To test the claim that the mean difference between the prices of the two models is $10, we can use a t-test for dependent samples.
Explanation:To test the claim that the mean difference between the prices of the two models is $10, we can use a t-test for dependent samples.
The null hypothesis (H0) is that the mean difference is equal to $10, while the alternative hypothesis (Ha) is that the mean difference is not equal to $10.We calculate the sample mean difference and the standard deviation of the differences.We calculate the t-statistic using the formula: t = (sample mean difference - hypothesized mean difference) / (standard deviation of the differences / sqrt(n)), where n is the number of pairs of observations.We compare the t-statistic to the critical value from the t-distribution with n-1 degrees of freedom at a significance level of 0.05.If the absolute value of the t-statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean difference is not equal to $10. Otherwise, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim that the mean difference is not equal to $10.Ice-cream palace has received an order for 3 gallons of ice cream the shop packages its ice cream in 1 quart containers
The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 2.4 cars and a standard deviation of 4. The number of cars running the red light was observed on 100 randomly chosen days and the mean number of cars calculated. Describe the sampling distribution of the sample mean.
Answer:
The sampling distribution of the sample mean is:
[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample means is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the distribution of sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
Let X = number of cars running a red light in a day, at a given intersection.
The information provided is:
[tex]E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100[/tex]
The sample selected is quite large, i.e. n = 100 > 30.
The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.
The mean of the sampling distribution of the sample mean is:
[tex]\mu_{\bar x}=\mu=2.4[/tex]
The standard deviation of the sampling distribution of the sample mean is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40[/tex]
The sampling distribution of the sample mean is:
[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]
Final answer:
The sampling distribution of the sample mean is approximately normally distributed, with a mean of 2.4 cars and a standard deviation of 0.4 cars.
Explanation:
The sampling distribution of the sample mean can be described as follows:
The sampling distribution of the sample mean is approximately normally distributed.The mean of the sampling distribution of the sample mean is equal to the mean of the population, which is 2.4 cars in this case.The standard deviation of the sampling distribution of the sample mean, also known as the standard error, is equal to the standard deviation of the population divided by the square root of the sample size. In this case, the standard deviation of the sampling distribution of the sample mean is 4 / sqrt(100) = 0.4 cars.Confidence interval precision: We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion? Group of answer choices We can select a higher confidence level and increase the sample size. We can select a higher confidence level and decrease the sample size. We can select a lower confidence level and increase the sample size. We can select a lower confidence level and decrease the sample size.
Answer:
We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]
In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.
Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.
Therefore, We can select a lower confidence level and increase the sample size.
The best option that could be used to produce higher precision in our estimates of the population proportion is;
Option C; We can select a lower confidence level and increase the sample size.
Formula for confidence interval is given as;
CI = p^ ± z√(p^(1 - p^)/n)
Where;
p^ is the sample proportion
z is the critical value at given confidence level
n is the sample size
Now, the margin of error from the CI formula is:
MOE = z√(p^(1 - p^)/n)
Now, the lesser the margin of error, the narrower the confidence interval and thus the more precise is the estimate of the population proportion.
Now, looking at the formula for MOE, two things that could change aside the proportion is;
z and n.
Now, the possible values of z are;
At CL of 99%; z = 2.576
At CL of 95%; z = 1.96
At CL of 90%; z = 1.645
We can see that the higher the confidence level, the higher the critical value and Invariably the higher the MOE.
Thus, to have a narrow CI, we need to use a lower value of CL and increase the sample size.
Read more at; https://brainly.com/question/14225622
The moods of U.S. Marines following a month-long training exercise conducted at cold temperature and at high altitudes were assessed. Negative moods, including fatigue and anger, increased substantially during the training and lasted up to three month after the training ended. The scores for 5 of the Marines were 14, 10, 13, 10, 11. The mean mood score was compared to population norms for college men; the population mean anger score for college men is 8.90. a) Test the null hypothesis that the population mean is 8.90 against the alternative that the population mean is greater than 8.90 at α=.05. Show all 6 steps. b) Interpret the results. What did we learn about the Marines’ negative moods?
Answer:
Step-by-step explanation:
Sample mean = (14 + 10 + 13 + 10 + 11)/5 = 11.6
Sample standard deviation,s = √(summation(x - mean)/n
Summation(x - mean) = (14 - 11.6)^2 + (10 - 11.6)^2 + (13 - 11.6)^2 + (10 - 11.6)^2 + (11 - 11.6)^2 = 13.2
s = √(13.2/5) = 1.62
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 8.9
For the alternative hypothesis,
µ > 8.9
it is a right tailed test because of >.
Since the number of samples is 5 and no population standard deviation is given, the distribution is a student's t.
Since n = 5,
Degrees of freedom, df = n - 1 = 5 - 1 = 4
t = (x - µ)/(s/√n)
Where
x = sample mean = 11.6
µ = population mean = 8.9
s = samples standard deviation = 1.62
t = (11.6 - 8.9)/(1.62/√5) = 3.73
We would determine the p value using the t test calculator. It becomes
p = 0.01
Since alpha, 0.05 > than the p value, 0.01, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that mean anger score of the marines is greater than that of college men.
Therefore, the marines's negative moods increased and it is higher than that of college men.
Min's mother spent $3.96 on ground coffee that costs $0.45 per ounce. How many ounces of ground coffee did she buy?
Answer:
8.8 ounces
Step-by-step explanation:
3.96/0.45=8.8
3) A state policeman has a theory that people who drive red cars are more likely to drive too fast. On his day off, he borrows one of the department’s radar guns, parks his car in a rest area, and measures the proportion of red cars and non-red cars that are driving too fast (he decided ahead of time to define "driving too fast" as exceeding the speed limit by more than 5 miles per hour). To produce a random sample, he rolls a die and only includes a car in his sample if he rolls a 5 or a 6. He finds that 18 out of 28 red cars are driving too fast and 75 of 205 other cars are driving too fast. Is this convincing evidence that people who drive red cars are more likely to drive too fast, as the policeman has defined it?
Answer:
Yes, it is convincing evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.
Step-by-step explanation:
From the, we wish to first test;
H0; P_r - P_o = 0
And; H0; P_r - P_o > 0
Where; P_r and P_o are the proportion of red cars and other cars, respectively, who are driving too fast.
We will use a significance level of a = 0.05.
Thus;
The procedure is a two-sample z-test for the difference of proportions.
For, Random Conditions: The policemen chose cars randomly by rolling a die.
10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply.
Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10.
So, P_r = 18/28 = 0.64
P_o = 75/205 = 0.37
P_c = (18 + 75)/(28 + 205) = 0.4
Thus:
z = [(0.64 - 0.37) - 0]/√[[(0.4 x 0.6)/28] + [(0.4 x 0.6)/205]]
z = 2.73
From the one tailed z-score calculator online, I got P value = 0.003167
Thus, the P-value of 0.0032 is less than a = 0.05, so we reject H0. We
have sufficient evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zero determinant: SL2(R) is the subset of GL2(R) with determinant 1. We know that multiplication is a binary operation on M2(R); show that it is an induced operation on the other 2 sets (You may freely use known facts from Math 3A for this).
Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0[/tex].
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1[/tex].
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
3 to the power of 5 and (3) to the power of 5
Answer:
243
Step-by-step explanation:
3x3=9
9x3=27
27x3=81
81x3=243
Evaluate the expression. 23 [ 14 + 4(36 ÷ 12)]
Answer:
598
Step-by-step explanation:
Answer:
598
Step-by-step explanation:
[tex]23[14 + 4(36 \div 12)] \\ = 23[14 + 4(3)] \\ = 23[14 + 12] \\ = 23 \times 26 \\ = 598 \\ [/tex]
A chord is 16 units from the center of a circle. The radius of the circle is 20 units. What is the length of the chord?
How is the measure of a central angle and the corresponding chord related to the measure of the arc intercepted by the chord?
Answer:
Length of the chord: 24 units
Angles are equal
Step-by-step explanation:
Drop a Perpendicular from the centre onto the chord. It will divide the chord into two equal parts, d units each
d² + 16² = r²
d² = 20² - 16²
d² = 144
d = 12
Chord = 2d = 24 units
Measure of central angle and the corresponding chord related to the measure of the arc intercepted by the chord are the same, they're equal
Answer:
The entire chord length is 12*2 = 24
The degree measure of a minor arc is equal to the measure of the central angle that intercepts it.
Step-by-step explanation:
We can make a right triangle to solve for 1/2 of the chord length. The hypotenuse is 20 and one of the legs is 16
a^2+b^2 = c^2
16^2 + b^2 = 20^2
256 +b^2 = 400
Subtract 256 from each side
b^2 = 400-256
b^2 =144
Take the square root of each side
b = 12
That means 1/2 of the chord length is 12
The entire chord length is 12*2 = 24
6. A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 568 consumers who recognize the Dull Computer Company name? Find the mean and standard deviation to answer this question. Must show work and explain why or why not.
Given Information:
Population = n = 800
Probability = p = 68% = 0.68
Answer:
We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.
Step-by-step explanation:
Let us first find out the mean and standard deviation.
mean = μ = np
μ = 800*0.68
μ = 544
standard deviation = σ = √np(1-p)
σ = √800*0.68(1-0.68)
σ = 13.2
we know that 68% of data fall within 2 standard deviations from the mean
μ ± 2σ = 544-2*13.2, 544+2*13.2
μ ± 2σ = 544-26.4 , 544+26.4
μ ± 2σ = 517.6, 570.4
μ ± 2σ = 518, 570
We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.
Select all the expression that
Answer:
2nd, 4th and last one are correct answers
Step-by-step explanation:
[tex] {16}^{ \frac{5}{2} } \\ = ( {4}^{2} )^{ \frac{5}{2} } \\ = {4}^{{2} \times \frac{5}{2} } \\ = {4}^{5} \\ [/tex]
Options: 2nd, 4th and last one are correct answers
A big ship drops its anchor.
E represents the anchor's elevation relative to the water's surface (in meters) as a function of time t (in seconds).
E=−2.4t+75
How far does the anchor drop every 5 seconds?
The anchor drops 63 meters in the first 5 seconds.
The given function E(t) = -2.4t + 75 represents the elevation of the ship's anchor relative to the water's surface at any given time t. To find how far the anchor drops every 5 seconds, substitute t = 5 into the function:
E(5) = -2.4(5) + 75
E(5) = -12 + 75
E(5) = 63
Therefore, after 5 seconds, the anchor has dropped 63 meters relative to the water's surface. This indicates the change in elevation during this time period. The negative coefficient of t in the function implies a downward motion, and the constant term (75) represents the initial height of the anchor above the water. So, the anchor drops 63 meters in the first 5 seconds.
Use Euler’s formula for exp(ix) and exp(-ix) to write cos(x) as a combination of exp(ix) and exp(-ix)
Answer = (cos(x) = (exp(ix)+exp(-ix))/2)
For real a and b, use the previous answer to find write both cos(a+b) and cos(a)cos(b) in terms of exp. Throughout the rest you will probably use exp(x+y)=exp(x)exp(y).
Answer:
[tex]cos(a+b)=\frac{e^{i(a-b)}+e^{i(-a+b)}}{2}[/tex]
Step-by-step explanation:
[tex]cos(x)=\frac{e^{ix}+e^{-ix}}{2}[/tex]
[tex]cos(a+b)[/tex]
We need to expand cos(a+b) using the cos addition formula.
[tex]cos(a+b)=cos(a)cos(b)-sin(a)sin(b)[/tex]
We know that we also need to use Euler's formula for sin, which is:
[tex]sin(x)=\frac{e^{ix}-e^{-ix}}{2}[/tex] (you can get this from a similar way of getting the first result, of simply just expanding [tex]e^{ix}=cosx+isinx[/tex] and seeing the necessary result)
We can now substitute our cos's and sin's for e's
[tex]cos(a+b)=(\frac{e^{ia}+e^{-ia}}{2})(\frac{e^{ib}+e^{-ib}}{2})-(\frac{e^{ia}-e^{-ia}}{2})(\frac{e^{ib}-e^{-ib}}{2})[/tex]
Now lets multiply out both of our terms, I'm using the exponent multiplication identity here ([tex]e^{x+y}=e^xe^y[/tex])
[tex]cos(a+b)=\frac{e^{i(a+b)} + e^{i(a-b)}+e^{i(-a+b)} + e^{i(-a-b)}}{4}-\frac{e^{i(a+b)} - e^{i(a-b)}-e^{i(-a+b)}+e^{i(-a-b)}}{4}[/tex]
Now we can subtract these two terms.
[tex]cos(a+b)=\frac{2e^{i(a-b)}+2e^{i(-a+b)}}{4}[/tex]
This is starting to look a lot tidier, let's cancel the 2
[tex]cos(a+b)=\frac{e^{i(a-b)}+e^{i(-a+b)}}{2}[/tex]
Using Euler's formula, we can write cos(x) as the average of exp(ix) and exp(-ix). Further, we demonstrated how to express cos(a+b) and cos(a)cos(b) in terms of exponential functions, utilizing the properties of Euler's formula and complex exponentials.
Using Euler's formula, exp(ix) = cos(x) + i sin(x) and exp(-ix) = cos(x) - i sin(x), we can represent cos(x) as a combination of exp(ix) and exp(-ix). By adding these two equations, we eliminate the sin(x) terms due to their opposite signs, leading us to the formula for cos(x):
cos(x) = (exp(ix) + exp(-ix)) / 2
To express cos(a+b), use the expansion:
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
Using Euler's formula, this expands to:
cos(a+b) = [exp(ia) + exp(-ia)]/2 * [exp(ib) + exp(-ib)]/2 - [exp(ia) - exp(-ia)]/2i * [exp(ib) - exp(-ib)]/2i
Similarly, to express cos(a)cos(b), we again use the representation of cos(x) in terms of exp:
cos(a)cos(b) = [exp(ia) + exp(-ia)]/2 * [exp(ib) + exp(-ib)]/2
"A movie data base claims that the average length of movies is 117 minutes. A researcher collected a random sample of 160 movies released during 2010–2015. The mean length of those movies is 118.44 minutes and the standard deviation is 8.82. The researcher wonders if the actual mean length of movies released during 2010-2015 is more than the data base value and wants to carry out a hypothesis test. What are the null and alternative hypothesis?"
Answer:
We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 117[/tex]
Alternative hypothesis:[tex]\mu > 117[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=118.44[/tex] represent the sample mean
[tex]s=8.82[/tex] represent the sample standard deviation
[tex]n=160[/tex] sample size
[tex]\mu_o =117[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 117[/tex]
Alternative hypothesis:[tex]\mu > 117[/tex]
Ximena spent 72 of the day
shopping at the mall. She spent
74 of this time trying on jeans.
What fraction of the day did
Charlotte spend trying on jeans?
Answer:
Step-by-step explanation:
1/8
1/2 multiply 1/4
The amount in milligrams of a drug in the body t hours after taking a pill is given by A(t) = 25(0.85)t a. What is the initial dose given? b. What percent of the drug leaves the body each hour? c. What is the amount of drug left after 10 hours? (Write answer using function notation)
Answer:
(a)25 Milligrams
(b)15%
(c)[tex]A(10) = 25(0.85)^{10}[/tex]
Step-by-step explanation:
The amount in milligrams of a drug in the body t hours after taking a pill is given by the model:
[tex]A(t) = 25(0.85)^t[/tex]
(a)Comparing this with the exponential decay model, [tex]A(t)=A_0(\frac{1}{2})^{\frac{t}{t_{1/2}} }[/tex], the initial dose given is 25 milligrams.
(b)From the model,
[tex]A(t) = 25(0.85)^t\\A(t) = 25(1-0.15)^t[/tex]
We can also use this method:
[tex]r = a - 1 = 0.85 - 1 = -0.15=-15\%[/tex]
We can see that for every hour, 15% of the drug leaves the body.
(c)After 10 hours
When t=10
[tex]A(10) = 25(0.85)^{10}[/tex]
The amount of drug left after 10 hours is given above in function notation.
a. The initial dose given is 25 milligrams. b. 85(0.85)^t percent of the drug leaves the body each hour. c. The amount of drug left after 10 hours is 0.2147 milligrams.
Explanation:a. The initial dose given can be found by substituting t = 0 into the function A(t) = 25(0.85)^t. This gives A(0) = 25(0.85)^0 = 25(1) = 25 milligrams.
b. To find the percent of the drug that leaves the body each hour, we need to find the rate of change of A(t) with respect to time. Taking the derivative of A(t) gives dA/dt = 25(0.85)^t * ln(0.85) = 21.25(0.85)^t. This represents the rate of change of A(t) with respect to time. To find the percent, we can divide this rate by the initial dose and multiply by 100: (21.25(0.85)^t / 25) * 100 = 85(0.85)^t percent.
c. To find the amount of drug left after 10 hours, we substitute t = 10 into the function A(t) = 25(0.85)^t: A(10) = 25(0.85)^10 = 25(0.0859) = 0.2147 milligrams.
Learn more about Drug dosage and elimination here:https://brainly.com/question/8728922
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Which of these statements is best? The errors in a regression model are assumed to have an increasing mean. The regression model assumes the error terms are dependent. The errors in a regression model are assumed to have zero variance. The regression model assumes the errors are normally distributed.
Answer:
[tex] \epsilon = Y -X\beta[/tex]
And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:
[tex] Cov (\epsilon) = \sigma^2 I[/tex]
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Step-by-step explanation:
Assuming that we have n observations from a dependent variable Y , given by [tex] Y_1, Y_2,....,Y_n[/tex]
And for each observation of Y we have an independent variable X, given by [tex] X_1, X_2,...,X_n[/tex]
We can write a linear model on this way:
[tex] Y = X \beta +\epsilon [/tex]
Where [tex]\epsilon_{nx1}[/tex] i a matrix for the error random variables, and for this case we can find the error ter like this:
[tex] \epsilon = Y -X\beta[/tex]
And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:
[tex] Cov (\epsilon) = \sigma^2 I[/tex]
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Final answer:
The best statement is that the regression model assumes the errors are normally distributed. In regression analysis, it is essential that the errors are independent, normally distributed, and have constant variance, which supports the validity of the model's predictions.
Explanation:
The correct statement among the provided options is that the regression model assumes the errors are normally distributed. This is a fundamental assumption of linear regression analysis, where it's assumed that the residuals or errors of the regression model are randomly distributed about an average of zero. These error terms must be independent, normal, and have constant variance (homoscedasticity) across all levels of the independent variables.
According to the theoretical foundation of regression, it is not assumed that errors have an increasing mean, nor that they have zero variance, as some diversity in errors is expected. Additionally, the assumption that errors are indeed dependent would violate the principles of ordinary least squares (OLS) regression, making the model invalid.
Normality, independence, and equal variance are key premises in regression analysis to ensure the validity of the model's inferences. Indeterminate errors that affect the dependent variable 'y' are assumed to be normally distributed and independent of the independent variable 'x'. This maintains the integrity of the regression model.
Does the sum of 1/5 + 1/5 equals 2/5 make sense in situations
Answer:
It equals 2/5.
Step-by-step explanation:
Basically you ignore the 5 because it stays the same. All you have to add are the numerators. 1+1=2. So it would be 2/5.
A scenario:
2 pizzas cut into 5th's. Everyone ate 4 in each and kept one slice. When you add those 2 together you would be 2 slices.
The speed S of blood that is r centimeters from the center of an artery is given below, where C is a constant, R is a radius of the artery, and S is measured in centimeters per second. Suppose a drug is administered and the artery begins to dilate at a rate of dR/dt. At a constant distance r, find the rate at which s changes with respect to t for C = 1.32 times 10^5, R = 1.3 times 10^-2, and dR/dt = 1.0 times 10^-5. (Round your answer to 4 decimal places.) S = C(R^2 - r^2) dS/dt =
Answer:
dS/dt ≈ 0.0343
Step-by-step explanation:
We are given;
C = 1.32 x 10^(5)
R = 1.3 x 10^(-2)
dR/dt = 1.0 x 10^(-5)
The function is: S = C(R² - r²)
We want to find dS/dt when r is constant.
Thus, let's differentiate since we have dR/dt;
dS/dR = 2CR
So, dS = 2CR.dR
Let's accommodate dt. Thus, divide both sides by dt to obtain;
dS/dt = 2CR•dR/dt
Plugging in the relevant values to get;
dS/dt = 2(1.32 x 10^(5))x 1.3 x 10^(-2) x 1.0 x 10^(-5)
dS/dt = 3.432 x 10^(-2)
dS/dt ≈ 0.0343
What is the value for x 2x+3=x-4
Answer:
x = -7
Step-by-step explanation:
2x + 3 = x - 4
→ Minus x from both sides to isolate -4
x + 3 = -4
→ Minus 3 from both sides to isolate x and henceforth find the value of x
x = -7
help need full solution i &ii
Step-by-step explanation:
[tex]1)\:v = 2 {e}^{3t} + 5 {e}^{ - 3t} \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ \frac{dv}{dt} = \frac{1}{dt} (2 {e}^{3t} + 5 {e}^{ - 3t} ) \\ = 2 \times \frac{1}{dt} {e}^{3t} + 5 \times \frac{1}{dt} {e}^{ - 3t} \\ \\ = 2 \times {e}^{3t} \times 3 + 5 \times {e}^{ - 3t} \times ( - 3) \\ = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \frac{dv}{dt} = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3 \times 1} - 15 {e}^ { - 3 \times 1} \\ \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3} - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg(\frac{dv}{dt} \bigg) _{t=1} = \bigg(\frac{6 {e}^{6} - 15}{ {e}^{3} } \bigg) \: m {s}^{ - 2} }}} \\ \\ 2) \: let \:s \: be \: the \: total \: distance \: travelled \\ \therefore \: s = v \times t \\ \therefore \: s= (2 {e}^{3t} + 5 {e}^{ - 3t}) \times t \\ \therefore \: (s)_{t=2} = (2 {e}^{3 \times 2} + 5 {e}^{ - 3 \times 2}) \times 2 \\ \therefore \: (s)_{t=2} = (2 {e}^{6} + 5 {e}^{ - 6}) \times 2 \\ \therefore \: (s)_{t=2} = 4 {e}^{6} + 10{e}^{ - 6} \\ \red{ \boxed{ \bold{\therefore \: (s)_{t=2} = \bigg(\frac{4 {e}^{12} + 10}{{e}^{ 6}} \: \bigg)m}}}\\ [/tex]
Determine the value of so that the area under the standard normal curve a. in the right tail is Round your answer to two decimal places. b. in the left tail is Round your answer to two decimal places. c. in the left tail is Round your answer to two decimal places. d. in the right tail is Round your answer to two decimal places. Click if you would like to Show Work for this ques
Answer:
a) 2.81 b)-2.33 c) -2.88 d)3.09
Step-by-step explanation:
The complete question is:
Determine the value of z so that the area under the standard normal curve
a. in the right tail is 0.025 Round your answer to two decimal places.
b. in the left tail is 0.01 Round your answer to two decimal places.
c. in the left tail is 0.002 Round your answer to two decimal places.
d. in the right tail is 0.01 Round your answer to two decimal places.
a)P( Z> ???)= 0.0025
P(Z> ???)= 1-P(Z<???)
P(Z<???)= 1-0.0025
P(Z<??)= 0.9975
From Z distribution table,
Z = 2.81
b) P(Z<???)= 0.01
From Z distribution table
Z= -2.33
c) P(Z< ??? ) = 0.002
From Z distribution table
Z= -2.88
d) P( Z> ???)= 0.001
P(Z> ???)= 1-P(Z<???)
P(Z<???)= 1-0.001
P(Z<??)= 0.999
From Z distribution table,
Z=3.09
implify the product. 2p(–3p2 + 4p – 5)
Answer:
=2p(-2p-5)
Step-by-step explanation:
2p(–3p2 + 4p – 5)
=2p(4p-6p-5)
=2p(-2p-5)
The "cold start ignition time" of an automobile engine is being investigated by a gasoline manufacturer. The following times, in seconds, were obtained for a test vehicle. 0.17, 1.94, 2.62, 2.35, 3.05, 3.15, 2.53, 4.81 and 1.92 Write answers as decimal values (i.e. 0.12) and round to two decimal places (if necessary)
Answer:
1. Sample Mean = 2.50
2. Sample Variance = 1.35
3. Standard Deviation = 1.16
Step-by-step explanation:
Note: As this question is not complete, similar can be found on internet where following are asked to calculate and here I will calculate these as well.
1. Sample Mean.
2. Sample Variance.
3. Sample Standard Deviation.
So, for sample mean to calculate from the given data points. We need to apply to following formula to calculate sample mean.
Sample Mean = (X1 + X2 + .... + XN) divided by Total number of data points.
Sample Mean = (0.17 + 1.94 + 2.62 + 2.35 + 3.05 + 3.15 + 2.53 + 4.81 + 1.92) divided by (9).
Sample Mean = 2.50
Likewise, from the answer of sample mean we can calculate sample variance by following steps.
Solution:
1. Subtract the obtained mean from each of the data point given.
(0.17 - 2.50) = -2.33
(1.94 - 2.50) = -0.56
(2.62 - 2.50) = 0.12
(2.35 - 2.50) = -0.15
(3.05 - 2.50) = 0.55
(3.15 - 2.50) = 0.65
(2.53 - 2.50) = 0.03
(4.81 - 2.50) = 2.31
(1.92 - 2.50) = -0.58
2. Square each of the differences obtained in step 1.
[tex]-2.33^{2}[/tex] = 5.43
[tex]-0.56^{2}[/tex] = 0.31
[tex]0.12^{2}[/tex] = 0.01
[tex]-0.15^{2}[/tex] = 0.02
[tex]0.55^{2}[/tex] = 0.30
[tex]0.65^{2}[/tex] = 0.42
[tex]0.03^{2}[/tex] = 0.01
[tex]2.31^{2}[/tex] = 5.34
[tex]-0.58^{2}[/tex] = 0.34
3. Sum all of these squares.
(5.43 + 0.31 + 0.01 + 0.02 + 0.30 + 0.42 + 0.01 + 5.34 + 0.34) = 12.18
4. Divide 12.18 by (n-1), where n = 9.
Sample Variance = 12.18/9 = 1.35
Now, by using sample variance, we can calculate standard deviation with easy simple steps.
So, in order to find standard deviation, we just need to take the square root of sample variance that we have already calculated above.
Standard Deviation = [tex]\sqrt{1.35}[/tex] = 1.16
Good Luck!
Please help me! I need this to complete my math hw
Answer:
1
Step-by-step explanation:
You "complete the square" by adding the square of half the x-term coefficient. Here, that is ...
((-2)/2)² = 1 . . . . value added to complete the square
If you want to keep 0 on the right, you must also subtract this value:
x² -2x -36 = 0
x² -2x +1 -36 -1 = 0 . . . . . . add and subtract 1 on the left
(x -1)² -37 = 0 . . . . . . . . . . . written as a square
Find the surface area of a triangular prism with measurements 8 cm 6 cm 5 cm and 3 cm
Answer:
86.96
Step-by-step explanation:
not sure what is height or base.
Answer:
86.96
Step-by-step explanation:
Which of the following is the equation of a line that passes through (-2, -1) and (-4, -3)?
Answer:
y = x+1
Step-by-step explanation:
You should use the point-slope form of y-y1 = m(x-x1) to solve this.
First you use the slope formula to find your slope. (y2-y1)/(x2-x1)
(-1 - -3)/(-2 - -4) = 2/2 = 1
With the slope of 1 found, just plug a point into the point-slope form.
y-(-1) = (1)(x-(-2)) ->
y+1 = x+2 ->
y = x+1
y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)
What is Slope of Line?The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.
The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.
The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is
m=y₂-y₁/x₂-x₁
The slope of line passing through (-2, -1) and (-4, -3)
m = -3+1/-4+2
=-2/-2= 1
Now let us find y intercept
-1=1(-2)+b
-1=-2+b
b=1
Hence, y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)
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The Price of kiwis can be deterrent by the equation P=1.15n where p is the price and n is the number of kiwis. What is the constant of proportionality (unit rate)?
Answer:
1.15
Step-by-step explanation:
The required constant of proportionality of equation P = 1.15 n is k = 1.15.
What is an equation?An equation is a combination of different variables, in which two mathematical expressions are equal to each other.
Given that,
The equation for the price of the kiwis is,
P = 1.15 n
Here, P, shows the price and n shows the number of kiwis.
The ratio that establishes a proportionate link between any two given values is referred to as the constant of proportionality.
Let a proportionality equation is,
y = k x,
k is the constant of proportionality here,
To find the constant of proportionality,
Compare the given equation P = 1.15 n with standard proportionality equation,
k = 1.15
The constant of proportionality is 1.15 unit.
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The mean weight of an adult is 6060 kilograms with a variance of 100100. If 118118 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 0.80.8 kilograms? Round your answer to four decimal places.
Answer:
The probability that the sample mean would differ from the population mean by greater than 0.8 kg is P=0.3843.
Step-by-step explanation:
We have a population with mean 60 kg and a variance of 100 kg.
We take a sample of n=118 individuals and we want to calculate the probability that the sample mean will differ more than 0.8 from the population mean.
This can be calculated using the properties of the sampling distribution, and calculating the z-score taking into account the sample size.
The sampling distribution mean is equal to the population mean.
[tex]\mu_s=\mu=60[/tex]
The standard deviation of the sampling distribution is equal to:
[tex]\sigma_s=\sigma/\sqrt{n}=\sqrt{100}/\sqrt{118}=10/10.86=0.92[/tex]
We have to calculate the probability P(|Xs|>0.8). The z-scores for this can be calculated as:
[tex]z=(X-\mu_s)/\sigma_s=\pm0.8/0.92=\pm0.87[/tex]
Then, we have:
[tex]P(|X_s|>0.8)=P(|z|>0.87)=2*(P(z>0.87)=2*0.19215=0.3843[/tex]