Answer:
a) -0.842
b) $0
c) 11,664 cases
Step-by-step explanation:
a)
The price elasticity of demand E at a given point [tex]\large (p_1,q_1)[/tex] is defined as
[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)[/tex]
and in this case, it would measure the possible response of tissues demand due to small changes in its price when the price is at [tex]\large p_1[/tex]
When the price is set at $32 the demand is
[tex]\large q=(108-32)^2=5,776 [/tex]
cases of tissues, so
[tex]\large (p_1,q_1)=(32,5776)\Rightarrow \frac{p_1}{q_1}=\frac{32}{5776}=0.00554[/tex]
Also, we have
[tex]\large \frac{\text{d}q}{\text{d}p}=-2(108-p)\Rightarrow \frac{\text{d}q}{\text{d}p}(32)=-2(108-32)=-152[/tex]
hence
[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)=0.00554(-152)=-0.842[/tex]
That would mean the demand is going down about 0.842% per 1% increase in price at that price level.
b)
When the price is $108 the demand is 0, so the price should always be less than $108.
On the other hand, the parabola
[tex]\large q=(108-32)^2=5,776 [/tex] is strictly decreasing between 0 and 108, that means the maximum demand would be when the price is 0.
c)
When the price is 0 the demand is
[tex]\large (108)^2=11,664[/tex] cases
The elasticity for p = 32 is E = -0.84.
The maximum revenue is obtained when p = 32, and the demand for that price is q = 5,184.
How to determine the price elasticity?
It is given by:
[tex]E = \frac{p}{q} *\frac{dq}{dp} (p)[/tex]
if p = $32, then:
q(32) = (108 − 32)^2 = 5,776
and:
dq/dp = -2*(108 - p)
Evaluating that in p = 32 we get:
-2*(108 - 32) = -152
Then the elasticity is:
E = (32/5,776)*-152 = -0.84.
How to maximize the revenue?The revenue is equal to the demand times the price, so we get:
R = p*q = p*(108 - p)^2 = p*(p^2 - 216p + 11,664)
To maximize this we need to find the zeros of the derivation, we have:
R' = 3p^2 - 2*216*p + 11,664
The zeros of that equation are given by:
[tex]p = \frac{432 \pm \sqrt{(-432)^2 - 4*3*11,664} }{2*3} \\\\p = (432 \pm 216)/6[/tex]
Notice that we need to use the smaller of these values, so the demand never becomes zero, then we use:
p = (432 - 216)/6 = 36
(the other root gives p = 108, so that is a minimum).
c) For this price, the number of cases demanded is:
q = (108 - 36)^2 = 5,184
If you want to learn more about economy, you can read:
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Include ALL the steps (statements/reasons) by writing a paragraph proof.
Answer:
The proof is given below.
Step-by-step explanation:
IMPORTANT: angle bisector theoremthis theorem says that, if a point is on the angular bisector of an angle, then it is equidistant from the sides of the angle.
here, since X lies on the angular bisector of angle MBC and BCN , it will be equidistant from the sides BM and CN ( by using above theorem) .since the line BM also passes through A and the line CN also passes through A, we can say that X is equidistant from the sides AM and AN also.converse of angle bisector theorem : if a point is equidistant from the sides of an angle, then it lies on the angular bisector of that angle.by using this converse of angle bisector theorem, we can say that X lies on the angular bisector of angle A.hence, it is proved that X is on the bisector of ANGLE A.(the proof of angle bisector theorem can be explained, but it is difficult to type the whole thing. so watch this video for the proof of this theorem : https://youtu.be/6GS4lS4btNI )
A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. All he knowsis that, for a large number of tires tested, the mean mileage was 25,000miles, and the standard deviation was 4000 miles. What interval wouldyou suggest?
The suggested mileage interval, excluding no more than 10% of the mileage on tires, is approximately 18,420 to 31,580 miles.
To determine the mileage interval that excludes no more than 10% of the mileage on tires, we can use the standard normal distribution and the properties of the normal curve. The mileage data is normally distributed with a mean [tex](\(\mu\))[/tex] of 25,000 miles and a standard deviation [tex](\(\sigma\))[/tex] of 4,000 miles.
To find the interval, we can use the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
To exclude no more than 10% of the mileage, we need to find the Z-score corresponding to the 5th percentile on each side of the mean, as 10% is split between the lower and upper tails of the distribution.
Using a standard normal distribution table or a calculator, the Z-score for the 5th percentile is approximately -1.645 (negative due to being in the lower tail).
Now, we can use the Z-score formula to find the values of [tex]\(X\)[/tex] (mileage) corresponding to these Z-scores:
[tex]\[ X_{\text{lower}} = \mu + Z \times \sigma \][/tex]
[tex]\[ X_{\text{upper}} = \mu - Z \times \sigma \][/tex]
Substitute the values:
[tex]\[ X_{\text{lower}} = 25,000 - (-1.645) \times 4,000 \][/tex]
[tex]\[ X_{\text{upper}} = 25,000 + (-1.645) \times 4,000 \][/tex]
Calculating these values:
[tex]\[ X_{\text{lower}} \approx 31,580 \][/tex]
[tex]\[ X_{\text{upper}} \approx 18,420 \][/tex]
Therefore, the suggested mileage interval is approximately 18,420 miles to 31,580 miles to exclude no more than 10% of the mileage on the tires.
A trucking firm suspects that the mean life of a certain tire. it uses is less than 33,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires in gets a mean lifetime of 32, 450 miles with a standard deviation of 1200 miles. At α = 0.05, test the trucking firms claim.
a. State Hypothesis and Identify Claim.
b. Identify level of significance.
c. Choose correct probability distribution, locate critical values.identify rejection region.
d. Calculate test statistic.
e. Make decision
f. Write conclusion.
SHOW ALL YOUR WORK
Answer:
We accept the alternate hypothesis. We conclude that the mean lifetime of tires is is less than 33,000 miles.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 33,000 miles
Sample mean, [tex]\bar{x}[/tex] = 32, 450 miles
Sample size, n = 18
Alpha, α = 0.05
Sample standard deviation, s = 1200 miles
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 33000\text{ miles}\\H_A: \mu < 33000\text{ miles}[/tex]
b) Level of significance:
[tex]\alpha = 0.05[/tex]
c) We use One-tailed t test to perform this hypothesis.
d) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{32450 - 33000}{\frac{1200}{\sqrt{18}} } = -1.9445[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom } = -1.7396[/tex]
Rejection area:
[tex]t < -1.7396[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
e) We fail to accept the null hypothesis and reject it as the calculated value of t lies in the rejection area.
f) We accept the alternate hypothesis. We conclude that the mean lifetime of tires is is less than 33,000 miles.
????⃗ (x,y)=(3x−4y)????⃗ +2x????⃗ F→(x,y)=(3x−4y)i→+2xj→ and ????C is the counter-clockwise oriented sector of a circle centered at the origin with radius 44 and central angle ????/3π/3. Use Green's theorem to calculate the circulation of ????⃗ F→ around ????C.
The circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex] is[tex]\( 3\pi \)[/tex], computed using Green's theorem by transforming the line integral into a double integral over the enclosed region.
To calculate the circulation of [tex]\( \vec{F} \)[/tex] around C, we first need to compute the line integral of [tex]\( \vec{F} \)[/tex] along C. Using Green's theorem, we can rewrite this line integral as a double integral over the region enclosed by C .
Given [tex]\( \vec{F}(x, y) = (3x - 4y)\vec{i} + 2x\vec{j} \)[/tex], we can compute the partial derivatives [tex]\( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)[/tex], where[tex]\( P = 3x - 4y \) and \( Q = 2x \).[/tex]
[tex]\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - (-4) = 6\][/tex]
Since the region enclosed by C is the counter-clockwise oriented sector of a circle with radius 2 and central angle [tex]\( \frac{\pi}{4} \)[/tex], we can parameterize the curve [tex]\( C \)[/tex] as [tex]\( x(t) = 2\cos(t) \) and \( y(t) = 2\sin(t) \)[/tex] where [tex]\( 0 \leq t \leq \frac{\pi}{4} \).[/tex]
Now, applying Green's theorem, we have:
[tex]\[\text{Circulation} = \iint_D (6) \, dA\][/tex]
Using polar coordinates, the double integral becomes:
[tex]\[\begin{aligned}\text{Circulation} &= \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} (6) \cdot r \, dr \, dt \\&= \int_{0}^{\frac{\pi}{4}} \left[3r^2\right]_{0}^{2} \, dt \\&= \int_{0}^{\frac{\pi}{4}} 3(2)^2 \, dt \\&= \int_{0}^{\frac{\pi}{4}} 12 \, dt \\&= [12t]_{0}^{\frac{\pi}{4}} \\&= 12\left(\frac{\pi}{4}\right) - 12(0) \\&= 3\pi\end{aligned}\][/tex]
Therefore, the circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex] is [tex]\( 3\pi \).[/tex]
The question probable maybe:
Given in the attachment
An alternating current E(t)=120sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L=0.37 henrys, a resistor of R=7ohms and a capacitor of capcitance C=0.037 farads.What is the amplitude of the current I?
Answer:
14.488 amperes
Step-by-step explanation:
The amplitude I of the current is given by
[tex]\large I=\displaystyle\frac{E_m}{Z}[/tex]
where
[tex]\large E_m[/tex] = amplitude of the energy source E(t).
Z = Total impedance.
The amplitude of the energy source is 120, the maximum value of E(t)
The total impedance is given by
[tex]\large Z=\sqrt{R^2+(X_L-X_C)^2}[/tex]
where
R= Resistance
L = Inductance
C = Capacitance
w = Angular frequency
[tex]\large X_L=wL[/tex] = inductive reactance
[tex]\large X_C=\displaystyle\frac{1}{wC}[/tex] = capacitive reactance
As E(t) = 120sin(12t), the angular frequency w=12
So
[tex]\large X_L=12*0.37=4.44\\\\X_C=1/(12*7)=0.012[/tex]
and
[tex]\large Z=\sqrt{7^2+(4.44-0.012)^2}=8.283[/tex]
Finally
[tex]\large I=\displaystyle\frac{E_m}{Z}=\frac{120}{8.283}=14.488\;amperes[/tex]
A bowl contains blueberries and strawberries. There are a total of 16 berries in the bowl. The ratio of blueberries to strawberries is 3:1. How many of each berry are in the bowl?
Answer:
21 I believe
Step-by-step explanation:
3 x 5 = 15
15 + 1 = 16
because its one for every three blueberry then there are 5 strawberries and 16 blueberries so 5 + 16 = 21
Answer: there are 12 blueberries and 4 strawberries in the bowl
Step-by-step explanation:
The total number of blueberries and strawberries contained in the bowl is 16. The ratio of blueberries to strawberries is 3:1
Total ratio will be sum of the proportion the blueberries to the proportion of strawberries.
Total ratio = 3+1 = 4
To determine how many of each berry are in the bowl,
Number of blueberries in the bowl will be
(Proportion of blueberries / total ratio ) × 100
This becomes
3/4 × 16 = 12 blueberries
Number of strawberries in the bowl will be
(Proportion of strawberries / total ratio ) × 100
This becomes
1/4 × 16 = 4 strawberries
Read the severe weather warning and answer the question. Warning Below freezing temperatures will be accompanied by strong winds. Heavy snowfall is expected to last for several days. What type of severe weather is being described? Blizzard Drought Hurricane Tornado
Answer:
Hi! The answer is A, Blizzard.
Recently did this test on FLVS
Hope this helped!
Have a terrific Tuesday!!
~Lola
Answer:
Tornado
Step-by-step explanation:
Its not blizzard or hurricane.
(a) A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 153, 103, 125, 149, 118, 109, 86, 122, 138, 99. Let μ denote the average gas usage during January by all houses in this area. Compute a point estimate of μ.therms(b) Suppose there are 25,000 houses in this area that use natural gas for heating. Let τ denote the total amount of gas used by all of these houses during January. Estimate τ using the data of part (a).therms(c) Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms.(d) Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a).therms
Answer:
a) [tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]
b) [tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]
c) [tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]
d) Median =120
Step-by-step explanation:
1) Some important concepts
The mean refers to the "average that is used to derive the central tendency of data analyzed. It is determined by adding all the data points in a population and then dividing the total by the number of points".
Method of moments "involves equating sample moments with theoretical moments". For example the first sample moment about the origin is defined as [tex]M_1=\frac{1}{n} \sum_{i=1}^{n}x_i =\bar X [/tex]
The median is "the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average".
When we are trying to estimate the population proportion, p.
All estimation is based on the fact that the normal can be used to approximate the binomial distribution when np and nq are both at least 5. Where p is the probability of success and q the probability of failure.
2) Part a
Using the method of the moments a point of estimate for the [tex]\mu[/tex] is:
[tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]
3) Part b
If [tex]\hat \mu[/tex] is an individual estimate for the average gas usage during January and [tex]\tau[/tex] represent "the total amount of gas used by all of these houses during January" then the estimation for the total would be given by:
[tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]
3) Part c
For this part we want to estimate p ="the proportion of all houses that used at least 100 therms". If X is the random variable who represent the number of houses that exceed the usage of 100, we see that 8 out of 10 values are above 100, so the random variable X would be distributed binomial
[tex]X \sim Bin(10,0.8)[/tex] where n=10 and
[tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]
4) Part d
In order to find the median we need to put the data in order first, like this:
86,99,103,109,118,122,125,138,149,153
Since we have 10 observations and this number is even the procedure that we need to use in order to find the median is:
a) Find the value at position[n/2]=[10/2]=[5] on the data set ordered. For this case the value at position [5] is 118
b) Find the value at position[n/2 +1]=[10/2 +1]=[6] on the data set ordered. For this case the value at position [6] is 122
c) Find the average from the values obtained on steps a) and b). for this case (118+122)/2=120
So the Median = 120
The mean of the data set is 120.2
The total gas used = 3005000
The proportion of at least therms = 0.8
The median of the set = 120
a. To get the average gas usage, we are asked to calculate the mean of the observation.
Average
[tex]\frac{ 153+103+125+149+118+109+86+122+138+99}{10} \\\\[/tex]
= 120.2
b. The question says that 25000 houses make use of natural gas, then
total gases used by these houses =
120.2 * 25000
= 3005000
c. The proportion of the houses that used above 100 therms,
The house above 100 here are, 125, 149, 118, 109, 122, 138
They are 8 in number.
8/10 = 0.8
0.8 is the proportion that used at least 100 therms.
d. We have to find the median for the set here. We arrange the details in ascending order.
86,99,103,109,118,122,125,138,149,153
Median = 118+122/2
= 240/2
= 120
The managers want to know how many boxes of 12 cookies can be filled with the 3,258 cookies that have been baked. Fatima starts by subtracting the largest number of boxes she can easily calculate. She knows that 100 boxes of 12 cookies can be put into one crate. How many crates can be filled from the total of 3,258 cookies?
Answer:
i know im late but it's 2 crates.
Step-by-step explanation:
To find the number of crates, divide the total number of cookies by the number of cookies in each box.
Explanation:To find the number of crates that can be filled with the 3,258 cookies, we divide the total number of cookies by the number of cookies in each box. In this case, there are 12 cookies in each box.
So, we divide 3,258 by 12:
3,258 ÷ 12 = 271.5
Since we can't have half of a crate, we round down to the nearest whole number:
271.5 ≈ 271
Therefore, 271 crates can be filled from the total of 3,258 cookies.
A random sample of n1 = 49 measurements from a population with population standard deviation σ1 = 3 had a sample mean of x1 = 12. An independent random sample of n2 = 64 measurements from a second population with population standard deviation σ2 = 4 had a sample mean of x2 = 14. Test the claim that the population means are different. Use level of significance 0.01.What distribution does the sample test statistic follow? Explain.
Answer:
We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.
Step-by-step explanation:
We have large sample sizes [tex]n_{1} = 49[/tex] and [tex]n_{2} = 64[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 12-14 = -2.
The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,
[tex]\sqrt{\frac{(3)^{2}}{49}+\frac{(4)^{2}}{64}}[/tex] = 0.6585.
We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-2}{0.6585} = -3.0372[/tex]. Because -3.0372 fall inside RR, we reject the null hypothesis.
The test statistic follow a standard normal distribution because we are dealing with large sample sizes.
In this scenario of comparing two independent samples and given that the sample sizes are large, the sample test statistic follows the Standard Normal distribution or Z-distribution. The Z-test statistic representing the difference in sample means (in units of standard error) is compared with critical values for a two-tailed test at 0.01 significance level to determine if there's sufficient evidence to reject the null hypothesis that the two population means are equal.
Explanation:The test in your question pertains to a hypothesis testing scenario featuring two independent samples. This scenario typically involves two population means given that population standard deviations are known. The distribution followed by the sample test statistic in such cases is the Standard Normal distribution or Z-distribution, as the sample sizes (n1 = 49, n2 = 64) are sufficiently large. To test the claim that population means are different (at a significance level of 0.01), you'd typically construct a Z-test statistic that represents the difference in sample means (x1 - x2) in units of its standard error. The Z-test statistic is calculated as follows:
[tex]Z = \frac{x_1 - x_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}[/tex]Here, x1 and x2 are the sample means, σ1 and σ2 are the population standard deviations and n1 and n2 are the samples sizes. The resulting Z-score can be compared with critical Z-scores for a two-tailed test at the given level of significance (0.01) to determine whether or not the null hypothesis (two population means are equal) can be rejected.
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Employees at an office were surveyed about what types of hot beverages they drank that day. It was found that 50% of them drank coffee and 25% drank tea. The survey showed that 10% of the employees drank both coffee and tea.
For these employees, are events "drank coffee" and "drank tea" mutually exclusive?
Choose one answer: Yes or No?
Find the probability that a randomly selected person from this group drank coffee OR drank tea?
Answer:
65%
Step-by-step explanation:
No, the events "drank coffee" and "drank tea" are not mutually exclusive, as there are 10% of employees drank both coffee and tea.
If there are 50% drank coffee and 10% of them enjoy both, then there are 40% of the employees enjoy only coffee.
Similarly, there are 15% of employees who only enjoy tea.
Then the probability of selecting a person who only enjoy tea or coffee is
40% + 15% = 65%
Answer:
No.
0.65 or 65%
A country has two political parties, the Demonstrators and the Repudiators. Suppose that the national senate consists of 100 members, 44 of which are Demonstrators and 56 of which are Repudiators.
(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?
(b)Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?
Answer:
(a) 4.148 x 10^(12) ways
(b) 5,827,360 ways
Step-by-step explanation:
Number of Demonstrators (D) = 44
Number of Repudiators (R) = 56
(a)
5 senate members must be Repudiators and 5 must be demonstrators, assuming that the order at which they are selected is irrelevant:
[tex]N= C^{D}_{5} * C^{R}_{5}\\N=\frac{56!}{5!(56-5)!} *\frac{44!}{5!(44-5)!} \\N=3,819,816*1,086,008\\N=4.148 *10^{12}[/tex]
(b)
Since there are two different positions, (speaker and vice speaker), order is important in this situation, and the total number of ways to select two senators from each party is:
[tex]N= P^{D}_{2} * P^{R}_{2}\\N=\frac{56!}{(56-2)!} *\frac{44!}{(44-2)!} \\N=3,080*1,892\\N=5,827,360[/tex]
The question involves applying the concept of combinations in mathematics to determine the number of ways committee members, speakers, and vice speakers can be selected from two different political parties.
Explanation:The subject of the question involves two main components of mathematics: combinatorics and probability. This involves calculating the number of ways certain events can occur given a certain number of possibilities.
Let's first solve part (a) of your question. We are asked how many ways there are to select a committee of 10 senate members with the same number of Demonstrators and Repudiators. We want five senators from each party. Given there are 44 Demonstrators and 56 Repudiators, the number of ways we can pick a committee is the product of comb(44,5) and comb(56,5) which are the combinations of picking 5 out of 44 Demonstrators and 5 out of 56 Repudiators, respectively.
In part (b) of your question, we are asked how many ways there are for two speakers and two vice speakers to be selected, one from each party. This is simply comb(44,1) multiplied by comb(43,1) multiplied by comb(56,1) multiplied by comb(55,1). This is because we first choose 1 out of 44 Demonstrators for a speaker position, then 1 out of the remaining 43 Demonstrators for a vice speaker position, then 1 out of 56 Repudiators for a speaker position, and finally 1 out of the remaining 55 Repudiators for a vice speaker position.
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A shirt and a tie together cost $48 the shirt costs $22 more than the tie what is the cost of the shirt
Answer:
35$
Step-by-step explanation:
Let the shirt be = X
And the tie =Y
X + Y= 48$
X = 22 + Y (The shirt costs $22 more than the tie)
22 + 2y = 48
2y = 26
y = 13
X= 48 – 13
X = 35
therefore, the cost of the shirt is $35
and the cost of the tie is $13
The cost of the shirt is 35$ such the shirt and tie together cost $48.
How to form an equation?Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.
In other words, an equation is a set of variables that are constrained through a situation or case.
Let's say the shirt cost is S while the tie is T
Together;
S + T = 48
And,
S = 22 + T
By substituting
22 + T + T = 48
2T = 26
T = 13
So,
S = 48 - 13 = 35
Hence "The cost of the shirt is 35$ such the shirt and tie together cost $48".
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Which of the following is not a property of a chi-square distribution?
a. ????2 is skewed to the right.
b. The number of degrees of freedom defines the shape of the distribution of ????2 .
c. ????2 can have both positive and negative values.
d. All of these choices are properties of the ????2 distribution.
Answer:
c) Is not a property (hence (d) is not either)
Step-by-step explanation:
Remember that the chi square distribution with k degrees of freedom has this formula
[tex]\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2[/tex]
Where N₁ , N₂m .... [tex] N_k [/tex] are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.
Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true
The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.
Look at the 95% confidence interval and say whether the following statement is true or false. ""This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."" Be sure to explain your answer.
Answer:
False
Step-by-step explanation:
Confidence intervals provide a range for a population parameter at a given significance level. The parameter can be mean, standard deviation etc.
In this example population is the prices of the rents of all the unfurnished one-bedroom apartments in the Boston area
significance level is 95%. Thus, the chance being the true population parameter in the given interval is 95%.
But, "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." statement is false because the population parameter is missing. Confidence interval may describe population mean for example but it does not describe the whole population.
9. A judge hears the following arguments in a murder hearing. The DNA test that places the accused at the murder scene has a true positive rate of 90% (i.e. the probability that the test returning positive given that the accused was actually present at the scene is 0.9). Similarly, the DNA test has a false negative rate of 80% (i.e. the probability that the test returns negative given that the accused was not present at the scene is 0.8). Everyone in the town has a equal probability of being at the murder scene, and the town has a population of 10,000. Given the fact that the DNA test returned a positive result for the accused, what is the probability that the accused was at the murder scene?
Answer:
0.0004498
Step-by-step explanation:
Let us define the events:
A = The test returns positive.
B = The accused was present.
Since everyone in the town has an equal probability of being at the murder scene, and the town has a population of 10,000
P(B) = 1/10000 = 0.0001
We have that the probability the test returning positive given that the accused was actually present at the scene is 0.9
P(A | B) = 0.9
and the probability that the test returns negative given that the accused was not present at the scene is 0.8
[tex]\large P(A^c|B^c)=0.8[/tex]
where
[tex]\large A^c,\;B^c[/tex] are the complements of A and B respectively.
We want to determine the probability that the DNA test returned a positive result given that the accused was at the murder scene, that is, P(B | A).
We know that P(A | B) = 0.9, so
[tex]\large \frac{P(A\cap B)}{P(B)}=0.9\Rightarrow P(A\cap B)=0.9P(B)=0.9*0.0001\Rightarrow\\\\P(A\cap B)=0.00009[/tex]
Now, we have
[tex]\large P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.00009}{P(A)}[/tex]
So if we can determine P(A), the result will follow.
By De Morgan's Law
[tex]\large A^c\cap B^c=(A\cup B)^c[/tex]
so
[tex]\large 0.8=P(A^c|B^c)= \frac{P(A^c\cap B^c)}{P(B^c)}=\frac{P((A\cup B)^c)}{P(B^c)}=\frac{1-P(A\cup B)}{1-P(B)}\Rightarrow\\\\\frac{1-P(A\cup B)}{1-0.0001}=0.8\Rightarrow P(A\cup B)=1-0.8(1-0.0001)\Rightarrow\\\\P(A\cup B)=0.20008[/tex]
Using the formula
[tex]\large P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]
and replacing the values we have found
[tex]\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007[/tex]
and finally, the desired result is
[tex]\large P(B|A)=\frac{0.00009}{P(A)}=\frac{0.00009}{0.20007}\Rightarrow\\\\\boxed{P(B|A)=0.0004498}[/tex]
Suppose that you construct a 95% confidence interval for the population mean, using some sample values, and you obtain the range of 50 to 70. Then, which of the following might be the 90% confidence interval using the same sample values. 50 to 100
70 to 90
60 to 80
55 to 95
65 to 85
The 90% confidence interval using the same sample values is E. 65 to 85.
How to depict the confidence interval?
The 90% confidence interval would be narrower than the 95% confidence interval but the middle point always remains the same.
The middle point there should be (60 + 90)/2 = 75 for the confidence interval. The confidence interval width for a 95% confidence interval width is 30.
For the 65 to 85 confidence interval, the width is 20, therefore this can be true because 20 < 30, therefore 65 to 85 could be the possible confidence interval required here.
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Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090% of all males. (Accommodating 100% of males would require very wide seats that would be much tooexpensive.) Men have hip breadths that are normally distributed with a mean of 14.5 in. and a standard deviation of 1.2 in. Find Upper P90. That is, find the hip breadth for men that separates the smallest 90% from the largest 10%.
The hip breadth for men that separates the smallest 90% from the largest 10% is P90__in.
(Round to one decimal place as needed.)
Answer:
Hip breadths less than or equal to 16.1 in. includes 90% of the males.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 14.5
Standard Deviation, σ = 1.2
We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.10.
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 1.282) = 0.90[/tex]
[tex]\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1[/tex]
Hence, hip breadth of 16.1 in. separates the smallest 90% from the largest 10%.
That is hip breaths greater than 16.1 in. lies in the larger 10%.
A poll was taken of 1000 residents in county. The residents sampled were asked whether they think their local government did a good job overall. 750 responded "yes". Let p denote the proportion of all residents in that county who think their local government did a good job. Construct a 95% confidence interval for p. Round off to two decimal places. a) (0.72, 0.78) b)(0.70, 0.86) c (0.68, 0.92) d) (0.10,1.56) e)(0.79, 0.87)
Answer:
a) (0.72, 0.78)
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Description in words of the parameter p
[tex]p[/tex] represent the real population proportion of all residents in that county who think their local government did a good job
[tex]\hat p[/tex] represent the estimated proportion of all residents in that county who think their local government did a good job
n=1000 is the sample size required
[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Numerical estimate for p
In order to estimate a proportion we use this formula:
[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.
[tex]\hat p=\frac{750}{1000}=0.75[/tex] represent the estimated proportion of all residents in that county who think their local government did a good job
Confidence interval
The confidence interval for a proportion is given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.750 - 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.72[/tex]
[tex]0.750 + 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.78[/tex]
And the 95% confidence interval would be given (0.72;0.78).
We are confident at 95% that the true proportion of people who think their local government did a good job is between (0.72;0.78).
a) (0.72, 0.78)
A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a 15 day waiting period to purchase a gun?"" Results from a random sample of US citizens showed that 318 of the 520 men who were surveyed supported this proposed law while 379 of the 460 women sampled said ‘‘yes"". Use this information to find a 95% confidence interval for the difference in the two proportions
the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).
To find the 95% confidence interval for the difference in proportions, we can use the formula:
[tex]\[\text{CI} = (\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{{\hat{p}_1(1 - \hat{p}_1)}}{n_1} + \frac{{\hat{p}_2(1 - \hat{p}_2)}}{n_2}}\][/tex]
Where:
- [tex]\(\hat{p}_1\) and \(\hat{p}_2\)[/tex] are the sample proportions.
- [tex]\(n_1\) and \(n_2\)[/tex] are the sample sizes.
- z is the z-score corresponding to the desired level of confidence.
Given:
- [tex]\(n_1 = 520\), \(n_2 = 460\)[/tex]
- [tex]\(\hat{p}_1 = \frac{318}{520}\), \(\hat{p}_2 = \frac{379}{460}\)[/tex]
- z = 1.96 for a 95% confidence interval
Let's plug in the values and calculate:
[tex]\[\hat{p}_1 = \frac{318}{520} \approx 0.6115\]\[\hat{p}_2 = \frac{379}{460} \approx 0.8239\]\[\text{CI} = (0.8239 - 0.6115) \pm 1.96 \times \sqrt{\frac{{0.6115 \times (1 - 0.6115)}}{520} + \frac{{0.8239 \times (1 - 0.8239)}}{460}}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{\frac{{0.6115 \times 0.3885}}{520} + \frac{{0.8239 \times 0.1761}}{460}}\][/tex]
[tex]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000457 + 0.000368}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000825}\]\[\text{CI} = (0.2124) \pm 1.96 \times 0.0287\]\[\text{CI} = (0.2124) \pm 0.0562\][/tex]
Thus, the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).
The marginal cost of drilling an oil well depends on the depth at which you are drilling; drilling becomes more expensive, per meter, as you dig deeper into the earth. The fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and, if x is the depth in meters, the marginal costs are C' (x) = 4000 + 10x (Riyals/meter).Find the total cost of drilling a 500-meter well.
To calculate the total cost of drilling a 500-meter well, add the fixed cost to the sum of the marginal costs for every meter drilled. The total cost comes out to 3,500,000 riyals.
Explanation:The total cost of drilling a 500-meter well comprises both fixed costs and the marginal cost per meter of depth. We are given that the fixed costs amount to 1,000,000 riyals. The marginal cost function is given as C'(x) = 4000 + 10x, which means the cost per additional meter drilled increases linearly with depth.
To find the total cost of drilling a 500-meter well, we need to compute the integral (i.e., the area under the curve) of the marginal cost function from 0 to 500 and add the fixed costs. This calculation represents the sum of the increasing cost per meter for every meter drilled.
By evaluating the integral ∫ (4000 + 10x) dx from 0 to 500, we get 2,500,000 riyals. This is the total variable cost of drilling a 500m well. Adding it to the fixed cost (1,000,000 riyals), the grand total comes out to be 3,500,000 riyals.
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x+y+z=1
-2x+4y+6z=2
-x+3y-5z=11
Solved using elimination
Answer: x = 0
y = 2
z = -1
Step-by-step explanation:
The system of equations are
x+y+z=1 - - - - - - - - - - 1
-2x+4y+6z=2 - - - - - - - - - 2
-x+3y-5z=11 - - - - - - - - - 3
Step 1
We would eliminate x by adding equation 1 to equation 3. It becomes
4y -4z = 12 - - - - - - - - - 4
Step 2
We would multiply equation 1 by 2. It becomes
2x + 2y + 2z = 2 - - - - - - - - - 5
We would add equation 2 and equation 5. It becomes
6y + 8z = 4 - - - - - - - - - 6
Step 3
We would multiply equation 4 by 6 and equation 6 by 4. It becomes
24y - 24z = 72 - - - - - - - - 7
24y + 32z = 16 - - - - - - - - 8
We would subtract equation 8 from equation 7. It becomes
-56z = 56
z = -56/56 = -1
Substituting z = -1 into 7, it becomes
24y - 24×-1 = 72
24y + 24 = 72
24y = 72 - 24 = 48
y = 48/24 = 2
Substituting y = 2 and z = -1 into equation 1, it becomes
x + 2 - 1 = 1
x = 1 - 1 = 0
Researchers measured skulls from different time periods in an attempt to determine whether interbreeding of cultures occurred. Results are given below. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.
n x s
4000 B.C. 30 131.62 mm 5.19 mm
A.D. 150 30 136.07 mm 5.35 mm
What are the null and alternative hypotheses?Identify the test statistic, F=?The P-value is ?What is the concluion for this hypothesis test?A. Fail to reject Upper H0. There is sufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.B. Reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.C. Fail to reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.D. Reject Upper H 0. There is sufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.
Answer:
C. Fail to reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150
Step-by-step explanation:
Hello!
You have two different independent samples and are asked to test if the population variances of both variables are the same.
Sample 1 (4000 B.C)
X₁: Breadth of a skull from 4000 B.C. (mm)
X₁~N(μ₁;σ₁²)
n₁= 30 skulls
X[bar]₁= 131.62 mm
S₁= 5.19 mm
Sample 2 (A.D. 150)
X₂: Breadth of a skull from 150 A.D. (mm)
X₂~N(μ₂;σ₂²)
n₂= 30 skulls
X[bar]₂= 136.07 mm
S₂= 5.35 mm
Since you want to test the variances, the proper test to do is an F-test for the population variance ratio. The hypothesis can be established as equality between variances or as a quotient between them.
The hypothesis is:
H₀: σ₁²/σ₂² = 1
H₁: σ₁²/σ₂² ≠ 1
Remember, when you express the hypothesis as a quotient of variances, if it's true that they are the same, the result will be 1, this is the number you'll use to replace in the F-statistic.
α: 0.05
F= (S₁²/S₂²) * (σ₁²/σ₂²) ~F[tex]_{n1-1;n2-1}[/tex]
F= (5.19/5.35)*1 = 0.97
The p-value = 0.5324
Since the p-value is greater than the level of significance, the decision is to not reject the null hypothesis.
Using critical values:
Left: F[tex]F_{n1-1;n2-1;\alpha /2} = \frac{1}{F_{n2-1;n1-1;1-\alpha /2} } = \frac{1}{F_{29;29;0.95} } = \frac{1}{2.10} } =0.47[/tex]
Right: [tex]F_{n1-1; n2-1; 1-\alpha /2} = F_{29; 29; 0.975} = 2.10[/tex]
The calculated F-value (0.97) is in the not rejection zone (0.47<F<2.10) ⇒ Don't reject the null hypothesis.
I hope this helps!
The hypotheses are that there is no significant difference (null) or that there is a significant difference (alternative) in skull breadths from 4000 B.C. and A.D. 150. An F-test is used to test these via the comparison of sample variances. The conclusion depends on the P-value: if it is greater than 5%, the null is accepted (option C), and if less, rejected (option D).
Explanation:The null and alternative hypotheses for this question can be stated as follows:
Null Hypothesis (H0):
There is no significant difference in the variation of maximal skull breadths in 4000 B.C. and A.D. 150.
Alternative Hypothesis (H1):
There is a significant difference in the variation of maximal skull breadths in 4000 B.C. and A.D. 150.
To test these hypotheses, we use the F-test for the equality of two variances. The test statistic (F) is calculated by taking the ratio of the sample variances, which in this case would be (5.19^2) / (5.35^2).
The P-value associated with the F statistic is then used to determine the significance of the evidence against the null hypothesis. If the P-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
The conclusion of the hypothesis test depends on the calculated P-value. If P-value is less than 0.05, we conclude that there is a significant difference in the variation of skull breadths, thereby rejecting the null hypothesis (option D). If the P-value is greater than 0.05, we fail to reject the null hypothesis, concluding that the skull variations are not significantly different (option C).
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An object moves in simple harmonic motion with period 7 seconds and amplitude 3cm. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a negative direction. Give the equation modeling the displacement d as a function of time t.
The equation modeling the displacement d as a function of time t for an object in simple harmonic motion with a period of 7 seconds and amplitude of 3cm is d = -3cos(2pi/7 * t + pi).
Explanation:The displacement d of an object moving in simple harmonic motion can be modeled by the equation d=Acos(wt).
Given that the period T is 7 seconds, we can use the formula T=2pi/w to solve for the angular frequency w. Rearranging the equation, we have w = 2pi/T. Plugging in the given period T=7, we get w = 2pi/7.
Since the object initially moves in a negative direction, we would have a phase shift of pi in the cosine function. Therefore, the equation modeling the displacement d as a function of time t is d = -3cos(2pi/7 * t + pi).
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If 4 items are chosen at random without replacement from 7 items, in how many ways can the 4 items be arranged, treating each arrangement as a different event (i.e., if order is important)?
A. 35
B. 840
C. 5040
D. 24
Answer:
840
Step-by-step explanation:
a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Find the P-value and state a conclusion. Round the answer to four decimal places.
Answer:
p-value: 0.6527
Step-by-step explanation:
Hello!
You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:
Sample 1 (Boys aged 6-11)
n₁= 546
x₁= 89
^p₁= x₁/n₁ = 89/546 ≅0.16
Sample 2 (girls aged 6-11)
n=508
x= 74
^p= x/n = 74/508 ≅ 0.15
If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:
H₀: ρ₁ = ρ₂
H₁: ρ₁ ≠ ρ₂
This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:
Z= (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂) ≈ N(0;1)
√(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))
ρ`= x₁ + x₂ = 89+74 = 0.154 ≅ 0.15
n₁ + n₂ 546 + 508
Z⁰ᵇ = (0.16-0.15) - (0)
√(0.15 * (1 - 0.15) * (1/546 + 1/508))
Z⁰ᵇ = 0.45
I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:
P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527
p-value: 0.6527
Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05
Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.
I hope it helps!
Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or parametric description of the surface. F=〈x,y,z〉across the slanted face of the tetrahedron z=10−2x−5y in the first octant; normal vectors point upward.
Answer:
50
Step-by-step explanation:
Please see attachment .
What is the greatest common factor (GCF) of 48 and 56? A. 168 B. 8 C. 4 D. 336
Answer:
The answer is B.8.
Step-by-step explanation:
This is the answer because 8 is the greatest factor that will go into both 48 and 56. 8x6=48, 8x7=56.
The greatest common factor (GCF) of 48 and 56 is 8. It is the highest number that divides both numbers without leaving a remainder. Thus, the correct answer is Option B. 8.
The greatest common factor (GCF) of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF of 48 and 56, follow these steps:
List the factors:Therefore, the correct answer is Option B. 8. This makes 8 the highest number that can evenly divide both 48 and 56.
The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with mean 26 mpg and standard deviation 12 mpg. If a random sample of 36 such cars are chosen and tested, what is the probability the average mpg is less than 28 mpg?
Answer:
The probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean [tex]\mu=26[/tex] mpg and standard deviation [tex]\sigma=12[/tex] mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
[tex]z=\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The portability is given by, [tex]P(X<28)[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{28-26}{\frac{12}{\sqrt{36}}})[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{2}{\frac{12}{6}})[/tex]
[tex]=P(z<\dfrac{2}{2})[/tex]
[tex]=P(z<1)[/tex]
Using z-table,
[tex]=0.8413[/tex]
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Answer:
he probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean mpg and standard deviation mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
The portability is given by,
Using z-table,
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
A grade school teacher has developed several ideas about how to improve her students’ learning outcomes, and now she needs to pick one to implement. Which of the following tools will NOT help to determine the most useful solution idea? A. PICK chart B. Prioritization matrix C. Nominal group technique D. Pareto chart
Answer:C
Step-by-step explanation: