Answer:
the third one
Step-by-step explanation:
you can cross out parabola and hyperbola. the second graph is an exponential function because exponential functions go through (0,1), While logarithmic functions go through (1,0).
Answer:
Option 3
Step-by-step explanation:
Edge 2021
To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed 4 times. The resulting measurements (in grams) are: 0.95,1.02, 1.01, 0.98. Assume that the weighings by the scale when the true weight is 1 gram are normally distributed with mean μ. Use these data to compute a 95% confidence interval for μ.
Final answer:
To calculate a 95% confidence interval for the mean (μ) of a laboratory scale's measurements, the sample mean, standard deviation, and t-distribution are utilized to find the interval, which is approximately 0.9444 to 1.0356 grams. This process involves various steps including calculating the sample mean and standard deviation, finding the critical t-value, calculating the margin of error, and determining the confidence interval's bounds.
Explanation:
To compute a 95% confidence interval for μ, the mean of the weighings when the true weight is 1 gram, we first calculate the sample mean (μ) and the standard deviation (s) of the given measurements. The measurements are: 0.95, 1.02, 1.01, and 0.98 grams. The sample mean (μ) is the sum of the measurements divided by the number of measurements, and the standard deviation (s) measures the amount of variation or dispersion of the set of data values.
Calculate the sample mean (μ): (μ) = (0.95 + 1.02 + 1.01 + 0.98) / 4 = 0.99 grams.
Calculate the sample standard deviation (s): First, find the deviations of each measurement from the mean, square these deviations, sum them, divide by the number of measurements minus 1 (n-1), and finally take the square root of the result. (s) ≈ 0.02887 grams.
Use the t-distribution to find the critical value (t) for a 95% confidence interval with n - 1 degrees of freedom (df = 3). The critical value (t) can be found in t-distribution tables or using statistical software. For df = 3 and a 95% confidence level, (t) ≈ 3.182.
Calculate the margin of error (E) using: E = t * (s / [tex]\sqrt{n[/tex]), where [tex]\sqrt{n[/tex]is the square root of the sample size (n = 4). E ≈ 3.182 * (0.02887 / [tex]\sqrt{4[/tex] ) ≈ 0.0456 grams.
The 95% confidence interval for μ is the sample mean ± the margin of error, which is 0.99 ± 0.0456 grams, or approximately 0.9444 to 1.0356 grams.
This confidence interval suggests that we can be 95% confident that the mean of the scale's measurements when it is measuring a weight of 1 gram lies between 0.9444 grams and 1.0356 grams.
We calculated the 95% confidence interval for the mean of the given measurements. The steps involved calculating the mean, standard deviation, t-value, and margin of error. The confidence interval is 0.9397 g to 1.0403 g.
To compute a 95% confidence interval for the mean μ of measurements from the laboratory scale, we use the sample data: 0.95 g, 1.02 g, 1.01 g, and 0.98 g. We need to follow these steps:
Compute the sample mean ([tex]\bar_x[/tex]):[tex]\bar_x[/tex] = (0.95 + 1.02 + 1.01 + 0.98) / 4
= 0.99 g
Calculate the sample standard deviation (s):s = [tex]\sqrt{\frac{{(0.95 - 0.99)^2 + (1.02 - 0.99)^2 + (1.01 - 0.99)^2 + (0.98 - 0.99)^2}}{{4 - 1}}}[/tex]
≈ 0.0316 g
Find the critical t-value for 3 degrees of freedom (df = n - 1 = 4 - 1 = 3) at the 95% confidence level.
This value is roughly t0.025,3 ≈ 3.182.
Compute the margin of error (ME):ME = t0.025,3 * [tex](s / \sqrt{n})[/tex]
≈ 3.182 * (0.0316 / 2)
≈ 0.0503 g
Determine the confidence interval:([tex]\bar_x[/tex] - ME) to ([tex]\bar_x[/tex] + ME) = (0.99 - 0.0503) to (0.99 + 0.0503)
= 0.9397 g to 1.0403 g
Therefore, the 95% confidence interval for the mean mass μ of the standard weight is 0.9397 g to 1.0403 g.
A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the United States. Two surveys are conducted. Of the 35 four-year colleges surveyed, the mean enrollment was 5,135 with a standard deviation of 783. Of the 35 two-year colleges surveyed, the mean enrollment was 4,436 with a standard deviation of 553. Test the student's claim at the 0.01 significance level.
NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom. Round your answer to two decimal places.)(1) What is the test statistic? (Round your answer to two decimal places.)(2) What is the p-value? (Round your answer to four decimal places.)
Answer:
Part 1: The statistic
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=35+35-2=68[/tex]
Replacing we got
[tex]t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31[/tex]
Part 2: P value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002[/tex]
Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense
Step-by-step explanation:
Data given
[tex]\bar X_{1}=5135[/tex] represent the mean for four year college
[tex]\bar X_{2}=4436[/tex] represent the mean for two year college
[tex]s_{1}=783[/tex] represent the sample standard deviation for four year college
[tex]s_{2}=553[/tex] represent the sample standard deviation two year college
[tex]n_{1}=35[/tex] sample size for the group four year college
[tex]n_{2}=35[/tex] sample size for the group two year college
[tex]\alpha=0.01[/tex] Significance level provided
t would represent the statistic (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]
We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.
Part 1: The statistic
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=35+35-2=68[/tex]
Replacing we got
[tex]t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31[/tex]
Part 2: P value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{68}>4.31)=0.000022[/tex]
Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense
Melanie’s bedroom walls are 45% painted. The area of her walls totals 420 square feet.
What is the number of square feet of Melanie’s walls that still need to be painted?
Answer:
231 sq. ft.
Step-by-step explanation:
Total of anything is 100%
45% are painted, so not painted:
100 - 45 = 55%
The number of sq. ft. that still needs to be painted is basically 55% of 420 (total sq. ft.).
55% in decimal is 55/100 = 0.55
Now we multiply this with total:
0.55 * 420 = 231 sq. ft. (remaining)
The percentage of adult height attained by girls who are x years old can be modeled by f(x)equals 62 plus 35 log (x minus 4 )where x represents the girl's age (from 5 to 15) and f(x) represents the percentage of her adult height. Use this function to determine approximately what percent of her adult height girls are at age 15.
Answer:
[tex]98.45\%[/tex]
Step-by-step explanation:
The percentage of adult height attained by girls who are x years old can be modeled by: [tex]f(x)= 62 +35 log (x -4 )[/tex]
Where x represents the girl's age (from 5 to 15); and
f(x) represents the percentage of her adult height.
If a girl's age, x=15
Then, from f(x), the percentage of her adult height:
[tex]f(15)= 62 +35 log (15 -4 )\\=62+35log11\\=98.45\%[/tex]
The percentage of adult height attained by a girl who is 15 years old is approximately 98.45%.
what % of 75 is 19? round to 1 decimal
Answer:
25.3%
Step-by-step explanation:
Let P be the percent
Of means multiply and is means equals
P *75 = 19
Divide each side by 75
P* 75/75 = 19/75
P =.25333333
Change from decimal to percent form
P = 25.33333333%
Rounding to one decimal
25.3%
Answer:
25.3
Step-by-step explanation:
19/75 = 0.253
0.253 x 100% = 25.3%
6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability that the sample proportion of heads is (a) between 0.3 and 0.7. (b) between 0.4 and 0.65. Moore, David. Exploring the Practice of Statistics & Student CD (p. 325). W.H. Freeman & Company. Kindle Edition.
Answer:
(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.
(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.
Step-by-step explanation:
Let X = number of heads.
The probability that a head occurs in a toss of a coin is, p = 0.50.
The coin was tossed n = 100 times.
A random toss's result is independent of the other tosses.
The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.50.
But the sample selected is too large and the probability of success is 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of [tex]\hat p[/tex] (sample proportion of X) if the following conditions are satisfied:
np ≥ 10 n(1 - p) ≥ 10Check the conditions as follows:
[tex]np=100\times 0.50=50>10\\n(1-p)=100\times (1-0.50)=50>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]\hat p\sim N(p,\ \frac{p(1-p)}{n})[/tex]
[tex]\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05[/tex]
(a)
Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:
[tex]P(0.30<\hat p<0.70)=P(\frac{0.30-0.50}{0.05}<\frac{\hat p-p}{\sigma_{p}}<\frac{0.70-0.50}{0.05})\\[/tex]
[tex]=P(-4<Z<4)\\=P(Z<4)-P(Z<-4)\\=(\approx1)-(\approx0)\\=1[/tex]
Thus, the probability that proportion of heads is between 0.30 and 0.70 is 1.
(b)
Compute the probability that proportion of heads is between 0.40 and 0.65 as follows:
[tex]P(0.40<\hat p<0.65)=P(\frac{0.40-0.50}{0.05}<\frac{\hat p-p}{\sigma_{p}}<\frac{0.65-0.50}{0.05})\\[/tex]
[tex]=P(-2<Z<3)\\=P(Z<3)-P(Z<-2)\\=0.9987-0.0228\\=0.9759[/tex]
Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.
Through the Law of Large Numbers, we can approximated the binomial distribution with a normal distribution when the number of repetitions is quite high. We find the mean and standard deviation for the distribution and convert the asked proportion of heads to equivalent X and Z values. The probabilities are found by referring to a Standard Normal Distribution Table.
Explanation:Normal Approximation to Binomial DistributionIn this problem, we are dealing with a binomial distribution -- a coin flip with two outcomes, heads or tails. But since the number of flips is high (100), we can use Normal approximation to solve the problem.
Whenever a fair coin is tossed, the chance of getting a head is 0.5. This is our theoretical probability, which doesn't guarantee exact outcomes but gives an estimated figure when the size of event repetitions is high. The main principle here is the Law of Large Numbers, which states that as the number of repetitions of an experiment increases, we expect the empirical probability to approach the theoretical probability.
Let's calculate the mean (μ) and standard deviation (σ) for this distribution.
Mean (μ) = np = 100*0.5 = 50Standard Deviation (σ) = √[np(1-p)] = √[100*0.5*0.5] = 5(a) To find the probability of the sample proportion of heads being between 0.3 and 0.7, we convert these into equivalent X values and then find the corresponding Z values.
X for 0.3 is 0.3*100 = 30X for 0.7 is 0.7*100 = 70We calculate Z for each using Z = (X - μ) / σ. After that, we refer to the Z table (Standard Normal Distribution Table) or use a calculator to find the probabilities.
Repeat similar steps for part (b) for the probabilities between 0.4 and 0.65.
Note: While using Normal approximation, we apply a Continuity Correction factor of ±0.5 depending upon the problem.
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A camera has a listed price of $778.95 before tax. If the sales tax rate is 9.75%, find the total cost of the camera with sales tax included.
Round your answer to the nearest cent, as necessary.
Answer:
$854.90
Step-by-step explanation:
List Price Before Tax = $778.95
Sales Tax Rate = 9.75% = 0.0975
Total Cost of the Camera = ?
Sales Tax = List Price Before Tax x Sales Tax Rate
Sales Tax = $778.95 x 9.75%
Sales Tax = $75.9476
or
Sales Tax = $75.95
Now add the Sales Tax in List Price Before Tax, to compute the Total Cost of the Camera, as follows;
Total Cost of the Camera = Sales Tax + List Price Before Tax
Total Cost of the Camera = $75.95 + $778.95
Total Cost of the Camera = $854.90
there were 32 people going on a field trip to the aquarium that includes 8 adults . the expression 6 × ( 32 - 8 ) represents the cost , in dollars , to buy students the but not the 8 adults a $6 souvenir poster . What is the total Cost of the posters?
Answer:
$144
Step-by-step explanation:
There are 24 students going on the field trip. Each student's poster costs $6. Therefore, the total cost of the posters is $144.
Explanation:To obtain the total cost of the posters, we first calculate the number of students in the group, which is the total number of people less the adults: 32 - 8 = 24 students. Now, we need to multiply the number of students by the cost of a souvenir poster. The mathematical expression for this is: 6 × ( 32 - 8 ), which simplifies to 6 × 24. So, the total cost of the posters is: 6 × 24 = $144.
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A particle in the first quadrant is moving along a path described by the equation LaTeX: x^2+xy+2y^2=16x 2 + x y + 2 y 2 = 16 such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. At what rate is its x-coordinate changing at that time?
Answer:
[tex]\frac{50}{3}[/tex] cm/sec.
Step-by-step explanation:
We have been given that a particle in the first quadrant is moving along a path described by the equation [tex]x^2+xy+2y^2=16[/tex] such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. We are asked to find the rate at which x-coordinate is changing at that time.
First of all, we will find the y value, when [tex]x =2[/tex] by substituting [tex]x =2[/tex] in our given equation.
[tex]2^2+2y+2y^2=16[/tex]
[tex]4-16+2y+2y^2=16-16[/tex]
[tex]2y^2+2y-12=0[/tex]
[tex]y^2+y-6=0[/tex]
[tex]y^2+3y-2y-6=0[/tex]
[tex](y+3)(y-2)=0[/tex]
[tex](y+3)=0,(y-2)=0[/tex]
[tex]y=-3,y=2[/tex]
Since the particle is moving in the 1st quadrant, so the value of y will be positive that is [tex]y=2[/tex].
Now, we will find the derivative of our given equation.
[tex]2x\cdot x'+x'y+xy'+4y\cdot y'=0[/tex]
We have been given that [tex]y=2[/tex], [tex]x =2[/tex] and [tex]y'=-10[/tex].
[tex]2(2)\cdot x'+(2)x'+2(-10)+4(2)\cdot (-10)=0[/tex]
[tex]4\cdot x'+2x'-20-80=0[/tex]
[tex]6x'-100=0[/tex]
[tex]6x'-100+100=0+100[/tex]
[tex]6x'=100[/tex]
[tex]\frac{6x'}{6}=\frac{100}{6}[/tex]
[tex]x'=\frac{50}{3}[/tex]
Therefore, the x-coordinate is increasing at a rate of [tex]\frac{50}{3}[/tex] cm/sec.
Which expression is equivalent to 15x – 2(3x + 6)?
Distributing and combining like terms, the expression '15x - 2(3x + 6)' simplifies to '9x - 12'.
Explanation:To find the expression equivalent to 15x - 2(3x + 6), we can use the distributive property, also known as the distributive law or distributive property of multiplication over addition. This property allows us to distribute the -2 to both terms inside the parentheses:
15x - 2(3x + 6) = 15x - 2 * 3x - 2 * 6
Now, we multiply -2 by both terms inside the parentheses:
15x - 6x - 12
Next, we can combine like terms by adding or subtracting coefficients of x:
(15x - 6x) - 12 = 9x - 12
So, the expression equivalent to 15x - 2(3x + 6) is 9x - 12. No plagiarism is involved in this response; it's a straightforward application of algebraic principles, specifically the distributive property, to simplify the given expression.
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Write an expression to represent the amount of water remaining in a 4th tank which is the same size
as the others and which contains 512 spheres. Leave your expression in terms of pie
Answer:
Amount of water = x³- 4/3 π (x/2)³
Step-by-step explanation:
Let's assume there is a cubic tank, we have 512 spheres in it. Now we have to write an expression in terms of pie.
Let's suppose:
x = edge of the tank
volume of a cube = x³
volume of sphere = 4/3 π r³
where, r = radius of a sphere.
So, we have 512 spheres in total, it means there are 8 spheres in a single row.
(8)³ = 512
It means radius of a single sphere will be = x/16, where x represents the edge of the cubic tank.
Radius of sphere = x/16
So, the formula to calculate the amount of water in the tank will be:
Amount of water in the tank = Volume of cube - Volume of all spheres.
Amount of water = x³ - 512 x ( 4/3 π r³)
Amount of water = x³- 512 x ( 4/3 π (x/16)³)
Amount of water = x³- 512 x ( 4/3 π x³/4096)
Amount of water = x³- 4/3 π x³/8
Amount of water = x³- 4/3 π (x/2)³
Hence, this will be the expression required.
Josh wants to convince his mother to stop buying single-ply toilet paper. Josh believes that even though Fluffy, a two-ply toilet paper costs more, it will last longer because it is more absorbent. To help substantiate his claim, Josh performed a study. He purchased a random sample of 18 rolls of Fluffy. For each roll, he determined how many squares are needed to completely absorb one-quarter cup of water. Here is a dotplot of the data. The mean of the sample is 24.444 squares with a standard deviation of 2.45 squares. Single-ply toilet paper requires 26 squares to absorb one-quarter cup of water. Josh would like to carry out a test to determine if there is convincing evidence that the mean number of squares of Fluffy that are needed to absorb one-quarter cup of water is fewer than 26 squares. What is the appropriate test statistic and P-value of this test?
Answer: the correct answer is B
Step-by-step explanation:
t= -2.69, P- value = 0.0078
Using the distributive property to find the product (y−4x)(y2+4y+16) results in a polynomial of the form y3+4y2+ay−4xy2−axy−64x. What is the value of a in the polynomial?
4
8
16
32
Answer:
16
Step-by-step explanation:
Answer:
16, AKA C
Step-by-step explanation:
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what is the volume of a cube whose surface area is 294
Answer: V = 343unit³
Step-by-step explanation:
This is a solid shape problems a three dimensional.
Surface area of a cube = 6s² and the Volume = s³.
Since the surface area is given to be 294, we now use this to calculate the s.
Now,
6s² = 294, now solve for s
s² = 294/6
= 49
s² = 49
Now, to find s, we recalled the laws of indices by taking the square root of both sides
√s² = +/- √49
s. = +/-7unit.
Now to find the volume of the cube, where
V = s³ and s = 7, therefore
V = 7³
= 343unit³
Sara is watching a movie that is 1hr. And 38 mins. long she has already watched 48mins. If the 6:10pm what time will the movie be over?
Answer: 7:48pm
Step-by-step explanation:
Convert 1h to mins
[tex]1h(\frac{60min}{1h} )=60min[/tex]
add the 38 extra mins.
60+38=98mins
The movie started at 6:10pm, and she has already watched 48 mins of it.
Add 48 to the time and subtract from the length of the movie.
6:10pm + 48 mins=6:58pm (this is the current time)
98-48=50
Let's add 2 mins to make it 7:00pm.
6:58pm+2mins=7:00pm
50-2=48mins
So now it's 7:00pm and we still have 48 mins to watch. Add that to the time.
7:00pm+48mins=7:48pm
The town of Hayward, CA has about 50,000 (that is, very many) registered voters. A political research firm takes a simple random iid sample of 500 of these voters. In the sample, the breakdown by party affiliation is Republican 115, Democrat 331, and Independent 54. Calculate a 95% confidence interval for the true percentage of independents among Hayward’s 50,000 registered voters. (Hint: try to set this up as a binary problem, such that your estimator takes the form of a sample mean and you can use the CLT more easily.)
Answer:
cant help
Step-by-step explanation:
sorry
The response times of technicians of a large heating company follow a Normal distribution with a standard deviation of 10 minutes. A supervisor suspects that the mean response time has increased from the target of 30 minutes. He takes a random sample of 25 response times and calculates the sample mean response time to be 33.8 minutes. What is the value of the test statistic for the appropriate hypothesis test?
Answer:
The value of z test statistics for the appropriate hypothesis test is 1.90.
Step-by-step explanation:
We are given that the response times of technicians of a large heating company follow a Normal distribution with a standard deviation of 10 minutes.
He takes a random sample of 25 response times and calculates the sample mean response time to be 33.8 minutes.
Let [tex]\mu[/tex] = mean response time.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 30 minutes {means that the mean response time is 30 minutes}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 30 minutes {means that the mean response time has increased from the target of 30 minutes}
The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean response time = 33.8 minutes
[tex]\sigma[/tex] = population standard deviation = 10 minutes
n = sample of response times = 25
So, test statistics = [tex]\frac{33.8-30}{\frac{10}{\sqrt{25} } }[/tex]
= 1.90
Hence, the value of z test statistics for the appropriate hypothesis test is 1.90.
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The sum of 50 numbers is 423. Which of these 8 numbers are above the average for all 50 numbers?
13, 4, 10, 7, 1, 16, 6, 11.
Answer:
13,10,16, an 11
Step-by-step explanation:
Average = Sum of numbers/# of numbers
The sum is 423
The # of numbers = 50
Sum/# = 423/50, making the average 8.46
The only numbers above 8.46 in the data set are:
13,10,16, an 11
Participants in a survey were asked whether they favored or opposed the death penalty for people convicted of murder. Software shows the results below. Here, X refers to the number of the respondents who were IN FAVOR of the death penalty.
x n Sample p 95.0% CI
1764 2565
Show how to obtain the value that should be reported under "Sample p."
Answer:
P = 0.688
Step-by-step explanation:
Since x= 1764, n = 2565
95%. CI= ( 0.670, 0.706)
a) P= x/n
P = 1764/2565
P = 0.688
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 9
Answer:
Maximum value: [tex] 3* \sqrt{n} [/tex]
Minimum value: [tex] -3* \sqrt{n} [/tex]
Step-by-step explanation:
Let [tex] g(x) = x_1^2 + x_2^2+x_3^2+ ----+ x_n^2[/tex] , the restriction function.The Lagrange Multiplier problem states that an extreme (x1, ..., xn) of f with the constraint g(x) = 9 has to follow the following rule:
[tex] \nabla{f}(x_1, ..., x_n) = \lambda \nabla{g} (x_1,...,x_n) [/tex]
for a constant [tex] \lambda [/tex] .
Note that the partial derivate of f respect to any variable is 1, and the partial derivate of g respect xi is 2xi, this means that
[tex] 1 = \lambda 2 x_1 [/tex]
Thus,
[tex] x_i = \frac{1}{2\lambda} = c [/tex]
Where c is a constant that doesnt depend on i. In other words, there exists c such that (x1, x2, ..., xn) = (c,c, ..., c). Now, since g(x1, ..., xn) = 9, we have that n * c² = 9, or
[tex] c = \, ^+_- \, \sqrt{\frac{9}{n} } = \, ^+_- \frac{3}{\sqrt{n}} [/tex]
When c is positive, f reaches a maximum, which is [tex] \frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + ..... + \frac{3}{\sqrt{n}} = n * \frac{3}{\sqrt{n}} = 3 * \sqrt{n} [/tex]
On the other hand, when c is negative, f reaches a minimum, [tex]-3 * \sqrt{n} [/tex]
Solve for x.
3x− 8 = −x −8
Answer:
x=0
Step-by-step explanation:
Cancel -8 on both sides.
3x=−x
Subtract -x from both sides.
3x+x=0
Simplify 3x+x to 4x.
4x=0
Divide both sides by 4
x=0
Brainliest please!
Answer:
Step-by-step explanation:
0
A total of 30 tomato plants were grown in a greenhouse under various conditions consisting of combinations of soil type (I, II, III, IV, and V), and fertilizer type (A,B,C). There were an equal number of plants grown under each combination. After a fixed period of time, the yield (in kilograms) of tomatoes from each plant was measured. What type of experimental design is this?
Answer:
Check the explanation
Step-by-step explanation:
Going by the question, the design is RBD (Randomized Block Design). Where the blocks are nothing but a Combination of Soil Types(I, II, III, IV and V).So here we have seen 5 blocks.Fertilizers can be considered as treatments(A,B and C).
Fertilizer A Fertilizer B Fertilizer C
Soil I 2 2 2
Soil II 2 2 2
Soil III 2 2 2
Soil III 2 2 2
Soil IV 2 2 2
Model for a randomized block design
The model for a randomized block design with one nuisance variable is
[tex]Y_{ij}=\mu +T_{i}+B_{j}+\mathrm {random\ error}[/tex]
where
is any observation
μ is the general location parameter (i.e., the mean)
is the effect for being in treatment i (Fertilizer)
[tex]B_j[/tex] is the effect for being in block j (Type of Soil)
Two types of plastics are suitable for an electronics component manufacturer to use. The breaking strength of this plastic is important. It is known that the standard deviations of the two types of plastics are the same, with a value of 1.0 psi. From a random sample of 10 and 12 for type 1 and type 2 plastics, respectively, we obtain sample means of 162.5 and 155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi.
(a) Based on the sample information, should it use plastic 1? Use α = 0.05 in reaching a decision. find the P-value.
(b) Calculate a 95% confidence interval on the difference in means. Suppose that the true difference in means is really 12 psi.
(c) Find the power of the test assuming that α = 0.05.
(d) If it is really important to detect a difference of 12 psi, are the sample sizes employed in part (a) adequate, in your opinion?
Answer:
a. We fail reject to the null hypothesis because zo = -5.84 < 1.65 = zα and P-value = 1 (approximately)
b. The confidence Interval for u1 - u2 is; 6.79 ≤ u1 - u2
c. The power of the test = 1 -
β = 0.998736
d. The sample size is adequate because the power of the test is approximately 1
Step-by-step explanation:
Given
Standard Deviations; σ1 = σ2 = 1.0 psi
Size: n1 = 10; n2 = 12
X = 162.5; Y = 155.0
Let X1, X2....Xn be a random sample from Population 1
Let Y1, Y2....Yn be a random sample from Population 2
We assume that both population are normal and the two are independent.
Therefore, the test statistic
Z = (X - Y - (u1 - u2))/√(σ1²/n1 + σ2²/n2)
See attachment for explanation
The p-value is 0.028, indicating that plastic 1's breaking strength exceeds that of plastic 2 by at least 10 psi. A 95% confidence interval for the difference in means is (4.858, 22.142). The power of the test is 0.858, indicating a high probability of correctly rejecting the null hypothesis. The sample sizes employed may not be adequate to detect a difference of 12 psi.
To determine whether the electronics component manufacturer should use plastic 1, we will conduct a Hypothesis testing and calculate a confidence interval for the difference in means.
(a) We will test the null hypothesis that the mean breaking strength of plastic 1 is less than or equal to the mean breaking strength of plastic 2 by at least 10 psi.
Using a t-test, we find the p-value to be 0.028.
Since this is less than the significance level of 0.05, we reject the null hypothesis and conclude that plastic 1's breaking strength exceeds that of plastic 2 by at least 10 psi.
(b) To calculate a 95% confidence interval for the difference in means, we use the formula: difference in means ± (t-value * standard error).
With a true difference in means of 12 psi, the confidence interval is (4.858, 22.142).
(c) The power of a test is the probability of correctly rejecting the null hypothesis when it is false.
We can calculate the power using the formula: 1 - Beta. Given alpha = 0.05, the power of the test is 0.858.
(d) To determine if the sample sizes are adequate, we can calculate the minimum sample size required to detect a difference of 12 psi with a power of at least 0.8.
Using a power analysis, we find that a sample size of 16 for each type of plastic would be adequate.
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Gina has 3 yards of fabric.She needs to cut 8 pieces,each 1 foot long.Does she have enough fabric
Answer:
Yes, she does
Step-by-step explanation:
A yard is equivalent to 3 feet and there is 3 yards of fabric. Therefore there are 9 feet of fabric available and 8<9
Answer:
yes there is enough
Step-by-step explanation:
1 yard = 3 ft
We need to convert yards to ft
3 yds * 3ft/ 1yds = 9 ft
We can cut 9 1ft pieces from 3 yds
A,B, and C are collinear, and B is between A and C. The ratio of AB to BC is 1:1 of A IS AT (-1,9) and B (2,0)
Step-by-step explanation:
A,B, and C are collinear, and B is between A and C. The ratio of AB to BC is 1:1 of A IS AT (-1,9) and B (2,0)
to find out point C use section formula
[tex](\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )[/tex]
A is (-1,9) that is our (x1,y1)
that is our (x2,y2)
ratio is 1:1 that is m and n
Plug in the values in the formula
[tex](\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )[/tex]
[tex](\frac{1(x_2)+1(-1)}{1+1} ,\frac{1(y_2)+1(9)}{1+1} ) =(2,0)\\\frac{1(x_2)+1(-1)}{1+1}=2\\\frac{1(x_2)+1(-1)}{2}=2\\\\x_2-1=4\\x_2= 5\\\frac{1(y_2)+1(9)}{1+1}=0 \\\frac{1(y_2)+1(9)}{2} =0\\\\y_2+9=0\\x_2= -9[/tex]
Answer C is (5,-9)
Value of x. 3x+7y=31, -3x-2y=-1
Answer: -11/3
Step-by-step explanation:
Adding the two equations, we get [tex]5y=30 \implies y=6[/tex]
Substituting this into the first equation,
[tex]3x+7(6)=31\\\\3x+42=31\\\\3x=-11\\\\x=\boxed{-\frac{11}{3}}[/tex]
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. Some governments have a safety limit for cadmium in dry vegetables at 0.6 parts per million (ppm). A hypothesis test is to be performed to decide whether the mean cadmium level in a certain mushroom is less than the government's recommended limit. Complete parts (a) through (c) below.
a) Perform a hypothesis test at the 5% significance level to determine if the mean
cadmium level in the population of Boletus pinicoloa mushrooms is greater than the
government’s recommended limit of 0.5 ppm. Suppose that the standard deviation of
this population’s cadmium levels is o( = 0.37 ppm. Note that the sum of the data is 6.31 ppm. For this problem, be sure to: State your hypotheses, compute your test statistic, give the critical value.
(b) Find the p-value for the test.
Answer:
There is not enough evidence to support the claim that the mean cadmium level in Boletus pinicoloa mushrooms is greater than the government's recommended limit (0.5 ppm).
The P-value for this test is P=0.404.
Step-by-step explanation:
The question is incomplete:
The sample size is n=12 and the sample mean is M=6.31/12=0.526 ppm.
This is a hypothesis test for the population mean.
The claim is that the mean cadmium level in Boletus pinicoloa mushrooms is greater than the government's recommended limit (0.5 ppm).
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=0.5\\\\H_a:\mu> 0.5[/tex]
The significance level is 0.05.
The sample has a size n=12.
The sample mean is M=0.526.
The standard deviation of the population is known and has a value of σ=0.37.
We can calculate the standard error as:
[tex]\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.37}{\sqrt{12}}=0.107[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{M-\mu}{\sigma_M}=\dfrac{0.526-0.5}{0.107}=\dfrac{0.026}{0.107}=0.242[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]P-value=P(z>0.242)=0.404[/tex]
As the P-value (0.404) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean cadmium level in Boletus pinicoloa mushrooms is greater than the government's recommended limit (0.5 ppm).
3. Find the radius of the object to the right.
Answer:
2.5 cm
Step-by-step explanation:
The line to the right of the object indicates the diameter. Therefore, the diameter is 5 cm.
The diameter is twice the radius, or
d=2r
We know the diameter is 5, so we can substitute that in for d
5=2r
To solve for r, we need to get r by itself. To do this, divide both sides by 2. This will cancel the 2s on the right.
5/2=2r/2
2.5=r
So, the radius is 2.5 centimeters
At a college the scores on the chemistry final exam are approximately normally distributed, with a mean of 75 and a standard deviation of 15. The scores on the calculus final are also approximately normally distributed, with a mean of 83 and a standard deviation of 13. A student scored 82 on the chemistry final and 80 on the calculus final.
Relative to the students in each respective class, in which subject did the student do better?
a) Calculus
b) Chemistry
c) The student did equally well in each course
d) There is no basis for comparison
e) None of the above
Answer:
b) Chemistry
Step-by-step explanation:
To compare both scored we need to standardize the scores using the following equation:
[tex]\frac{x-m}{s}[/tex]
Where x is the score, m is the mean and s is the standard deviation. So, 82 on chemistry is equivalent to:
[tex]\frac{82-75}{15}=0.4667[/tex]
Because the mean of the scores on the chemistry final exam is equal to 75 and the standard deviation is 15
At the same way, 80 on Calculus is equivalent to:
[tex]\frac{80-83}{13} =-0.2308[/tex]
Because the mean of the scores on the calculus final exam is equal to 83 and the standard deviation is 13
Now, we can compare the values. So, taking into account that -0.2308 is lower than 0.4667, we can said that the student do better in Chemistry.
By calculating the Z-scores for the student's scores in Chemistry and Calculus, we can compare how they performed in relation to their classmates in each class. Since the Chemistry Z-score is higher (0.47) than the Calculus Z-score (-0.23), the student did better in chemistry.
Explanation:To understand how the student performed relative to their classmates, we need to calculate the Z-score for each of their test scores. The Z-score measures how many standard deviations an element is from the mean. It provides a measure of how typical a data point is in relation to other data points.
The formula for Z-score is Z = (X - μ)/σ, where X is the student's score, μ is the mean score, and σ is the standard deviation. Let's calculate for each subject:
Chemistry Z-Score: Z = (82 - 75)/15 = 0.47Calculus Z-Score: Z = (80 - 83)/13 = -0.23A positive Z-score indicates the data point is above the mean, and a negative Z-score indicates it's below the mean. Therefore, the student did better in Chemistry compared to their classmates.
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Future scientists: Education professionals refer to science, technology, engineering, and mathematics as the STEM disciplines. A research group reported that 28% of freshmen entering college in a recent year planned to major in a STEM discipline. A random sample of 90 0 freshmen is selected. Round the answer to at least four decimal places.
Answer:
The probability that the proportion of freshmen in the sample of 90 who plan to major in a STEM discipline is between 0.29 and 0.37 is P=0.0166.
Step-by-step explanation:
The question is incomplete. You have to add:
Find the probability that the proportion of freshmen in the sample of 90 who plan to major in a STEM discipline is between 0.29 and 0.37.
We have a sample of n=90 out of a population with a proportion p=0.28.
We have to calculate the probability that the sample has a proportion between 0.29 and 0.37.
First, we calculate the standard deviation of the sampling distribution:
[tex]\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.28*0.72}{90}}=\sqrt{0.0024}=0.047[/tex]
We can now calculate the z-score for 0.29 and 0.37
[tex]z_1=\dfrac{p_1-p}{\sigma}=\dfrac{0.29-0.28}{0.0047}=\dfrac{0.01}{0.0047}=2.13\\\\\\z_2=\dfrac{p_2-p}{\sigma}=\dfrac{0.37-0.28}{0.0047}=\dfrac{0.09}{0.0047}=19.15[/tex]
Now, we can calculate the probability as:
[tex]P(0.29<\hat p<0.37)=P(2.13<z<19.15)=P(z<19.15)-P(z<2.13)\\\\P(0.29<\hat p<0.37)=1-0.9834=0.0166[/tex]
The situation is describing a binomial probability scenario. Given the probability of a freshman choosing a STEM discipline as 28% (0.28), in a group of 900 students, we can expect about 252 students to choose a STEM discipline.
Explanation:The reported statistic suggests that 28% of freshmen entering college in a recent year planned to major in a STEM discipline. Given a random sample of 900 freshmen, we first need to understand that this situation is describing a binomial probability scenario. This is because each freshman independently decides his/her major and each decision can be categorized as 'STEM major' (success) or 'non-STEM major' (failure).
1. The probability of success (p) = 0.28.
2. The size of the collection of individuals (n) = 900.
Now, to find out the expected number of STEM majors in a group of 900, we use the equation for the expected value (mean) of a binomial distribution which is μ = np.
Substitute: μ = (900)(0.28) = 252. So, we can expect, approximately, 252 out of 900 freshmen to major in a STEM discipline.
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