Answer:
97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].
Step-by-step explanation:
We are given that a random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded;
32, 33, 28, 37, 29, 30, 22, 35, 23, 28, 30, 36.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average number of miles = [tex]\frac{\sum X}{n}[/tex] = 30.25
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 4.71
n = sample of tires = 12
[tex]\mu[/tex] = population average number of miles
Here for constructing a 97% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.
So, 97% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.55 < [tex]t_1_1[/tex] < 2.55) = 0.97 {As the critical value of t at 11 degrees of
freedom are -2.55 & 2.55 with P = 1.5%}
P(-2.55 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.55) = 0.97
P( [tex]-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.97
P( [tex]\bar X-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.97
97% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]30.25-2.55 \times {\frac{4.71}{\sqrt{12} } }[/tex] , [tex]30.25+2.55 \times {\frac{4.71}{\sqrt{12} } }[/tex] ]
= [26.78 miles, 33.72 miles]
Therefore, 97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].
The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $595 . The $14,000 is invested at a 0.5% higher rate of interest than the $7000 . What is the interest rate of each investment?
Answer:
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
Step-by-step explanation:
Let interest rate of $14000 be x%
and Interest rate for $7000 be y %
According to the first condition
14000 * x% - 7000 * y% = 595
multiply by 100
14000x-7000y = 59500
/700
20x-10y=85.................(1)
II condition
x%=y%+0.5%
x=y+0.5
x-y=0.5..................................(2)
solve (1) & (2)
20 x -10 y = 85 .............1
Total value
1 x -1 y = 0.50 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by -10
20.00 x -10.00 y = 85.00
-10.00 x + 10.00 y = -5.00
Add the two equations
10.00 x = 80.00
/ 10.00
x = 8.00
plug value of x in (1)
20.00 x -10.00 y = 85.00
160.00 -10.00 y = 85.00
-10.00 y = 85.00 -160.00
-10.00 y = -75.00
y = 7.50
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
The mean number of hours of part-time work per week for a sample of 317 teenagers is 29. If the margin of error for the population mean with a 95% confidence interval is 2.1, construct a 95% confidence interval for the mean number of hours of part-time work per week for all teenagers.
Answer:
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Step-by-step explanation:
Information given
[tex]\bar X= 29[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex] ME= 2.1[/tex] represent the margin of error
n represent the sample size
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
And this formula is equivalent to:
[tex] \bar X \pm ME[/te]x
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Determine whether the underlined value is a parameter or a statistic. In a national survey of high school students (grades 9 to 12), Modifying 25 % of the students who responded reported that someone had offered, sold, or given them an illegal drug on school property.
Answer:
I am assuming the underlined value is 25%. It is a parameter
Step-by-step explanation:
The value is is a parameter. This is because the parameter is a value that describes the population.
The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.
The underlined 25% value is the value that summarizes the entire population of high school students
Alan has reached 25% of his weekly exercise time goal so far this week. If he has exercised for a total of 42 minutes this week, what is his weekly exercise time goal in minutes
Answer:
His weekly exercise time goal is 168 minutes.
Step-by-step explanation:
This question can be solved using a rule of three.
42 minutes is 25% = 0.25 of the total
x minutes is 100% = 1 of the total.
Then
42 minutes - 0.25
x minutes - 1
[tex]0.25x = 42[/tex]
[tex]x = \frac{42}{0.25}[/tex]
[tex]x = 168[/tex]
His weekly exercise time goal is 168 minutes.
Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously. (a) What is the formula for A(t), the balance after t years? (b) What differential equation is satisfied by A(t), the balance after t years? (c) How much money will be in the account after 5 years? (d) When will the balance reach $3000? (e) How fast is the balance growing when it reaches $3000?
Answer:
a)[tex]A(t)=2000e^{0.085t}[/tex]
b)[tex]A'(t)=170e^{0.085t}[/tex]
c)$3059.1808
d)t=4.77 years
e) The balance growing is $254.99/year
Step-by-step explanation:
We are given that Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously.
Principal = $2000
Rate of interest = 8.5%
a) What is the formula for A(t), the balance after t years?
Formula [tex]A(t)=Pe^{rt}[/tex]
So,[tex]A(t)=2000e^{0.085t}[/tex]
B)What differential equation is satisfied by A(t), the balance after t years?
So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]
[tex]A'(t)=170e^{0.085t}[/tex]
c)How much money will be in the account after 5 years?
Substitute t = 5 in the formula "
[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]
d)When will the balance reach $3000?
Substitute A(t)=3000
So, [tex]3000=2000e^{0.085t}[/tex]
t=4.77
The balance reach $3000 in 4.77 years
e)How fast is the balance growing when it reaches $3000?
Substitute the value of t = 4.77 in derivative formula :
[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]
Hence the balance growing is $254.99/year
A school librarian purchases a novel for her library. The publisher claims that the book is written at a 5th grade reading level, but the librarian suspects that the reading level is higher than that. The librarian selects a random sample of 40 pages and uses a standard readability test to assess the reading level of each page. The mean reading level of these pages is 5.2 with a standard deviation of 0.8. Do these data give convincing evidence at the = 0.05 significance level that the average reading level of this novel is greater than 5?
Answer:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.
Step-by-step explanation:
Information provided
[tex]\bar X=5.2[/tex] represent the sample mean
[tex]s=0.8[/tex] represent the sample standard deviation
[tex]n=40[/tex] sample size
[tex]\mu_o =5[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is higher than 5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.
Please answer this question I give brainliest thank you! Number 14
The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is $600 or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. (a) Which form of the hypotheses should be used to test the manager's claim? H0: - Select your answer - Ha: - Select your answer - The member of the hotel's accounting staff suspects that the total charges for guest bills have Select in recent months. To test the manager’s claim, the staff member will conduct Select test of the population Select . (b) What conclusion is appropriate when H0 cannot be rejected? When H0 cannot be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent months. (c) What conclusion is appropriate when H0 can be rejected? When H0 can be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent m
Answer:
a) Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Step-by-step explanation:
a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:
Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
What statistical test and how many of them will need to be used to explore the difference in average cholesterol within each group before and after the treatment (i.e., how much did cholesterol level change within each group as result of the treatment)?
Answer:
The statistical test to be used is the paired t-test.
Step-by-step explanation:
The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference between these two means.
We use the paired t-test if we have two measurements on the same item, person or thing. We should also use this test if we have two items that are being measured with a unique condition.
For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.
Similarly, in this case a paired t-test would be used to deter whether there was any changes in the cholesterol level within each group as result of the treatment.
Thus, the statistical test to be used is the paired t-test.
Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
The equation of a circle is x2 + y2 = 56.25. Find the radius of the circle?
Answer:
r = 7.5
Step-by-step explanation:
Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!
An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly more than 10 years. In order to test this claim, 9 car transmissions are randomly selected and their useful lifetimes are recorded. The sample mean lifetime is 13.5 years and the sample standard deviation is 3.2 years. Assuming that the useful lifetime of a typical car transmission has a normal distribution, based on these sample result, the correct conclusion at 1% significance level for this testing hypotheses problem is: Group of answer choices
Answer:
Step-by-step explanation:
The question is incomplete. The missing information is the group of answer choices. The group of answer choices are
a) none of the above
b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).
c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).
d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).
e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).
Solution:
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,
H0: µ ≥ 10
For the alternative hypothesis,
µ < 10
This is a left tailed test.
Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.
Since n = 9,
Degrees of freedom, df = n - 1 = 9 - 1 = 8
t = (x - µ)/(s/√n)
Where
x = sample mean = 13.5
µ = population mean = 10
s = samples standard deviation = 3.2
t = (13.5 - 10)/(3.2/√9) = 3.28
Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.
1 - α = 1 - 0.01 = 0.99
The negative critical value is - 2.896
Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.
By using probability value,
We would determine the p value using the t test calculator. It becomes
p = 0.0056
Level of significance = 1%
Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years
The correct option is
a) none of the above
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
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Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
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Which is the solution to the equation 0.5 x + 4.2 = 5.9? Round to the nearest tenth if necessary. 0.9 3.4 5.1 20.2
Answer:
x = 3.4
Step-by-step explanation:
Step 1: Isolate x
0.5x = 1.7
Step 2: Divide both sides by 0.5
x = 3.4
And we have our final answer!
Answer: x=3.4
Step-by-step explanation:
[tex]0.5x+4.2=5.9[/tex]
multiply both sides by 10
[tex]5x+42=59[/tex]
subtract 42 on both sides
[tex]5x=17[/tex]
divide 5 on both sides
[tex]x=\frac{17}{5}[/tex] or
Simplify
x= 3.4
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Answer:
97% Confidence interval = (12.62, 18.98)
Step-by-step explanation:
Complete Question
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Solution
We first compute the sample mean and standard deviation for this sample distribution
Sample mean = (Σx)/N = (158/10) = 15.8
Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 15.8
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 10 - 1 = 9
Significance level for 97% confidence interval
(100% - 97%)/2 = 1.5% = 0.015
t (0.015, 9) = 2.9982 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 3.36
n = sample size = 10
σₓ = (3.36/√10) = 1.0625
97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 15.8 ± (2.9982 × 1.0625)
CI = 15.8 ± 3.1809882246
97% CI = (12.6190117754, 18.9809882246)
97% Confidence interval = (12.62, 18.98)
Hope this Helps!!!
Black walnut trees contain chemicals that inhibit the growth of other plants. In a simple experiment to test whether this is true, you grow several tomato plants in soil with and without decomposing leaves from a black walnut tree. You collect data on plant height as a measure of growth. In this experiment, __________ is the independent variable, __________ is the dependent variable, and __________ is the control.
Answer:
Height of tomato plant is the dependent variable
Presence of walnut leaves in the soil is the independent variable
Tomato plants grown without walnut leaves is the control
Step-by-step explanation:
An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.
A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.
A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
Answer:
Option A is correct.
A uniform distribution.
Step-by-step explanation:
Complete Question
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
A) Uniform Distribution
B) Continuous Distribution
C) Poisson Distribution
D) Relative Frequency Distribution
Solution
A uniform distribution is one in which all the variables have the same probability of occurring.
It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.
And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.
Hope this Helps!!!
Someone flips five coins, but you don’t see the outcome. The person reports that no tails are showing. What is the probability that the person flipped 5 heads?
Answer:
0.03125 = 3.125% probability that the person flipped 5 heads
Step-by-step explanation:
For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. 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.
Five coins:
This means that n = 5.
Fair coin:
Equally as likely to be heads or tails, so p = 0.5.
What is the probability that the person flipped 5 heads?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]
0.03125 = 3.125% probability that the person flipped 5 heads
Polygon ABCD is plotted on a coordinate plane and then rotated 90 clockwise about point C to form polygon A’B’C’D Match each vertex of polygon ABCD to its coordinates.
Answer:
A' - (8,2)
B' - (5,1)
C' - (4,2)
D' - (4,5)
Step-by-step explanation:
See attachment for the missing figure.
We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)
Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then
A'(8,2);
B'(5,1);
C' is the same as C, thus, C'(4,2);
D'(4,5).
The graph of the function f(x) =4 square root x is shown. what is the domain of the function?
Answer:
[0, positive infinity)
Step-by-step explanation:
The domain is all x values a graph inputs. In a square root function, you cannot have negative inputs as it will turn out imaginary numbers. Therefore, your domain is all values of x above and including 0.
Answer: d on Ed
Step-by-step explanation:
Just took the test
Any help would be great
Answer:
2/5
Step-by-step explanation:
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
after allowing 20%discount on the marked price of a computer, 13 %vat was leived on it if its price become rs36160. what amount wsaeveid in the vat
Answer: Rs. 40,000
Step-by-step explanation:
Let say Marked price of the Watch
= M Rs
Discount = 20 %
Discount = (20/100)M = 0.2M Rs
Price after Discount = M - 0.2M =
Rs 0.8M
13 % Value added tax
=> VAT = (13/100) * 0.8M = 0.104M Rs
Value of Watch = 0.8M + 0.104M = 0.904M Rs
0.904M = 36160
=> M = 40000
Marked Price of Watch = Rs 40,000
Consider three consecutive positive integers, such that the sum of the
squares of the two larger integers is 5 more than 40 times the smaller
one. Find the smaller integer.
Answer:
17
Step-by-step explanation:
Let x represent the smaller integer. Then we have ...
(x +1)² +(x +2)² = 40x +5
2x² +6x +5 = 40x +5
x² -17x = 0 . . . . . subtract (40x+5), divide by 2
x(x -17) = 0 . . . . . factor
The solution of interest is x = 17.
The smaller integer is 17.
can someone help me again please im giving 20 points
Answer:
I answered this one for you already I think.
What is the domain of the following set of ordered pairs (-2,-5),(-3,8),(12,6),(8,3),(4,0),(-5,7)
Answer:
domain = {-5, -3, -2, 4, 8, 12}
Step-by-step explanation:
The domain is the set containing the x-coordinates of all ordered pairs.
domain = {-2, -3, 12, 8, 4, -5}
If you'd like, you can put the numbers in ascending order:
domain = {-5, -3, -2, 4, 8, 12}
Please help Solving linear and quadratic equations
Answer: B.
x ≈2.5
Step-by-step explanation:
[tex]-\left(u\right)^{-1}-6=-u+10[/tex]
[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]
[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]
x=2.52...
Answer:
x=2.5
Step-by-step explanation:
) For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.) x = (No Response) For which values of x is f '(x) positive? (Enter your answer using interval notation.) (No Response) For which values of x is f '(x) negative? (Enter your answer using interval notation.) (No Response) What do these values mean? f is (No Response) when f ' > 0 and f is (No Response) when f ' < 0. (b) For which values of x is f ''(x) zero? (Enter your answers as a comma-separated list.)
Answer:
Check below, please
Step-by-step explanation:
Step-by-step explanation:
1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)
When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say
[tex]f'(x)=0\: at \:x=2, and\: x=-2[/tex]
2. For which values of x is f '(x) positive?
Whenever we have
[tex]f'(x)>0[/tex]
then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.
[tex]f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)[/tex]
3. For which values of x is f '(x) negative?
On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So
[tex]f'(x) <0 \: at\: [-2,2][/tex]
4.What do these values mean?
[tex]f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)<0[/tex]
5.(b) For which values of x is f ''(x) zero?
In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly
at x=-4 and x=4
The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.
Required:
a. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength?
b. If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap?
Answer:
a. 0.0668
b. 0.9545
Step-by-step explanation:
We have the following information:
mean (m) = 10000
standard deviation (sd) = 100
(a)
We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:
P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]
P (x> 10150) = P (z> 1.5)
P (x> 10150) = 1 - P (z <1.5)
P (x> 10150) = 1 - 0.9332 (attached table)
P (x> 10150) = 0.0668
Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668
(b)
We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:
P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]
P (9800 <x <10200) = P (-2 <z <2)
P (9800 <x <10200) = P (z <2) - P (z <-2)
P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)
P (9800 <x <10200) = 0.9545
the proportion of pieces that would expect to scrap is 0.9545