8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
Step-by-step explanation:
We are given the following in the question:
Chamomile tea:
Unit cost = 18.20 per pound
Amount = x pounds
Total cost =
[tex]18.20\times x = 18.20x[/tex]
Orange tea:
Unit cost = 12.25 per pound
Amount = 12 pounds
Total cost =
[tex]12.25\times 12 = 147[/tex]
Total mixture:
Unit cost = 14.63 per pound
Amount = (12+x) pounds
Total cost =
[tex]14.63\times (12+x) = 175.56 + 14.63x[/tex]
We can write the equation that cost of mixture is equal to cost of chamomile tea and orange tea.
[tex]18.20x + 147 = 175.56 + 14.63x\\\Rightarrow 18.20x- 14.63x = 175.56-147\\\Rightarrow 3.57x = 28.56\\\Rightarrow x = 8[/tex]
Thus, 8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
The weight, in grams, of beans in a tin is normally distributed with mean m and standard deviation 7.8 grams. Given that 10% of tins contain less than 200 grams, explain why
Answer:
[tex]P(X<200)=P(\frac{X-\mu}{\sigma}<\frac{200-\mu}{\sigma})=0.1[/tex]
[tex]P(z<\frac{200-\mu}{\sigma})=0.1[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.28<\frac{200-\mu}{7.8}[/tex]
And if we solve for [tex]\mu[/tex] we got
[tex]\mu=200 +1.28*7.8=209.984[/tex]
Step-by-step explanation:
Assuming this question "The weight, in grams, of beans in a tin is normally distributed with mean m and standard deviation 7.8 grams. Given that 10% of tins contain less than 200 grams, Find the mean m. explain why. ?
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the lifetimes of TV tubes of a population, and for this case we know the distribution for X is given by:
[tex] X \sim N (\mu =m, 7.8)[/tex]
Where [tex]\mu =m[/tex] and [tex] \sigma =7.8[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we know the following condition:
[tex] P(X>200) =0.9[/tex] (a)
[tex] P(X<200) = 0.1[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value m.
As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28.. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9
If we use condition (b) from previous we have this:
[tex]P(X<200)=P(\frac{X-\mu}{\sigma}<\frac{200-\mu}{\sigma})=0.1[/tex]
[tex]P(z<\frac{200-\mu}{\sigma})=0.1[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.28<\frac{200-\mu}{7.8}[/tex]
And if we solve for [tex]\mu[/tex] we got
[tex]\mu=200 +1.28*7.8=209.984[/tex]
The mean weight of the beans in a tin m must be greater than 200 grams.
The problem states that the weight of beans in a tin is normally distributed with a mean m and a standard deviation of 7.8 grams.
It is also given that 10% of the tins contain less than 200 grams.
Since the distribution is normal, we can use the properties of the normal distribution to solve for the mean weight m
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
The fact that 10% of the tins weigh less than 200 grams indicates that 200 grams is less than one standard deviation below the mean.
The 10% corresponds to the bottom 10% of the distribution, which is less than the mean by a certain number of standard deviations.
The standard normal distribution table (or Z-table) tells us that a Z-score of approximately -1.28 corresponds to the 10th percentile (since 10% of the distribution is to the left of this Z-score).
Using the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
[tex]\[ \mu = X - Z \times \sigma \][/tex]
[tex]\[ \mu = 200 - (-1.28) \times 7.8 \][/tex]
[tex]\[ \mu = 200 + 10.064 \][/tex]
mu = 210.064
Therefore, the mean weight m (denoted by mu in our calculations) must be greater than 200 grams, specifically approximately 210.064 grams, to ensure that only 10% of the tins weigh less than 200 grams.
The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is ________.
Answer: The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596
Step-by-step explanation:
Since the weights of catfish are assumed to be normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = weights of catfish.
µ = mean weight
σ = standard deviation
From the information given,
µ = 3.2 pounds
σ = 0.8 pound
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as
P(x ≤ 3 ≤ 5.4)
For x = 3
z = (3 - 3.2)/0.8 = - 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.401
For x = 5.4
z = (5.4 - 3.2)/0.8 = 2.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.997
Therefore,.
P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596
To determine the probability that a catfish will weigh between 3 and 5.4 pounds, we calculate the z-scores for these weights and find the corresponding probabilities. The probability is approximately 0.5957 or 59.57%.
Explanation:The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds can be calculated using the standard normal distribution.
First, we convert the weights to z-scores using the formula:
Z = (X - μ) / σ
For 3 pounds:
Z = (3 - 3.2) / 0.8 = -0.25
For 5.4 pounds:
Z = (5.4 - 3.2) / 0.8 = 2.75
Next, we look up these z-scores on the z-table or use a calculator with normal distribution functions to find the probabilities.
P(Z < 2.75) = 0.9970 (rounded to four decimal places)
P(Z < -0.25) = 0.4013 (rounded to four decimal places)
Then we find the difference to determine the probability of a catfish weighing between these two values:
Probability = P(Z < 2.75) - P(Z < -0.25)
Probability = 0.9970 - 0.4013 = 0.5957
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is approximately 0.5957 or 59.57%.
A local Honda dealership collects data on customers. Here is data from 311 customers who purchased a Honda Civic. Hybrid Honda Civic Standard-engine Honda Civic Row Totals Male 77 117 194 Female 34 83 117 Column Totals 111 200 311 What does the data suggest about the relationship between gender and engine type? Group of answer choices Women are more likely to purchase a Honda Civic with a standard engine than men. omen are less likely to purchase a Honda Civic with a standard engine than men. Women and men are equally likely to purchase a Honda Civic with a standard engine.
Both male and female customers show a preference for the standard-engine Honda Civic, but the data suggests that women are slightly more likely to purchase this model than men are.
Explanation:According to this data, out of the total 311 customers, 194 are male and 117 are female. Among the male customers, 77 purchased a Hybrid Honda Civic and 117 purchased a Standard-engine Honda Civic. While among the female customers, 34 purchased a Hybrid Honda Civic and 83 purchased a Standard-engine Honda Civic. If we compare the proportions, we can see that approximately 60% (117 out of 194) of male customers and approximately 70% (83 out of 117) of female customers purchased a Honda Civic with a standard engine. Hence, we can conclude that both women and men have shown a preference for the standard-engine Honda Civic, but the data suggest that women are somewhat more likely to purchase a standard engine Honda Civic than men are.
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The data suggests that women are more likely to purchase a standard-engine Honda Civic compared to men, and men are more likely to choose a hybrid engine compared to women.
The data from the local Honda dealership regarding customer choices between Hybrid Honda Civic and Standard-engine Honda Civic suggests a relationship between gender and engine type choice. To determine this, we look at the proportions of each gender choosing each type of engine relative to their total gender group. Specifically, out of 194 males, 77 chose the hybrid (which is approximately 39.7%), and 117 chose the standard (approximately 60.3%). For females, 34 out of 117 chose the hybrid (approximately 29.1%), and 83 chose the standard (approximately 70.9%).
By comparing these percentages, it is clear that women are less likely to purchase a Hybrid Honda Civic than men because a smaller percentage of women chose the hybrid compared to men. Conversely, a larger percentage of women chose the standard engine compared to men, suggesting that women are more likely to purchase a Honda Civic with a standard engine than men.
Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is γ = 3.5 and that the excess X − 3.5 over the minimum has a Weibull distribution with parameters α = 2 and β = 1.5.
(a) What is the cdf of X?
F(x) = 0 x < 3.5
1−e^−((x−3.5)2.5)2 x ≥ 3.5
(b) What are the expected return time and variance of return time? [Hint: First obtain
E(X − 3.5)
and
V(X − 3.5).]
(Round your answers to three decimal places.)
E(X) = 10^−1 weeks
V(X) = (10^−1 weeks)2
(c) Compute
P(X > 6).
(Round your answer to four decimal places.)
Answer:
Step-by-step explanation: see attachment for solution
If X is the time in 10 to 1 week of the shipment of a defective product until the customer returns the product.
Now suppose the mini returns y = 3.5 and the excess X is 3.5 over the mini has a Weibull distribution with parameters Alfa 2 and Beta 1.5.The shipment of the defective product will be
A. 1-e x(25)2. B. 5/4 .Expected value is E (X) = 5.7125.
C. P(X>5) = 0.3679.Learn more about the shipment of a defective product
brainly.com/question/17125592.
A parking lot consists of a single row containing n parking spaces (n ≥ 2). Mary arrives when all spaces are free. Tom is the next person to arrive. Each person makes an equally likely choice among all available spaces at the time of arrival. Describe the sample space. Obtain P(A), the probability the parking spaces selected by Mary and Tom are at most 2 spaces apart.
The sample space includes all possible arrangements of n parking spaces. The probability (P(A)) that Mary and Tom select spaces at most 2 apart is [tex]\(\frac{3}{n-1}\)[/tex], assuming equal likelihood of choice.
The sample space consists of all possible arrangements of cars in the parking lot, given that there are n parking spaces. Each arrangement is equally likely.
For Mary and Tom to select parking spaces that are at most 2 spaces apart, we consider the following scenarios:
1. Mary selects any space, and Tom selects the same space or an adjacent space.
2. Mary selects any space, and Tom selects any space 1 space away.
3. Mary selects any space, and Tom selects any space 2 spaces away.
The number of favorable outcomes is 3n, and the total number of possible outcomes is [tex]\(n \times (n-1)\)[/tex] (since Mary and Tom can choose from n spaces each). Therefore, the probability P(A) is given by:
[tex]\[ P(A) = \frac{3n}{n \times (n-1)} \][/tex]
Simplify the expression:
[tex]\[ P(A) = \frac{3}{n-1} \][/tex]
A poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life. What is the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life. Report your answer as a decimal accurate to 3 decimal places.
Answer:
The probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.
Step-by-step explanation:
Let X = number of Americans who believe that statistics teachers know the true meaning of life.
The probability of the random variable X is,
P (X) = 0.574
The event of a person not believing that statistics teachers know the true meaning of life is the complement of the event X.
The probability of the complement of an event, E is the probability of that event not happening.
[tex]P(E^{c})=1-P(E)[/tex]
Compute the value of [tex]P(X^{c})[/tex] as follows:
[tex]P(X^{c})=1-P(X)\\=1-0.574\\=0.426[/tex]
Thus, the probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is 0.426.
Answer:
Probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life = 0.426 .
Step-by-step explanation:
We are given that a poll showed that 57.4% of Americans say they believe that statistics teachers know the true meaning of life.
Let the above probability that % of Americans who believe that statistics teachers know the true meaning of life = P(A) = 0.574
Now, probability of randomly selecting someone who does not believe that statistics teachers know the true meaning of life is given by = 1 - Probability of randomly selecting someone who believe that statistics teachers know the true meaning of life = 1 - P(A)
So, required probability = 1 - 0.574 = 0.426 .
Recall that log 2 = 1 0 1 x+1 dx. Hence, by using a uniform(0,1) generator, approximate log 2. Obtain an error of estimation in terms of a large sample 95% confidence interval.
Answer:
∫101/(x+1)dx=(1−0)∫101/(x+1)dx/(1−0)=∫101/(x+1)f(x)dx=E(1/(x+1))
Where f(x)=1, 0
And then I calculated log 2 from the calculator and got 0.6931471806
From R, I got 0.6920717
So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.
Todor was trying to factor 10x^2-5x+15 He found that the greatest common factor of these terms was 5 and made an area model:
Answer:
Model area of dimensions
[tex]p=5[/tex]
[tex]q=2x^2-x+3[/tex]
Step-by-step explanation:
Model for the Area
Suposse we have a rectangle of measures p and q, its area is computed by
[tex]A=p.q[/tex]
Note the expression is a product which means if an equation is found for the area, we can guess the dimensions of the supossed rectangle.
The expression for the area is
[tex]A=10x^2-5x+15[/tex]
This polynomial can be factored as
[tex]A=5(2x^2-x+3)[/tex]
We have found an explicit product, now we can guess the dimensions of the rectangle are
[tex]p=5[/tex]
[tex]q=2x^2-x+3[/tex]
Or viceversa
16. A car depreciates in value at a rate of 10%. The car currently has a value of $12,000.
What will be its value in 10 years?
Answer: $4,184.14
Step-by-step explanation:
If a car depreciates, it mean the the car is losing it's marketable worth and sometimes at a constant rate. The worth of the car after some years does not remain the same.
The formula for this depreciation when at a constant rate is denoted as:
D = P × [1 - r/100]^n
Where:
D=the Depreciated value of the car after n period which is what is being determined.
P = initial value of the car before depreciation is considered and in this case, P = $12,000
r = constant Rate Of depreciation and in this case = 10%
n = period being considered, which in this case, n = 10years.
Hence,
D = 12,000 × [1 - 10/100]^10
D = $4,184.14
Please help!! It’s due @ midnight
How much should Marc deposit weekly into an account at 8% compounded weekly in order to have
$4500 available for a round trip plane ticket, hotel, and spending money for a trip to Sweden in 2 years?
Please give step by step!
Answer:
ooh i just learned this, not 100% sure but the amount he should have to deposit is $3835.12
if the weekly one means depositing money every week then it would be $36.88 I think.
Step-by-step explanation:
p=?
r=.08
n=52
t=2
4500 = P(1+.08/52)^(52 x 2)
divide both sides by (1+.08/52)^(52 x 2)
and you are left with $3835.12
if i take into account that Marc is depositing the money every week the i would divide it by 104 (that is 52 x 2 because 52 weeks in a year and it says 2 years) you would be left with $36.88.
Hope I was any help.
Answer: Marc should deposit $39.87 weekly.
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of interval payments.
n represents the total number of payments made.
From the information given,
S = $4500
Assuming there are 52 weeks in a year, then
r = 0.08/52 = 0.0015
n = 52 × 2 = 104
Therefore,
4500 = R[{(1 + 0.0015)^104 - 1)}/0.0015][1 + 0.0015]
4500 = R[{(1.0015)^104 - 1)}/0.0015][1.0015]
4500 = R[{(1.169 - 1)}/0.0015][1.0015]
4500 = R[{(0.169)}/0.0015][1.0015]
4500 = R[112.67][1.0015]
4500 = 112.839R
R = 4500/112.839
R = 39.87
A market research firm knows from historical data that telephone surveys have a 36% response rate. In a random sample of 280 telephone numbers, what is the probability that the response rate will be between 33.5% and 39%?
Answer:
0.6604
Step-by-step explanation:
Given that a market research firm knows from historical data that telephone surveys have a 36% response rate.
Sample size of random sample = 280
We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error
= [tex]\sqrt{\frac{pq}{n} }[/tex]
Substitute p = 0.36 and q = 1-0.36= 0.64
p follows N with mean = 0.36 and std dev = [tex]\sqrt{\frac{0.36*0.64}{\sqrt{280} } } \\=0.0287[/tex]
Using normal distribution values we can find\
[tex]P(33.5p.c. < p < 39pc)\\= P(0.335<p<0.39)\\= F(0.39)-F(0.335)\\= 0.852183-0.191735\\=0.660448[/tex]
Answer:
Probability that the response rate will be between 33.5% and 39% = 0.66176 .
Step-by-step explanation:
We are given that a market research firm knows from historical data that telephone surveys have a 36% response rate.
The probability criterion we will use here is;
[tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] ~ N(0,1)
Here, p = 0.36 and n = sample size = 280
Let [tex]\hat p[/tex] = response rate
So, P(0.335 <= [tex]\hat p[/tex] <= 0.39) = P([tex]\hat p[/tex] <= 0.39) - P([tex]\hat p[/tex] < 0.335)
P([tex]\hat p[/tex] <= 0.39) = P( [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] <= [tex]\frac{0.39- 0.36}{\sqrt{\frac{0.39 (1-0.39)}{280} } }[/tex] ) = P(Z <= 1.03) = 0.84849
P([tex]\hat p[/tex] < 0.335) = P( [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] < [tex]\frac{0.335- 0.36}{\sqrt{\frac{0.335 (1-0.335)}{280} } }[/tex] ) = P(Z < -0.89) = 1 - P(Z <= 0.89)
= 1 - 0.81327 = 0.18673
Therefore, P(0.335 <= [tex]\hat p[/tex] <= 0.39) = 0.84849 - 0.18673 = 0.66176
Hence, probability that the response rate will be between 33.5% and 39% is 0.66176 or 66.18 % .
The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leaves on time in 0.24. What is the probability that the train arrives on time, given that it leaves on time?
Answer:
0.9524
Step-by-step explanation:
Note enough information is given in this problem. I will do a similar problem like this. The problem is:
The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?
Solution:
This is conditional probability.
Given:
Probability train arrive on time and leave on time = 0.8 Probability train arrive on time = 0.84 Probability train leave on time = 0.86Now, according to conditional probability formula, we can write:
[tex]P(Leave \ on \ time | arrive \ on \ time)[/tex] = P(arrive ∩ leave) / P(arrive)
Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"
and
P(arrive) means probability arriving on time given as 0.84, so:
0.8/0.84 = 0.9524
This is the answer.
An inverted pyramid is being filled with water at a constant rate of 50 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 8 cm, and the height is 15 cm. Find the rate at which the water level is rising when the water level is 3 cm.
Step-by-step explanation:
Below is an attachment containing the solution.
Answer:
19.53125 cm/s
Step-by-step explanation:
h : s
15 : 8
s/h = 8/15
s = 8h/15
V = ⅓×s²×h
V = ⅓(8h/15)²×h = 64h³/675
dV/dh = 64h²/225
At h=3,
dV/dh = 64(3²)/225 = 64/25
dh/dV = 25/64
dV/dt = 50
dh/dt = dh/dV × dV/dt
= 25/64 × 50
= 19.53125 cm/s
A brochure claims that the average maximum height for a certain type of plant is 0.7 m. A gardener suspects that this is not accurate locally due to variation in soil conditions, and believes the local height is shorter. A random sample of 40 mature plants is taken. The mean height of the sample is 0.65 m with a standard deviation of 0.20 m. Test the claim that the local mean height is less than 0.7 m using a 5% level of significance.
Answer:
As [tex]Z<-Z_{\alpha}[/tex], it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
Step-by-step explanation:
1. Relevant data:
[tex]\mu=0.70\\N=40\\\alpha=0.05\\X=0.65\\s=0.20[/tex]
2. Hypotesis testing
[tex]H_{0}=\mu=0.70[/tex]
[tex]H_{1} =\mu< 0.70[/tex]
3. Find the rejection area
From the one tail standard normal chart, whe have Z-value for [tex]\alpha=0.05[/tex] is 1.56
Then rejection area is left 1.56 in normal curve.
4. Find the test statistic:
[tex]Z=\frac{X-\mu_{0} }{\sigma/\sqrt{n}}[/tex]
[tex]Z=\frac{0.65-0.70}{0.20/\sqrt{40}}\\Z=-1.58[/tex]
5. Hypotesis Testing
[tex]Z_{\alpha}=1.56\\Z=-1.58[/tex]
[tex]-1.58<-1.56[/tex]
As [tex]Z<-Z_{\alpha}[/tex], it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
A student was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sample of size n = 79. Which of the following is a correct interpretation of the interval 0.13 < p < 0.34?
Check all that are correct.
a. There is a 90% chance that the proportion of the population is between 0.13 and 0.34.
b. The proportion of all students who take notes is between 0.13 and 0.34, 90% of the time.
c. With 90% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.13 and 0.34.
d. There is a 90% chance that the proportion of notetakers in a sample of 79 students will be between 0.13 and 0.34.
e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.
Answer:
After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)
And the conclusion for this case would be:
e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.
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".
Description in words of the parameter p
[tex]p[/tex] represent the real population proportion of students who take notes
[tex]\hat p[/tex] represent the estimated proportion of students who take notes
n is the sample size required
[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Numerical estimate for p
In order to estimate a proportion we use this formula:
[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.
Confidence interval
The confidence interval for a proportion is given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)
And the conclusion for this case would be:
e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.
Answer: between 20.3 and 20.9
Step-by-step explanation:
Beth looked through an old
cloth sack that contained small
plastic bags of yarn. There
were 8 bags of pink yarn, 11
white. 5 yellow, 3 green and 9
blue. What is the probability
that when Beth closes her eyes
and pulls out a bag, she will
get blue yarn? please help I have no clue how to do this and the next one is. what is the probability of her drawing white yarn will mark brainest ty please help godbless
Answer: blue is 25% chance
Step-by-step explanation:
8+11+5+3+9=36 then blue would be 9/36 of all the yarn and 9/36 converted into a percentage is 25%
Beth has a total of 36 yarn bags. The probability of her drawing a blue yarn is 9 out of 36, or 0.25. Similarly, the probability of her drawing a white yarn is 11 out of 36, or 0.306.
Explanation:The subject of this question is probability. The total number of yarn bags is the sum of all the different colors bags Beth has, which is 8 + 11 + 5 + 3 + 9 = 36 bags. The probability of an event occurring is determined by dividing the number of preferred events by the total number of outcomes. So, if Beth wants to pick a blue yarn, the number of favorable outcomes is 9 (blue yarn bags) and the total number of outcomes is 36 (total bags).
So, the probability of Beth picking a blue yarn is calculated as such 9/36 = 0.25.
Similarly, the probability of Beth picking a white yarn bag is calculated as 11/36 = 0.306. Hence, out of all the bags, she has a slightly higher chance of picking a white yarn bag.
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Which expression can be simplified as 625/n^12 check all that apply
Answer:
Only options, A and E give 625/n¹² on simplification. The other options do not apply.
(5n⁻³)⁴ = (625/n¹²)
(25n⁻⁶)² = (625/n¹²)
Step-by-step explanation:
625/n¹²
a) (5n⁻³)⁴
According to the law of indices, this becomes
(5⁴)(n⁻³)⁴ = 625(n⁻¹²) = 625/n¹²
This applies!
b) (5n⁻³)⁻⁴
According to the law of indices, this becomes
(5⁻⁴)(n⁻³)⁻⁴ = (n¹²)/625 = n¹²/625
Does Not apply!
c) (5n⁻⁴)³
This becomes
(5³)(n⁻⁴)³ = 125n⁻¹² = 125/n¹²
Does Not apply!
d) (25n⁻⁶)⁻²
This becomes
(25⁻²)(n⁻⁶)⁻² = n⁻¹²/625 = 1/(625n¹²)
Does Not apply!
e) (25n⁻⁶)²
This becomes
(25²)(n⁻⁶)² = 625n⁻¹² = 625/n¹²
This applies!
Answer:
A AND E
Step-by-step explanation:
Help Please...
Katy can text 32 words per minute on her phone. If she sends 8 texts
averaging 80 words each, how long would it take her to send all 8 texts?
Answer: it will take her 20 minutes to send all 8 texts.
Step-by-step explanation:
If she sends 8 texts averaging 80 words each, it means that the number of words in the 8 texts would be
8 × 80 = 640 words
Katy can text 32 words per minute on her phone. Therefore, the time it will take her to send 8 texts or 640 words would be
640/32 = 20 minutes.
Some scientists believe that the average surface temperature of the world has been rising steadily. The average surface temperature can be modeled by T = 0.02t + 15.0. where T is temperature in ∘C and t is years since 1950.
(a) What do the slope and T-intercept represent?
(b) Use the equation to predict the average global surface temperature in 2050.
Answer:
(a)
The slope of the equation = 0.02
Therefore T-intercept equals 15
(b)
Therefore the average global surface temperature in 2050 is 17°C.
Step-by-step explanation:
If a equation is in the form
y= mx+c........(1)
Then the slope of the equation is m.
Slope: The tangent of the angle which is made with the positive x-axis.
If θ be the angle , Then slope (m)= tanθ.
If a equation in the form
[tex]\frac{x}{a} +\frac{y}{b} =1[/tex]............(2)
Then x-axis intercept equals a and y-axis intercept equals b.
Given equation is
T=0.02t+15.0
Comparing with equation (1)
The slope of the equation = 0.02
Again we can rewrite the equation as
[tex]T-0.02t=15[/tex]
[tex]\Rightarrow \frac{T}{15} -\frac{0.02t}{15}=1[/tex]
Comparing with (2)
Therefore T-intercept equals 15
(b)
Here t= 2050-1950 =100
Putting t=100 in the given equation
T=0.02(100)+15 = 2+15 =17
Therefore the average global surface temperature in 2050 is 17°C.
A golfer's bag contains 24 golf balls, 18 of which are ProFlight brand and the other 6 are DistMax brand. Find the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax.
Answer:
1. Assuming with replacement, the probability is 7.91%; or
2. Assuming without replacement, the probability is 8.64%
Step-by-step explanation:
Total number of golf balls = 24
Let Pr denotes probability, P denotes ProFlights, D denotes DistMax.
The probability of selecting 5 balls can be with or without replacement. Since the question is silent on this, the answers to the methods are provided as follows:
1. Assuming with replacement
Pr(4 P and 1 D) = (18/24) × (18/24) × (18/24) × (18/24) × (6/24)
= 0.75 × 0.75 × 0.75 × 0.75 × 0.25
= 0.0791 = 7.91%
2. Assuming without replacement
Here, we assume that 4 ProFlights are selected first before 1 DistMax is selected, and the probability is as follows:
Pr(4 P and 1 D) = (18/24) × (17/23) × (16/22) × (15/21) × (6/20)
= 0.7500 × 0.7391 × 0.7273 × 0.7143 × 0.3000
= 0.0864 = 8.64%
Therefore, the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax is 7.91% assuming with replacement or 8.64% assuming without replacement.
Final answer:
To find the probability, we use the concept of combinations.
Explanation:
To find the probability that the golfer randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax, we need to use the concept of combinations. The total number of ways to select 5 golf balls from a bag containing 24 balls is C(24,5). The number of ways to select 4 ProFlight balls and 1 DistMax ball from their respective totals is C(18,4) * C(6,1). Therefore, the probability is given by:
P = (C(18,4) * C(6,1)) / C(24,5)
The method of Lagrange multipliers assumes that the extreme values exist, but that is not always the case. Show that the problem of finding the minimum value of f(x, y) = x^2 + y^2 subject to the constraint can be solved using Lagrange multipliers, but does not have a maximum value with that constraint.
Answer:
Incomplete question check attachment for complete question
Step-by-step explanation:
Given the function,
F(x, y)=x²+y²
The La Grange is theorem
Solve the following system of equations.
∇f(x, y)= λ∇g(x, y)
g(x, y)=k
Fx=λgx
Fy=λgy
Fz=λgz
Plug in all solutions, (x,y), from the first step into f(x, y) and identify the minimum and maximum values, provided they exist and
∇g≠0 at the point.
The constant, λ, is called the Lagrange Multiplier.
F(x, y)=x²+y²
∇f= 2x i + 2y j
So, given the constraint is xy=1.
g(x, y)= xy-1=0
∇g= y i + x j
gx= y. And gy=x
So, here is the system of equations that we need to solve.
Fx=λgx; 2x=λy. Equation 1
Fy=λgy; 2y=λx. Equation 2
xy=1
Solving this
x=λy/2. From equation 1, now substitute this into equation 2
2y=λ(λy/2)
2y=λ²y/2
2y-λ²y/2 =0
y(2-λ²/2)=0
Then, y=0. Or (2-λ²/2)=0
-λ²/2=-2
λ²=4
Then, λ= ±2
So substitute λ=±2 into equation 2
2y=2x
Then, y=x
So inserting this into the constraint g will give
xy=1. Since y=x
x²=1
Therefore,
x=√1
x=±1
Also y=x
Then, y=±1
Therefore, there are four points that solve the system above.
(1,1) (-1,-1) (1,-1) and (-1,1)
The first two points (1,1) (-1,-1) shows the minimum points because they show xy=1
The other points does not give xy=1
They give xy=-1.
Now,
F(x, y)=x²+y²
F(1,1)=1²+1²
F(1,1)=2
F(-1,-1)= (-1) ²+(-1)²
F(-1,-1)=1+1
F(-1,-1)=2
Then
F(1,1)= F(-1,-1)=2 is the minimum point
This gives the same four points as we found using Lagrange multipliers above.
For safety reasons, 3 different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the 3 systems detects theft with a probability of 0.88 independently of the others. The bank, obviously, is interested in the probability that when a theft occurs,at least one of the 3 systems will detect it. What is the probability that when a theft occurs, at least oneof the 3 systems will detect it? 0.9959 Your answer should be rounded to 5 decimal places.
Answer:
0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.
Step-by-step explanation:
For each system, there are only two possible outcomes. Either it detects the theft, or it does not. The probability of a system detecting a theft is independent of other systems. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Each of the 3 systems detects theft with a probability of 0.88 independently of the others.
This means that [tex]n = 3, p = 0.88[/tex]
What is the probability that when a theft occurs, at least oneof the 3 systems will detect it?
[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{3,1}.(0.88)^{1}.(0.12)^{2} = 0.03802[/tex]
[tex]P(X = 2) = C_{3,2}.(0.88)^{2}.(0.12)^{1} = 0.27878[/tex]
[tex]P(X = 3) = C_{3,3}.(0.88)^{3}.(0.12)^{0} = 0.68147[/tex]
[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3) = 0.03802 + 0.27878 + 0.68147 = 0.99827[/tex]
0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.
The probability that at least one of three independent alarm systems, each with a detection probability of 0.88, will detect a theft is approximately 0.99827.
Explanation:The problem you're asking about falls under the subject of probability, an area of mathematics that measures the likelihood an event will occur. The question asks for the probability that at least one of three independent alarm systems will detect a theft. These systems each have a detection probability of 0.88.
To solve this, it's easier to calculate the probability that none of the systems detect the theft and then subtract that from 1. The likelihood that a system will not detect a theft is 1 - 0.88, which equals 0.12. Since the systems are independent, the probabilities multiply, so: (0.12)^3 = 0.001728. But we want the opposite of this, so we subtract it from 1: 1 - 0.001728 = 0.998272 which is approximately 0.99827 when rounded to five decimal places. That's the probability that at least one alarm system will detect a theft.
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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of their significant others or friends, and 440 said that they are more likely to complain to their dog than to a friend. Suppose that it is reasonable to consider this sample as representative of the population of dog owners.(a) Construct a 90% confidence interval for the proportion of dog owners who take more pictures of their dog than other cant others or friends (Use a table or technology. Round your answers to three decimal places.)(________,_________)Interpret the interval,1. We are 90% confident that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends fals directly in the middle of this interval 2. There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls directly in the middle of this interval We are 90% confident that the mean number of dog owners who take more pictures of their dog than of their significant others or friends fails within this interval 3. There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within this interval 4. We are 90% confident that the true proportion of dog owners who take more pictures of the dog than of their ugnificant others or friends fails within this interval(b) Construct a 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend. (Use a table or technology. Round your answers to three decimal places.)(_______,_______)Interpret the interval,1. There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within this interval 2. We are 95% confident that the true proportion of dog owners who are more likely to complain to their dog than to a friend within this interval 3. we are 95% confident that the mean number of dog owners who are more to complain to the dog than to a friend directly within this interva4. There is 95% a chance that the true proportion of o wners who are more to come to the dog than to a friends directly into this interval 5. we are 95%content at the true proportion of owners who are more to come to the dog than to a friends directly into this interval(c) Give two reasons why the confidence First, the confidence level in part(b) is wider than the part(a).First, the confidece level in part(b) is________ the confidence level in part(a) is, so the critical value of part(b) is______the critical value of part(a), Second the _______ in part(b) is ______than in part(a).
Answer:
(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).
(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).
(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).
Step-by-step explanation:
(a)
Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.
Given:
X = 610
n = 1000
Confidence level = 90%
The (1 - α)% confidence interval for population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The sample proportion is:
[tex]\hat p=\frac{X}{n}=\frac{610}{1000}=0.61[/tex]
The critical value of z for a 90% confidence level is:
[tex]z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645[/tex]
Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:
[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)[/tex]
Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).
Interpretation:
There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).
Correct option is (3).
(b)
Let X = number of dog owners who are more likely to complain to their dog than to a friend.
Given:
X = 440
n = 1000
Confidence level = 95%
The (1 - α)% confidence interval for population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The sample proportion is:
[tex]\hat p=\frac{X}{n}=\frac{440}{1000}=0.44[/tex]
The critical value of z for a 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:
[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)[/tex]
Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).
Interpretation:
There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).
Correct option is (1).
(c)
The confidence interval in part (b) is wider than the confidence interval in part (a).
The width of the interval is affected by:
The confidence levelSample sizeStandard deviation.The confidence level in part (b) is more than that in part (a).
Because of this the critical value of z in part (b) is more than that in part (a).
Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.
First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).
a. (a) With a 90% level of confidence, this is (0.60, 0.63) (3) is the correct interpretation.
b. (0.42, 0.46) represents the 95% confidence interval. (1) is the correct interpretation.
c. First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than in part(a).
a)
The total number of dog owners is: n = 1,000
The number of people said that they take more pictures of their dog than their significant others is: x = 610
Calculate the sample proportion of people said that they take more pictures of their dog than their significant others.
[tex]\bar p = \frac{x}{n}[/tex]
[tex]\bar p = \frac{610}{1000}[/tex]
= 0.61
From the Z-tabulated values, the Z-critical value at the 90% confidence level is: 1.645
Calculate the 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends.
90% C.I. = [tex]\bar p +Z_{critical}\sqrt{\frac{\bar p (1-\bar p)}{n }[/tex]
90% C.I. = 0.61 ± 1.645 [tex]\sqrt{\frac{0.61(1-0.61)}{1000} }[/tex]
90% C.I. = 0.64 ± 0.015
90% C.I. = (0.595 ; 0.625) ≈ (0.60, 0.63)
Hence, the required 90% confidence interval is: (0.60 ∠ p ∠ 0.63)
We are 90% confident that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within this interval.
b)
The total number of dog owners is: n = 1,000
The number of people said that they take more pictures of their dog than their significant others is: x = 440
Calculate the sample proportion of people said that they take more pictures of their dog than their significant others.
[tex]\bar p = \frac{x}{n}[/tex]
[tex]\bar p = \frac{440}{1000}[/tex]
= 0.44
From the Z-tabulated values, the Z-critical value at the 95% confidence level is: 1.96
Calculate the 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends.
90% C.I. = [tex]\bar p +Z_{critical}\sqrt{\frac{\bar p (1-\bar p)}{n }[/tex]
90% C.I. = 0.44 ± 1.96 [tex]\sqrt{\frac{0.44(1-0.44)}{1000} }[/tex]
90% C.I. = 0.44 ± 0.016
90% C.I. = (0.424 ; 0.456) ≈ (0.42, 0.46)
Hence, the required 95% confidence interval is: (0.42 ∠ p ∠ 0.46)
We are 95% confident that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within this interval.
C)
Compared to component (a), part (b)'s confidence interval is more expansive.
Several factors influence the interval's width:
The confidence levelSample sizeStandard deviationPart (b) has a higher degree of confidence than part (a).
As a result, component (b)'s critical value of z is greater than part (a)'s.
Additionally, part (b) has a margin of error of 0.016, which is greater than part (a)'s margin of error of 0.015.
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Solve using normalcdf
Let X be the random variable representing monthly trainee income. X is distributed with mean $1100 and standard deviation $150. You want to find the proportion of trainees that earn less than $900 per month, or Pr(X < 900).
Using normalcdf (on a TI calculator, for instance), you would compute
normalcdf(-1E99, 900, 1100, 150)
to get a proportion of approximately 0.09121, or 9.121%.
That is, the syntax for normalcdf is
normalcdf(lower limit, upper limit, mean, standard deviation)
In this case, you pick a very large negative number for "lower limit" in order to simulate negative infinity.
Let y be a random variable with a known distribution, and consider the square loss function `(a; y) = (a????y)2. We want to find the action a that has minimal risk, namely, to find a = arg mina E(a ???? y)2, where the expectation is with respect to y. Show that a = Ey, and the Bayes risk (i.e. the risk of a) is Var(y). In other words, if you want to try to predict the value of a random variable, the best you can do (for minimizing expected square loss) is to predict the mean of the distribution. Your expected loss for predicting the mean will be the variance of the distribution. You should use the fact that Var(y) = Ey2 ???? (Ey)2.
Answer/ Explanation:
Since X is exponentially distributed, its expected value is given by E[X]=1/λ=2.
Therefore, E[Y]=E[1−2X]=E[1]+E[−2X]=E[1]−2E[X]=1−2E[X]=1−2⋅2=−3.
Hence,
We define the moment-generating function of Y as MY(t). It is given by
MY(t)=E[etY]=E[et(1−2X)]=E[ete−2tX]=E[et]E[e−2tX].
If I give you the hint that E[g(Y)]=∫∞0g(y)fY(y)dy, where fY(y) is the probability density function of Y, can you also solve for the moment generating function of Y?
We have E[X2]=2/λ2=2/(0.5)2=8. Thus,
E[Y2]=E[(1−2X)2]=E[1−4X+4X2]=E[1]−4E[X]+4E[X2]=1−4⋅2+4⋅8=25.
So,
Var(Y)=E[Y2]−E[Y]2=25−(−3)2=16.
Continuing for the moment-generating function:
MY(t)=E[et]E[e−2tX]=etE[e−2tX]=et∫∞x=0e−2txfX(x)dx,
where fX(x) is the probability density function of X and thus satisfies fX(x)=λe−λx. Substituting yields
MY(t)=et∫∞x=0e−2txλe−λxdx=λet∫∞x=0e−x(2t+λ)dx=λet2t+λ.
It is also good to note that
If you are after expectation, variance or moment generating function of Y then it is not needed to find the PDF of Y (see the answer of Ritz).
This is not an answer on the question in the title, but one on the question in the body.
FY(y)=P(Y≤y)=P(1−2X≤y)=P(X≥0.5−0.5y)=1−FX(0.5−0.5y)
Note that the last equality demands that FX is continuous.
Differentating on both sides gives fY on LHS and an expression in fX on RHS.
Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.
Function Point
w = y3 − 9x2y
x = es, y = et
s = −5, t = 10
Evaluate each partial derivative at the given values of s and t.
Answer:
The value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex]The value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex]The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Step-by-step explanation:
Given that the Function point are [tex]w=y^3-9x^2y[/tex]
[tex]x=e^s[/tex], [tex]y=e^t[/tex] and s = -5, t = 10
To find [tex]\frac{\partial w}{\partial s}[/tex] and [tex]\frac{\partial w}{\partial t}[/tex]using the appropriate Chain Rule :[tex]w=y^3-9x^2y[/tex]
Substitute the values of x and y in the above equation we get
[tex]w=(e^t)^3-9(e^s)^2(e^t)[/tex]
[tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to s by using chain rule we have[tex]\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)[/tex]
[tex]=e^{3t}-18e^{2s}.(e^t)[/tex]
[tex]=e^{3t}-18e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex][tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to t by using chain rule we have[tex]\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)[/tex]
[tex]=3e^{3t}-9e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :
[tex]\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}[/tex]
[tex]=e^{3(10}-18e^{2(-5)+10}[/tex]
[tex]=e^{30}-18e^{-10+10}[/tex]
[tex]=e^{30}-18e^0[/tex]
[tex]=e^{30}-18[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex][tex]\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}[/tex]
[tex]=3e^{3(10)}-9e^{2(-5)+10}[/tex]
[tex]=3e^{30}-9e{-10+10}[/tex]
[tex]=3e^{30}-9e{0}[/tex]
[tex]=3e^{30}-9[/tex]
[tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]Using the Chain Rule, it is found that:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18[/tex]
[tex]\frac{\partial W}{\partial s}(-5,10) = 3e^{30} - 9[/tex]
w is a function of x and y, which are functions of s and t, thus, by the Chain Rule:
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
Then, the derivatives are:
[tex]\frac{\partial W}{\partial x} = -18xy[/tex]
[tex]\frac{\partial x}{\partial s} = e^s[/tex]
[tex]\frac{\partial W}{\partial y} = 3y^2 - 9x^2[/tex]
[tex]\frac{\partial y}{\partial s} = 0[/tex]
Then
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial s} = -18xy(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^se^t(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s}e^t[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s + t}[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18e^{2(-5) + 10} = -18[/tex]
Then, relative to t:
[tex]\frac{\partial x}{\partial t} = 0[/tex]
[tex]\frac{\partial y}{\partial t} = e^t[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
[tex]\frac{\partial W}{\partial t} = (3y^2 - 9x^2)e^t[/tex]
[tex]\frac{\partial W}{\partial t} = (3e^{2t} - 9e^{2s})e^t[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = (3e^{20} - 9e^{-10})e^{10} = 3e^{30} - 9[/tex]
A similar problem is given at https://brainly.com/question/10309252
In a study of the effects of acid rain, a random sample of 100 trees from a particular forest is examined. Forty percent of these show some signs of damage. Which of the following statements is correct?
a.None of the above
b. The sampling distribution of the proportion of damaged trees is approximately Normal
c.This is a comparative experiment.
d.If a sample of 1000 trees was examined, the variability of the sample proportion would be larger than in a sample of 100 trees
e.If we took another random sample of trees, we would find that 40% of these would show some signs of damage
Answer: B
Step-by-step explanation:
The sampling distribution of the proportion of damaged trees is approximately Normal
Answer:B. The sampling distribution of the proportion of damaged trees is approximately Normal.
Step-by-step explanation: Sampling distribution is a probability distribution of a statistic which is gotten through the collection of a large number of samples from the population of interest.
A normal distribution is a distribution of the samples symmetrically around the mean, when a Statistical metric shows that the outcome is distributed around the central region it shows that the distribution is normal.
FOR A FORTY PERCENT INCIDENCE, THE SAMPLING DISTRIBUTION OF THE PROPORTION OF DAMAGED TREES IS APPROXIMATELY NORMAL AS IT IS CLOSE TO 50% WHICH IS THE MEAN OF 100%.
Compute the area of triangle, if x equals 4 more than 6. A) 10 B) 50 C) 100 D) 200 E) 400
Answer:
76cm²
Step-by-step explanation:
The triangle from all indication is an isosceles triangle:
Let x represent the side
∴ x = 4 - 6 = -2
x = -2
x + 2 = 0
Using the fomular, as area of triangle, that is:
Area = √s(s - a)(s - b)(s - c)
s = a + b+ c/2, where a = x + 2; b = x+ 2; c = x
∴ s = x + 2 + x + 2 + x/2 = 3x + 4/2
Area = √3x + 4/2( 3x + 4/2 - x - 2)(3x + 4/2 - x - 2)(3x + 4/2 - x)
= √3x + 4/2( x/2)(x/2)(x + 4/2)
= √3x + 4/2(x²/4)(x + 4)/2
Let's assume x = 12
∴ Area = √5760 = 76cm²
Answer:
Its C.100
Area of a triangle b*h/2 (6+4=10*2=20) 20*10=200/2 will give you 100.
Step-by-step explanation:
Consider the four points used in Problem 1: (-9,13), (-3, 9), (3, 6) and (9, 1) Three out of four of these points lie on one line. Which one of the four points does not lie on the same line as the other three? Justify your answer using slope. b) Find the equation of the line that contains three out of the four points
Answer:
a) ( 3 , 6 )
b) y = -2*x / 3 + 7
Step-by-step explanation:
Given:
- The four points are:
(-9,13), (-3, 9), (3, 6) and (9, 1)
- Three points lie on the same line.
Find:
Which one of the four points does not lie on the same line as the other three?
Solution:
- Compute the slope between each pair of point:
(-9,13), (-3, 9)
m_1 = ( 9 - 13 ) / ( -3 + 9 ) = - 2/3
(-9,13), (3, 6)
m_2 = ( 6 - 13 ) / ( 3 + 9 ) = - 0.5833
(-9,13), (9, 1)
m_3 = ( 1 - 13 ) / ( 9 + 9 ) = - 2/3
- We see that m_1 = m_3 ≠ m_2 . Hence, point ( 3 ,6 ) does not lie on the same line.
- The equation of line is expressed as:
y = m*x + C
m = -2/3
y = -2*x / 3 + C
1 = -2*9 / 3 + C ....... ( 9 , 1 )
C = 7
- The equation of the line is:
y = -2*x / 3 + 7
Points A(-9, 13), B(-3, 9) and D(9, 1) are colinear.
b). Equation of the line passing through these points will be [tex]y=-\frac{2}{3}x+7[/tex]
Property for the colinear points:If three points are colinear, slope of the lines joining these points will be equal.Slope of a line joining two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by,
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Given points in the question,
A(-9, 13), B(-3, 9), C(3, 6) and D(9, 1)
If the slopes of the lines joining these points are same, points will be colinear.
Slope of AB = [tex]\frac{13-9}{-9+3}=-\frac{2}{3}[/tex]
Slope of AC = [tex]\frac{13-6}{-9-3}=-\frac{7}{12}[/tex]
Slope of AD = [tex]\frac{13-1}{-9-9}=-\frac{2}{3}[/tex]
Since, slopes of AB and AD are equal, points A, B and D will be colinear.
b). Let the equation of the line passing through A, B and D is,
y - y' = m(x - x')
Since, this line passes through D(9, 1) and slope = [tex]-\frac{2}{3}[/tex]
Equation of the line → [tex]y-1=-\frac{2}{3}(x-9)[/tex]
[tex]y=-\frac{2}{3}x+6+1[/tex]
[tex]y=-\frac{2}{3}x+7[/tex]
Therefore, points A(-9, 13), B(-3, 9) and D(9, 1) are colinear and equation of the line passing through these points will be [tex]y=-\frac{2}{3}x+7[/tex].
Learn more about the slope of a line here,
https://brainly.com/question/490077?referrer=searchResults
Two hikers come to a ravine and want to know how wide it is. They set up two similar triangles as shown in the diagram. How far is it across the ravine?
Answer:
d = 80 ft
Step-by-step explanation:
Start with angle BCA, tan(BCA) = 30/15 so BCA = 63.43 degrees.
BCA = ECD since they are opposite angles.
tan ECD = d/40
40 tan(63.4degrees) = d
d = 80 ft
Alternatively, you can use similar triangles since the angles are the same, BCA = ECD, CED = CAB, and CBA = EDC. In that case use proportions to get 30/15 = d/40 so 2 = d/40 and d = 80.