopping at at least one signal (0.75). Using these, we can find the probabilities for various scenarios.
A. Probability of stopping at both signals:
To find this, we use the formula: P(A and B) = P(A) + P(B) - P(A or B). Here, P(A or B) is the probability of stopping at least at one signal, which is given as 0.75. Thus, the calculation would be 0.35 + 0.55 - 0.75 = 0.15. So, there is a 15% chance of stopping at both signals.
B. Probability of stopping at the first signal but not the second one:
This can be calculated by subtracting the probability of stopping at both signals from the probability of stopping at the first signal: 0.35 - 0.15 = 0.20. Therefore, there is a 20% chance of stopping at the first signal but not the second.
C. Probability of stopping at exactly one signal:
This involves adding the probabilities of stopping only at the first signal or only at the second signal. We've already calculated the first part as 0.20. For the second part, subtract the probability of stopping at both signals from stopping at the second signal: 0.55 - 0.15 = 0.40. Adding these together, 0.20 + 0.40 = 0.60, there is a 60% chance of stopping at exactly one signal.
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:
in a multicriteria decision problem the decision maker must evaluate each alternative with respect to each criterion.
Step-by-step explanation:
Multiple-criteria decision-making (MCDM) is a sub-discipline of operations research that explicitly evaluates multiple conflicting criteria in decision making (both in daily life and in settings such as business, government and medicine e.t.c)
In multicriteria decision problems, each alternative must be evaluated separately based on each criterion. It's not impossible to choose a single decision path, but it can be complex and might require multiple stages of decision-making over time. All statements in the question are relatively true.
Explanation:In multicriteria decision problems, various factors or criteria come into play. It's important to note that all the provided statements have some truth in them. From a decision-maker's perspective, one must evaluate each alternative against each criterion. This ensures that the pros and cons of each option are meticulously considered. Contrary to option a, it's not impossible to select a single decision alternative. However, the decision-making process could become complicated due to differing priorities and preferences, ultimately delaying the selection of a single alternative. On the other hand, statement c is partially true; depending on the complexity and scale of the decision, it might require several rounds of decision-making over time. Hence, all of these statements are indeed relatively accurate.
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When Alice spends the day with the babysitter, there is a 0.6 probability that she turns on the TV and watches a show. Her little sister Betty cannot turn the TV on by herself. But once the TV is on, Betty watches with probability 0.8. Tomorrow the girls spend the day with the babysitter.a) What is the probability that both Alice and Betty watch TV tomorrow?b) What is the probability that Betty watches TV tomorrow?c) What is the probability that only Alice watches TV tomorrow?
Answer:
a) There is a 48% probability that both Alice and Betty watch TV tomorrow.
b) There is a 48% probability that Betty watches TV tomorrow.
c) There is a 12% probability that only Alice watches TV tomorrow.
Step-by-step explanation:
We have these following probabilities:
A 60% probability that Alice watches TV.
If Alice watches TV, an 80% probability Betty watches TV.
If Alice does not watch TV, a 0% probability that Betty watches TV, since she cannot turn the TV on by herself.
a) What is the probability that both Alice and Betty watch TV tomorrow?
Alice watches 60% of the time. Betty watches in 80% of the time Alice watches. So:
[tex]P = 0.6*0.8 = 0.48[/tex]
There is a 48% probability that both Alice and Betty watch TV tomorrow.
b) What is the probability that Betty watches TV tomorrow?
Since Betty only watches when Alice watches(80% of the time), this probability is the same as the probability of both of them watching. So
[tex]P = 0.6*0.8 = 0.48[/tex]
There is a 48% probability that Betty watches TV tomorrow.
c) What is the probability that only Alice watches TV tomorrow?
There is a 60% probability that Alice watches TV tomorrow. If she watches, there is an 80% probability that Betty watches and a 20% probability she does not watch.
So
[tex]P = 0.6*0.2 = 0.12[/tex]
There is a 12% probability that only Alice watches TV tomorrow.
The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the histogram is missing. What is its height?
Answer:
The height of the missing rectangle is 0.15Explanation:
The image attached has the mentioned histogram.
Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.
Only the rectangle for the class [3,4] is missing.
The height of each rectangle is the relative frequency of the corresponding class.
The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.
In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.
1. Sum of the known relative frequencies:
0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.852. Missing frequency:
1 - 0.85 = 0.153. Conclusion:
The height of the missing rectangle is 0.15You have two fair, six-sides dice. However, the dice have been modified so that instead or 1,2,3,4,5,6 the sides are numbered 1,2,2,2,3,4. (Write all answers as fractions, not decimals) When the two dice are thrown, what is the probability their total is 4
Answer:
30.56%
Step-by-step explanation:
Let the sides on each dice be labeled as 1, 2a, 2b, 2c, 3, 4.
The sample space for the sum of the values being 4 is:
S={1,3; 3,1; 2a,2a; 2a,2b; 2a,2c; 2b,2a; 2b,2b; 2b,2c; 2c,2a; 2c,2b; 2c,2c}
There are 11 possible sums out of the 36 possible outcomes that result in a sum of 4. Therefore, the probability their total is 4 is:
[tex]P(S) =\frac{11}{36}=0.3056 =30.56\%[/tex]
There is a 30.56% probability that their sum is 4.
Final answer:
The probability of getting a sum of 4 with two modified dice numbered 1,2,2,2,3,4 is 5/36. This is found by adding all possible combinations that total 4, considering the multiplicity of the number 2 on the dice.
Explanation:
To calculate the probability that the sum of two modified dice is 4, we must consider all possible combinations of rolls that could result in a total of 4. Each die is numbered with 1,2,2,2,3,4, so the outcomes that give us a sum of 4 are (1,3), (2,2), (3,1), and there are three different 2s on each die that can contribute to the sum.
Therefore, the probability of getting a sum of 4 with one die already showing 2 is the probability of rolling either a 1 or another 2 on the second die.
The total number of outcomes for one die is 6. To find the sum of 4:
(1,3) - There is 1 way to roll a 1 and 1 way to roll a 3.(2,2) - There are 3 ways to roll a 2 on the first die and 3 ways to roll a 2 on the second die, but since the outcome is indistinguishable (2,2) is considered once, making it 3 ways in total.(3,1) - There is 1 way to roll a 3 and 1 way to roll a 1.This results in 1 + 3 + 1 = 5 favorable outcomes. Since there are a total of 36 possible outcomes when rolling two dice, the probability is 5/36.
The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of a wedding is $29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600.a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
Answer:
a) 0.0392
b) 0.4688
c) At least $39,070 to be among the 5% most expensive.
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 = 29858, \sigma = 5600[/tex]
a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?
This is the pvalue of Z when X = 20000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20000 - 29858}{5600}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a pvalue of 0.0392.
So this probability is 0.0392.
b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?
This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.
X = 30000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30000 - 29858}{5600}[/tex]
[tex]Z = 0.02[/tex]
[tex]Z = 0.02[/tex] has a pvalue of 0.5080.
X = 20000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20000 - 29858}{5600}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a pvalue of 0.0392.
So this probability is 0.5080 - 0.0392 = 0.4688
c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 29858}{5600}[/tex]
[tex]X - 29858 = 5600*1.645[/tex]
[tex]X = 39070[/tex]
The wedding would have to cost at least $39,070 to be among the 5% most expensive.
Solve, graph, and give interval notation for the compound inequality:
−2x − 4 > −6 AND 3(x + 2) ≤ 18
Answer:
1. (-∞,1) 2. (-∞,4]
Step-by-step explanation:
-2x-4 > -6
-2x > -2
x < 1
3(x+2) ≤ 18
3x+6 ≤ 18
3x ≤ 12
x ≤ 4
Find the balance of $6,000 deposited at 6% compounded semi-annually for 3 years
Answer:
The balance will be $7,164.31.
Step-by-step explanation:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.
In this problem, we have that:
[tex]P = 6000, r = 0.06[/tex]
Semi-annually is twice a year, so [tex]n = 2[/tex]
We want to find A when [tex]t = 3[/tex]
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 6000(1 + \frac{0.06}{2})^{2*3}[/tex]
[tex]A = 7164.31[/tex]
The balance will be $7,164.31.
The length of a Texas Pee Wee football field is 218 feet greater than its width. The area of the field is 20,160 square feet
Please show work !!
Answer: [tex]9.616 ft[/tex]
Step-by-step explanation:
The last part of the question is: Find the value of the width
If the Texas Pee Wee football field has a rectangular shape (as shown in the figure), where the width is [tex]w[/tex] and the length is [tex]218 ft w[/tex]; its area [tex]A[/tex] is:
[tex]A=20,160 ft^{2}=(length)(width)[/tex]
[tex]20,160 ft^{2}=(218 w)(w)[/tex]
Isolating [tex]w[/tex]:
[tex]w=\sqrt{\frac{20,160 ft^{2}}{218}}[/tex]
Finally:
[tex]w=9.616 ft[/tex] This is the widht of the Texas Pee Wee football field
The Texas Pee Wee football field has a width of approximately 70 feet and a length of 288 feet, calculated by solving a quadratic equation derived from the given area and the relationship between length and width.
Step-by-Step Explanation:
Let the width of the field be denoted as w.
Therefore, the length of the field is w + 218 feet.
The area of the rectangle (football field) is given by the formula:
Area = length x width.
Substituting the given values into the formula, we have:
20,160 = w × (w + 218).
This results in a quadratic equation:
w² + 218w - 20,160 = 0.
We will solve this quadratic equation using the quadratic formula:
w = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = 218, and c = -20,160.
Calculate the discriminant:
b² - 4ac = 218² - 4 × 1 × (-20,160) = 47524 + 80640 = 128164.
Find the square root of the discriminant:
√128164 ≈ 357.94.
Substitute back into the quadratic formula:
w = (-218 ± 357.94) / 2.
This results in two potential solutions:
w = (357.94 - 218) / 2 ≈ 69.97 (approximately 70 feet) and w = (-218 - 357.94) / 2 (a negative value, which is not possible for width).
Thus, the width w is approximately 70 feet.
Substitute the width back into the length formula:
length = 70 + 218 = 288 feet.
Therefore, the dimensions of the Texas Pee Wee football field are approximately 70 feet in width and 288 feet in length.
The perimeter of a rhombus is 64 and one of its angles has measure 120. Find the lengths of the diagonals.
Answer:
8[tex]\sqrt{2}[/tex]
Step-by-step explanation:
64/4=16
so that is 16 on each side.
The diagonal of the rhombus create a right triangle.
We then use the Pythagorean theorem.
[tex]a^{2} +b^{2} =c^{2}[/tex]
[tex]16^{2} +16^{2} =c^{2}[/tex]
[tex]256+256=c^{2}[/tex]
[tex]512 =c^{2}[/tex]
[tex]\sqrt{512} =c[/tex]
8[tex]\sqrt{2}[/tex]
The length of the diagonals of a rhombus with perimeter of 64 and one of its angles as 120 degrees are 16 units and 27.71 units
Properties of a rhombusThe diagonals are angle bisectorsThe 4 sides are congruent.The diagonal are perpendicular bisectors
Therefore,
perimeter = 4l
where
l = length
64 = 4l
l = 64 / 4
length = 16
One of its angle is 120°. Therefore, let's use the angle to find the length of the diagonal.
Using trigonometric ratio,
cos 60° = adjacent / hypotenuse
cos 60° = x / 16
x = 16 × cos 60
x = 8
2(x) = diagonal
diagonal = 16 units
The second diagonal length
sin 60° = opposite / hypotenuse
sin 60 = y / 16
y = 16 × sin 60
y = 13.8564064606
y = 13.85
Therefore,
diagonal = 2(13.85) = 27.71 units
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At the dance recital, Ms. Deutsch needs seven parent volunteers to help students get on and off stage, plus four parent volunteers per room of students. On the day of the recital, Ms. Deutsch uses 39 parent volunteers. How many backstage rooms were there?
plz help Like now plz
Answer:
8 rooms
Step-by-step explanation:
39 - 7 = 32. 32/4 per room is 8 rooms.
Final answer:
The equation to find the number of backstage rooms needed is (Total volunteers - Stage volunteers) / Volunteers per room = Number of rooms. Using the values provided (39 - 7) / 4, we find that there were eight backstage rooms at the dance recital.
Explanation:
Ms. Deutsch needs seven parent volunteers to help students get on and off stage and four parent volunteers per backstage room. The total number of parent volunteers used on the day of the recital is 39. To find the number of backstage rooms, we subtract the seven parent volunteers required for stage assistance from the total, leaving us with 32 volunteers. We then divide this number by the four parent volunteers per room, resulting in eight backstage rooms.
Initial number of volunteers required for stage assistance: 7
Each room requires: 4 volunteers
Total volunteers: 39
Volunteers for rooms: 39-7 = 32
Number of rooms: 32 / 4 = 8
Dale and Betty go through a traffic light at the same time but Dale goes straight and Betty turns right. After two minutes Dale is 2000 yd from the intersection and Bettyis 750 yd from the intersection. Assuming the roads met at a right angle and both were perfectly straight, how far are Dale and Betty away from each other after two minutes?
Answer:
2,136 yards
Step-by-step explanation:
Since the roads met at a right angle, the distance between Dale and Betty can be interpreted as the hypotenuse of a right triangle with sides measuring 2000 yd and 750 yd. The distance between them is:
[tex]d^2=2000^2+750^2\\d=\sqrt{2000^2+750^2}\\d=2,136\ yards[/tex]
Dale and Betty are 2,136 yards away from each other after two minutes.
Determine the truth values of these statements:
(a) The product of x2 and x3 is x6 for any real number x.
(b) x2 > 0 for any real number x.
(c) The number 315 − 8 is even.
(d) The sum of two odd integers is even.
Final answer:
The product of x^2 and x^3 equals x^6 for any real number x is true. The statement that x^2 > 0 for any real number x is false, as it should state x^2 >= 0. 315 - 8 being even is false since it results in an odd number. The sum of two odd integers being even is true.
Explanation:
Determine the truth values of these statements:
The product of x2 and x3 is x6 for any real number x. This statement is true because according to the laws of exponents, when multiplying powers with the same base, you add the exponents. Therefore, x2 * x3 = x2+3 = x6.
x2 > 0 for any real number x. This statement is false because when x = 0, x2 = 0, not greater than 0. The correct statement should be x2 >= 0 for any real number x.
The number 315 − 8 is even. This statement is true because 315 − 8 = 307, and any number ending in 7 is odd. Therefore, the statement is false.
The sum of two odd integers is even. This statement is true because when you add two odd numbers, the sum is always even. For example, 3 + 5 = 8.
You have 200 dice in a bag. One of the dice has a six on all sides so it will land on a six every time you roll it. The other 199 are normal dice with six sides, each with a different number. You randomly pick one of the dice from the bag and roll it three times. It lands on six all three times. What is the probability it is the die that always lands on six and what is the probability it is a normal die?
Answer:
Step-by-step explanation:
There are 200 dice out of which 199 are fair
Prob for 6 in one special die = 1 and
Prob for 6 in other die = 1/6
A1- drawing special die and A2 = drawing any other die
A1 and A2 are mutually exclusive and exhaustive
P(A1) = 1/200 and P(A2) = 199/200
B = getting 6
i) Required probability
= P(A1/B) = [tex]\frac{P(A1B)}{P(A1B)+P(A2B)} \\[/tex]
P(A1B) = [tex]\frac{1}{200} *1 = \frac{1}{200}[/tex]
P(A2B) = [tex]\frac{199}{200}*\frac{1}{6}=\frac{199}{1200}[/tex]
P(B) = [tex]\frac{205}{1200} =\frac{41}{240}[/tex]
P(A1/B) = [tex]\frac{1/200}{41/240} =\frac{7}{205}[/tex]
P(A2/B) = [tex]\frac{199/1200}{41/240} =\frac{199}{205}[/tex]
If the New England Patriots get home-field advantage, you believe there is a 60% probability they will make it to the Super Bowl. If not, this probability is only 30%. Assuming a 70% probability that the Patriots get home-field advantage, what is the probability they will make it to the Super Bowl?
Answer:
There is a 51% probability they will make it to the super bowl.
Step-by-step explanation:
We have these following probabilities:
A 70% probability that the Patriots get homefield advantage.
A 30% probability that the Patriots does not get homefield advantage.
If they get homefield advantage, a 60% probability of making the Super Bowl.
If they do not get homefield advantage, a 30% probability of making the Super Bowl.
What is the probability they will make it to the Super Bowl?
This is 60% of 70%(when they get homefield advantage and make the super bowl), and 30% of 30%(no homefield, no super bowl). So
[tex]P = 0.6*0.7 + 0.3*0.3[/tex]
[tex]P = 0.51[/tex]
There is a 51% probability they will make it to the super bowl.
A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 10 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.
Answer:
the amount of water that is left in the tank after 10 min is 98 L
Step-by-step explanation:
since the water drains off at rate that is proportional to the water present
(-dV/dt) = k*V , where k= constant
(-dV/V) = k*dt
-∫dV/V) = k*∫dt
-ln V/V₀=k*t
or
V= V₀*e^(-k*t) , where V₀= initial volume
then since V₁=0.7*V₀ at t₁= 3 min
-ln V₁/V₀=k*t₁
then for t₂= 10 min we have
-ln V₂/V₀=k*t₂
dividing both equations
ln (V₂/V₀) / ln (V₁/V₀) =(t₂/t₁)
V₂/V₀ = (V₁/V₀)^(t₂/t₁)
V₂=V₀ * (V₁/V₀)^(t₂/t₁)
replacing values
V₂=V₀ * (V₁/V₀)^(t₂/t₁) = 200 L * (0.7)^(10min/5min) = 98 L
then the amount of water that is left in the tank after 10 min is 98 L
The problem represents a case of exponential decay. Initially, 30% of water, or 60 liters, leaks out in 5 minutes, leaving 140 liters in the tank. Assuming the same rate of leakage, another 30% of water or 42 liters will leak out in the next 5 minutes, leaving 98 liters in the tank after 10 minutes.
Explanation:In this problem, we are dealing with a situation involving exponential decay due to the water leakage which happens at a rate proportional to the amount of water present in the tank.
First, let's consider the 30% of water that leaks out in the first 5 minutes from a 200-liter tank. This amounts to 60 liters (200 * 0.30), which leaves 140 liters (200 - 60) of water in the tank after 5 minutes.
Now, since the decrease of water is proportional to the amount of water present, this implies an exponential decay over time. Given that the amount of water decreased by 30% in the first 5 minutes, it's reasonable to assume that it will decrease by the same percentage in the next 5 minutes as well.
So, the amount of water left in the tank after 10 minutes would be 98 liters (140 * 0.70).
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A beam of light from a monochromatic laser shines into a piece of glass. The glass has thickness L and index of refraction n=1.5. The wavelength of the laser light in vacuum is L/10 and its frequency is f. In this problem, neither the constant c nor its numerical value should appear in any of your answers.
Additional information to complete the question:
How long does it take for a short pulse of light to travel from one end of the glass to the other?
Express your answer in terms of some or all of the variables f and L. Use the numeric value given for n in the introduction.
T = ___________ s
Answer:
[tex]T = \frac{15}{f}[/tex]
Step-by-step explanation:
Given:
Thickness og glass = L
Index of refraction n=1.5
Frequency = f
[tex]Wavelength = \frac{L}{10}[/tex]
λ(air) [tex]= \frac{L}{10}[/tex]
λ(glass) = λ(air) / n
= [tex]\frac{\frac{L}{10}}{1.5}[/tex]
= [tex]\frac{L}{10} * \frac{1}{1.5}[/tex]
= [tex]\frac{L}{15}[/tex]
V(glass) = fλ(glass)
[tex]= f * \frac{L}{15}[/tex]
[tex]T = \frac{L}{V_{glass}} = \frac{15}{f}[/tex]
Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Find the probability that a randomly selected adult has an IQ greater than 131.5. Group of answer choices
Answer:
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(IQ greater than 131.5)
P(x > 131.5)
[tex]P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)[/tex]
[tex]= 1 - P(z \leq 2.1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
There are three workstations available having steady-state probabilities of 0.99, 0.95, 0.85 of being available on demand. What is the probability that at least two of the three will be available at any given time?
Answer:
99.065% probability that at least two of the three will be available at any given time.
Step-by-step explanation:
We have these following probabilities:
99% probability of the first workstation being available
95% probability of the second workstation being available
85% probability of the third workstation being avaiable
Two being available:
We can have three outcomes
First and second available, third not. So
0.99*0.95*0.15 = 0.141075
First and third available, second not. So
0.99*0.05*0.85 = 0.042075
Second and third available, first not. So
0.01*0.95*0.85 = 0.008075
Adding them all
P(2) = 0.141075 + 0.042075 + 0.008075 = 0.191225
Three being available:
P(3) = 0.99*0.95*0.85 = 0.799425
What is the probability that at least two of the three will be available at any given time?
P = P(2) + P(3) = 0.191225 + 0.799425 = 0.99065
99.065% probability that at least two of the three will be available at any given time.
The probability that at least two out of the three workstations are available is 0.967.
Explanation:We can find the probability that at least two out of the three workstations are available using the concept of complementary events. The probability of at least two workstations being available is equal to 1 minus the probability of none or only one workstation being available.
Let's calculate the probability of none or only one workstation being available:
Probability of none being available: 0.01 * 0.05 * 0.15 = 0.00075
Probability of only one being available: (0.99 * 0.05 * 0.15) + (0.01 * 0.95 * 0.15) + (0.01 * 0.05 * 0.85) = 0.03225
Now, subtracting this from 1:
1 - (0.00075 + 0.03225) = 0.967
Therefore, the probability that at least two out of the three workstations are available at any given time is 0.967.
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The newly elected president needs to decide the remaining 7 spots available in the cabinet he/she is appointing. If there are 13 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?
Answer: 8648640 ways
Step-by-step explanation:
Number of positions = 7
Number of eligible candidates = 13
This can be done by solving the question using the combination Formula for selection in which we use the combination formula to choose 7 candidates amomg the possible 13.
The combination Formula is denoted as:
nCr = n! / (n-r)! * r!
Where n = total number of possible options.
r = number of options to be selected.
Hence, selecting 7 candidates from 13 becomes:
13C7 = 13! / (13-7)! * 7!
13C7 = 1716.
Considering the order they can come in, they can come in 7! Orders. We multiply this order by the earlier answer we calculated. This give: 1716 * 7! = 8648640
Binomial Distribution. Surveys repeatedly show that about 40% of adults in the U.S. indicate that if they only had one child, they would prefer it to be a boy. Suppose we took a random sample of 15 adults and the number who indicated they preferred a boy was 8. This would be considered a rare event because the probability of 8 or more is so low.
True/False
Answer:
False
Step-by-step explanation:
We are given the following information:
We treat adult who prefer one child to be a boy as a success.
P(prefer one child to be a boy) = 40% = 0.4
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15 and x = 8
We have to evaluate:
[tex]P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131[/tex]
Since the probability of 8 or more is 0.2131 is not very small, thus, it is not a rare event.
Thus, the given statement is false.
Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 4.9 times that of particle B. The period of particle B is 2.4 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to
The ratio of the radius of particle A to that of particle B in uniform circular motion is approximately 2.437.
Explanation:To find the ratio of the radius of particle A to that of particle B in uniform circular motion, let's first consider the equations of motion for uniform circular motion. The centripetal acceleration, a_c, is given by the equation a_c = v^2/r, where v is the velocity and r is the radius of the circle. The period of motion, T, is the time it takes for one complete revolution around the circle. Given that the acceleration of particle A is 4.9 times that of particle B, we can write the equation a_A = 4.9 * a_B. Also, the period of particle B is 2.4 times the period of particle A, so we can write the equation T_B = 2.4 * T_A.
Next, we can use the equations of motion to express the velocity and period in terms of the acceleration and radius. From the equation a_c = v^2/r, we can rearrange it to solve for v: v = sqrt(a_c * r). By substituting this expression for v into the equation T = 2 * pi * r / v, we can solve for the period T in terms of a_c and r. Plugging these expressions for the velocities and periods of particles A and B into the equations a_A = 4.9 * a_B and T_B = 2.4 * T_A, we can form an equation that relates the radii of the two particles: sqrt(a_A * r_A) = 4.9 * sqrt(a_B * r_B) and 2 * pi * r_B / sqrt(a_B * r_B) = 2.4 * (2 * pi * r_A / sqrt(a_A * r_A)). Simplifying these equations, we can solve for the ratio of the radii r_A/r_B and find that it is approximately 2.437.
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To find the ratio of the radius of motion between particles A and B, we can use the equations of uniform circular motion and the given information. The ratio is approximately 2.21.
Explanation:To find the ratio of the radius of motion between particles A and B, we need to analyze the given information. Let's denote the acceleration of particle B as aB and the acceleration of particle A as aA. We're told that aA is 4.9 times aB, and the period of particle B is 2.4 times the period of particle A.
From the equations of uniform circular motion, we know that the acceleration is given by a = (4π2)/T2, where T is the period. Since aA = 4.9aB and TB = 2.4TA, we can set up the following equation:
(4π2)/TA2 = 4.9(4π2)/TB2
After canceling out common terms, we'll find that (TA2) / (TB2) = 4.9. Taking the square root of both sides, we get TA / TB = √4.9 = 2.21. Since the period is inversely proportional to the angular velocity, we can conclude that the ratio of the radii of motion is approximately 2.21.
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2.65 Consider the situation of Exercise 2.64. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not actually fail. Event A occurs with probability 0.20, and event B occurs with probability 0.35. (a) What is the probability that the component does not fail the test? (b) What is the probability that the component works perfectly well (i.e., neither displays strain nor fails the test)? (c) What is the probability that the component either fails or shows strain in the test?
Answer:
a) 0.80
b) 0.45
c) 0.55
Step-by-step explanation:
Given P(A) = 0.20 and P(B) = 0.35
Applying probability of success and failure; P(success) + P( failure) = 1
a) probability that the component does not fail the test = The component does not fail a particular test [P(success)] = 1 - P(A)
= 1 - 0.20 = 0.80
b) probability that the component works perfectly well
= P( the component works perfectly well) - P(component shows strain but does not fail test)
= 0.80 - 0.35 = 0.45
c) probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)
= 1 - 0.45 = 0.55
This question is based on the concept of probability. Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.
Given:
Event A occurs with probability P(A) = 0.20, and event B occurs with probability P(B) = 0.35.
According to the question,
Given P(A) = 0.20 and P(B) = 0.35,
As we know that, probability of success and failure,
⇒ P(success) + P( failure) = 1
a) Probability that the component does not fail the test = The component does not fail a particular test
= P(success) = 1 - P(A)
= 1 - 0.20 = 0.80
b) Probability that the component works perfectly well
= P( the component works perfectly well) - P(component shows strain but does not fail test)
= 0.80 - 0.35 = 0.45
c) Probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)
= 1 - 0.45 = 0.55
Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.
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what are the common factors for 54,24,18
Answer:
Step-by-step explanation:
We find what number we multiply by another number to get 54, 24, 18
54: 1, 2, 3, 6, 9, 18, 27, 54
24: 1, 2, 3, 4, 6, 8, 12, 24
18: 1, 2, 3, 6, 9, 18
Now the numbers that repeat in these sets are the common factors
We got 1, 2, 3, and 6, which make these our common factors
Answer:1,2,3,6
Step-by-step explanation:
The factors of 54,24 and 18 are 1,2,3,6
A group of five applicants for a pair of identical jobs consists of three men and two women. The employer is to select two of the five applicants for the jobs. Let S denote the set of all possible outcomes for the employer’s selection. Let A denote the subset of outcomes corresponding to the selection of two men and B the subset corresponding to the selection of at least one woman. List the outcomes in A, B, A ∪ B, A ∩ B, and A ∩ B. (Denote the different men and women by M1, M2, M3 and W1, W2, respectively.)
Answer:
A= {M1,M2},{M2,M3}, {M2,M3}
A U B = S
A n B = 0
A n B'= A
Step-by-step explanation:
A= ( Two males) = { (M1,M2), (M2,M3), (M2,M3)
B= (Atleast one female) = {M1,W1}, {M,W1}, {M3,W1}, {M1,W2} , {M2,W2}, {M3,W2}
Following are the solution to the required function:
Set function:Given that there are five applicants with three men and two women.
Let S be the subset of the set of all possible outcomes,
[tex]\{M_1, M_2\}, \{M_2, M_3\},\{M_3,M_1\},\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]
Let A denote the subset of outcomes corresponding to the selection of two men.
The possible outcomes of A are,
[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]
Let B be the subset corresponding to the selection of at least one woman.
[tex]\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]
Then [tex]\bar{B} =[/tex]
[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]
Find [tex]A\cup B\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cup \{ \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}, \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\[/tex]
Find [tex]A\cap B\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cap \{ \{W_1,M_1\},\{W_2,M_1\}, \{W_1, M_2\}, \\ \{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\ =\{\phi\}[/tex]
Find [tex]A\cap \bar{B}\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\cap\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\[/tex]
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A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance of 1000 (psi)2. A random sample of 12 specimens has a mean compressive strength of 3250 psi. (a) Construct a 95% confidence interval on mean compressive strength. (b) Suppose that the manufacturer of the concrete claims the average compressive strength is 3270 psi. Based on your answer in part (a), what would you say about this claim
Answer:
a) [tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]
[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]
So on this case the 95% confidence interval would be given by (3232.108;3267.892)
b) For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.
Step-by-step explanation:
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= 3250[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma^2 =1000[/tex]represent the sample standard variance
[tex] s = \sqrt{1000}[/tex] represent the sample deviation
n=12 represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
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 table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]
[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]
So on this case the 95% confidence interval would be given by (3232.108;3267.892)
Part b
For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.
Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency:30 = tall20 = shortNull hypothesis: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.What is the Chi-square value associated with the appropriate test of significance?
Answer:
Chi-square value = 2
Step-by-step explanation:
Given data:
Frequency: Tall Short
30 20
Null hypothesis ( as it is already given so there is no difference between observed and expected values)
Ratio: 1:1 (50% expected frequency of each tall and short pea-plant)
Solution:
Phenotype Observed Expected O-E (O-E)² (O-E)[tex]^{2/E}[/tex]
O E
Short 20 25 -5 25 1
TALL 30 25 5 25 1
TOTAL 2
So, from all these calculations using expected and observed values we get chi-square value equal to 2.
Final answer:
To test the null hypothesis that the deviation from a 1:1 phenotypic ratio is due to chance in a cross between a heterozygous tall and a homozygous short pea plant, the Chi-square value is calculated to be 2.
Explanation:
The question relates to a cross between a heterozygous tall pea plant and a homozygous short pea plant, with the intention to calculate the Chi-square value to test the null hypothesis that the observed deviation from a 1:1 ratio is due to chance. In this scenario, the expectation is a 1:1 phenotypic ratio, meaning 25 tall and 25 short plants out of 50 offspring.
To calculate the Chi-square (χ²) value, the formula is χ² = Σ ( (observed - expected)² / expected ), where Σ symbolizes the sum of calculations for each category. For tall plants, the calculation is ((30-25)² / 25) = (5² / 25) = 1 and for short plants, the calculation is ((20-25)² / 25) = (5² / 25) = 1. Therefore, the total χ² value is 1 + 1 = 2.
Find the volume of a cube with side length of 7 in.
147
343
49
215
Answer:
48
Step-by-step explanation:
to find erea you just multiply one number by the other
Answer:48
Step-by-step explanation:
Four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not. The candidates are interviewed at random, and the first qualified candidate interviewed will be hired. The outcomes are the sequences of candidates that are interviewed. So one outcome is 2, and another is 431.
a. List all the possible outcomes.
b. Let A be the event that only one candidate is interviewed. List the outcomes in A.
c. Let B be the event that three candidates are interviewed. List the outcomes in B.
d. Let C be the event that candidate 3 is interviewed. List the outcomes in C.
e. Let D be the event that candidate 2 is not interviewed. List the outcomes in D
f. Let E be the event that candidate 4 is interviewed. Are A and E mutually exclusive? How about B and E, C and E, D and E?
Answer:
a) [tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]
b) A={1, 2}
c) B={341, 342, 431, 432}
d) C={31, 32, 341, 342, 431, 432}
e) D={1, 31, 41, 341, 431}
f) E={41, 42, 341, 342, 431, 432}
Step-by-step explanation:
We know that four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not.
a) We get a set of all possible outcomes:
[tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]
b) Let A be the event that only one candidate is interviewed.
We get a set of all possible outcomes:
A={1, 2}
c) Let B be the event that three candidates are interviewed.
We get a set of all possible outcomes:
B={341, 342, 431, 432}
d) Let C be the event that candidate 3 is interviewed.
We get a set of all possible outcomes:
C={31, 32, 341, 342, 431, 432}
e) Let D be the event that candidate 2 is not interviewed.
We get a set of all possible outcomes:
D={1, 31, 41, 341, 431}
f) Let E be the event that candidate 4 is interviewed.
We get a set of all possible outcomes:
E={41, 42, 341, 342, 431, 432}
We conclude that the events A and E are mutually exclusive.
We conclude that the events B and E are not mutually exclusive.
We conclude that the events C and E are not mutually exclusive.
We conclude that the events D and E are not mutually exclusive.