Answer:
B)
Step-by-step explanation:
If you try to divide 135 with each number,every number is divisible by 135 XD
Jackson is testing different types of planting soil to determine which type is most effective for growing strawberry bushes. He purchases two different brands of planting soil from a local store. Jackson applies brand A to an area of the yard that receives full sunlight and brand B to another area of the yard that is in partial sunlight. He waters the area with brand B daily; he waters the area with brand A every other day. At the end of the study, Jackson concludes brand A is more effective in growing strawberry bushes. Is Jackson's conclusion valid? Explain why or why not.
Answer:
The answer to this question can be described as follows:
Step-by-step explanation:
In the outcome of Jackson isn't legitimate as product A and product B were n’t considered equal in watering the plants. It is the preparation for all two to be compared, Product A with only a region of the backyard, which receives direct energy, and Product B will be maintained regularly by the very same area of the yards, that is in minimal lighting.
No, The conclusion done by Jackson's is not valid for this situation.
What is effective soiling?Effective Soiling depth is the depth of soil material which is above the layer that impedes movement of air and water and growth in plant roots.
According to outcome of Jackson that isn't legitimate as product A and product B were considered equal in watering the plants. Then it is the preparation of two to be compare with product B, Product A with only a region of the backyard, which receives direct energy, and Product B will maintained regularly very same area of the yards, that is with minimal lighting.
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Find two matrices A and B such that 2A-3B= [5 0
-1 2]
(5 & 0 are on top; -1 & 2 are on the bottom— it’s a 2x2 matrix)
The easy answer: If, say, B is the zero matrix, i.e.
[tex]B=\begin{bmatrix}0&0\\0&0\end{bmatrix}[/tex]
then
[tex]2A=\begin{bmatrix}5&0\\-1&2\end{bmatrix}\implies A=\begin{bmatrix}\dfrac52&0\\\\-\dfrac12&1\end{bmatrix}[/tex]
Can anyone pls help me to solve question 2 f and g and pls provide me a explanation I’m with that questions for three days
Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Step-by-step explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
__
f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
__
g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).
An analysis of over 160 empirical studies conducted by Norris and her colleagues (2006) revealed evidence of severe to very severe impairment (interference with functioning) among survivors. What percent of survivors were incapacitated?
Answer:
15-25% of survivors were incapacitated
Step-by-step explanation:
According to the analysis conducted by Norris and her colleague in 2006, it was found that out of the 160 empirical studies conducted, 41% showed evidence of severe to very severe impairment (interference with functioning) among disaster survivors.
This in turn corresponds to 15-25% increase in demand for mental health services by the population affected by this disasters.
This percentage of people affected by disasters suffer from disaster's syndrome which include but not limited to shock, bewilderment and void of deep emotion.
Hence they become incapacitated.
The question is incomplete, as the required details are not given. However, I will give a general explanation on how to determine percentages.
To calculate the percentage of survivors that were incapacitated, we need:
The total number of survivors (n)The number of survivors that were incapacitated (k)Assume that:
[tex]n = 250[/tex] --- survivors
[tex]k = 100[/tex] --- survivors that were incapacitated
The percentage (p) is calculated as follows:
[tex]p = \frac kn \times 100\%[/tex]
So, we have:
[tex]p = \frac{100}{250} \times 100\%[/tex]
[tex]p = \frac{100\times 100}{250} \%[/tex]
[tex]p = \frac{10000}{250} \%[/tex]
[tex]p = 40 \%[/tex]
Using the assumed values, 40% of the survivors were incapacitated.
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PLEASE HELP! Exam is at 9:15 am!!! A bicyclist pedals a bicycle at 40 revolutions per minute resulting in a speed of 7 mile per hour. How fast will the bicyclist go if he pedals 60 revolution per minute.
Answer:
Step-by-step explanation:
Answer:
10.5m/h
Step-by-step explanation:
speed at 40 revolutions= 7/60 m/minute = 0.1167
circumference of wheel = 101167/40 = 7/2400
now,
at 60 pedals = speed = 60*7/2400 = 0.175m/minute
speed= 0.175*60 m/h = 10.5
which factorization is equivalent to the expression 30x+70
Answer:
10 (3x+7)
Step-by-step explanation:
factor out 10
10 (3x+7)
For the data set 2.5, 6.5, 9, 19, 20, 2.5. What is the mean?
Answer:
9.9166667
Step-by-step explanation:
All of the numbers equal 59.5 then divide by 6 equals the answer above.
To calculate the mean of the given data set, sum all the data values and divide by the number of values. The mean of this data set is 9.9167.
To find the mean of the given data set: 2.5, 6.5, 9, 19, 20, 2.5, follow these steps:
Sum of the data values: 2.5 + 6.5 + 9 + 19 + 20 + 2.5 = 59.5Number of data values: There are 6 values in the data set.Calculate the mean: Mean = Sum of data values / Number of data values = 59.5 / 6 = 9.9167Hence, the mean of the data set is 9.9167.
Orlando invests $1000 at 6% annual interest compounded daily and Bernadette invests $1000 at 7%
simple interest. After how many whole years will Orlando's investments be worth more than
Bernadette's investments?
Answer:
6 Years
Step-by-step explanation:
Orlando invests $1000 at 6% annual interest compounded daily.
Orlando's investment = [tex]A=1000(1+\frac{0.06}{365})^{(365\times t)}[/tex]
Bernadette invests $1000 at 7% simple interest.
Bernadette's investment = A = 1000(1+0.07×t)
By trail and error method we will use t = 5
Bernadette's investment will be after 5 years
1000(1 + 0.07 × 5)
= 1000(1 + 0.35)
= 1000 × 1.35
= $1350
Orlando's investment after 5 years
[tex]A=1000(1+\frac{0.06}{365})^{(365\times 5)}[/tex]
= [tex]1000(1+0.000164)^{1825}[/tex]
= [tex]1000(1.000164)^{1825}[/tex]
= 1000(1.349826)
= 1349.825527 ≈ $1349.83
After 5 years Orlando's investment will not be more than Bernadette's.
Therefore, when we use t = 6
After 6 years Orlando's investment will be = $1433.29
and Bernadette's investment will be = $1420
So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.
You have decided to stop drinking Starbucks coffee and invest that money in an IRA. If you
deposit $
Complete question :
You have decided to stop drinking Starbucks coffee and invest that money in an IRA. If you deposit $492 each month earning 6% interest, how much will you have in the account after 40 years?
Answer:
$250,329.60
Step-by-step explanation:
Given:
Principal, P= $492 per month
= $492 * 12 = $5,904 per year
Rate, R= 6%
Time, T= 40 years
Let's first find the amount after 40 years.
Amount = 5940 * 40 = $236,160
The interest after 40 years =
[tex] Interest = \frac{PRT}{100} = \frac{5904 * 6 * 40}{100}[/tex]
Interest = $14,169.60
The total amount I will have in my account after 40 years:
Amount + interest
= $236,160 + $14,169.60
= $250,329.60
(b) Based on the summary statistics, would it be more likely to obtain a yield
of 123 or more bushels per acre from a plot of GM corn or a plot of regular
corn? Justify your answer.
Answer: yes, GM is expected to provide performance greater than 123
Step-by-step explanation:
b) It is expected that the GM achieves a performance greater than 123. This situation occurs because 123 is less than the average for GM but is greater than the average for Regular. Thus we observe that P (performance> 123 | GM)> 0.5 and P (performance> 123 | Regular) <0.5
The following are quality control data for a manufacturing process at Kensport Chemical Company. The data show the temperature in degrees centigrade at five points in time during a manufacturing cycle. Sample x R 1 95.72 1.0 2 95.24 0.9 3 95.18 0.9 4 95.44 0.4 5 95.46 0.5 6 95.32 1.1 7 95.40 0.9 8 95.44 0.3 9 95.08 0.2 10 95.50 0.6 11 95.80 0.6 12 95.22 0.2 13 95.56 1.3 14 95.22 0.6 15 95.04 0.8 16 95.72 1.1 17 94.82 0.6 18 95.46 0.5 19 95.60 0.4 20 95.74 0.6 The company is interested in using control charts to monitor the temperature of its manufacturing process. Compute the upper and lower control limits for the R chart. (Round your answers to three decimal places.) UCL
Answer: The upper control limit is 1.40581
The lower control limit is 0.000
Explanation: Check attachment
The upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.
To compute the upper and lower control limits for the R chart, we need to calculate the average range (R-bar) and the control limits.
First, let's calculate the range (R) for each sample by subtracting the smallest value from the largest value:
Sample 1: 1.0
Sample 2: 0.9
Sample 3: 0.9
Sample 4: 0.4
Sample 5: 0.5
Sample 6: 1.1
Sample 7: 0.9
Sample 8: 0.3
Sample 9: 0.2
Sample 10: 0.6
Sample 11: 0.6
Sample 12: 0.2
Sample 13: 1.3
Sample 14: 0.6
Sample 15: 0.8
Sample 16: 1.1
Sample 17: 0.6
Sample 18: 0.5
Sample 19: 0.4
Sample 20: 0.6
Next, calculate the average range (R-bar) by summing up all the ranges and dividing by the number of samples:
R-bar = (1.0 + 0.9 + 0.9 + 0.4 + 0.5 + 1.1 + 0.9 + 0.3 + 0.2 + 0.6 + 0.6 + 0.2 + 1.3 + 0.6 + 0.8 + 1.1 + 0.6 + 0.5 + 0.4 + 0.6) / 20
R-bar = 0.735
To calculate the upper control limit (UCL) for the R chart, multiply the R-bar by a constant factor. For small sample sizes like this (n=5), the constant factor is typically 2.704:
UCL = R-bar * 2.704
UCL = 0.735 * 2.704
UCL = 1.986
Finally, to calculate the lower control limit (LCL) for the R chart, multiply the R-bar by another constant factor. For small sample sizes like this, the constant factor is typically 0:
LCL = R-bar * 0
LCL = 0
Therefore, the upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.
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The most expensive type of life insurance
The most expensive type of life insurance is cash-value (whole) life insurance due to its integrated savings component, investment aspect, and the management involved. These features add complexity and cost to the policy, making it more expensive than term life insurance, which only offers a death benefit.
The most expensive type of life insurance is typically cash-value (whole) life insurance. This kind of policy not only provides a death benefit to beneficiaries upon the insured individual's death but also includes a savings component known as the cash value. Over time, the policy accumulates a cash value that the policyholder can borrow against or even withdraw. Part of the premium goes towards life insurance, while another part is invested by the company, often in a conservative mix of investments. This investment aspect and the management involved make whole life insurance more expensive compared to term life insurance which only provides a death benefit without a savings component.
It is also worth noting that individuals purchasing life insurance are likely to have more knowledge about their own health risks than the insurance company. This information asymmetry can impact the pricing of life insurance policies. Moreover, the fundamental law of insurance states that the average person's premium payments over time must cover the claims, the insurer's costs, and profits. Therefore, insurance premiums, particularly for policies with cash values, reflect the cost of investment management and the integrated savings aspects, which can contribute to their higher costs.
M(8, 7) is the midpoint of RS. The coordinates of S are (9, 5). What are the coordinates of R?
Answer:
M(Mx, My), R(Rx, Ry), S(Sx, Sy)
M is midpoint of RS
=> Rx + Sx = 2Mx => Rx = 2Mx - Sx = 2 x 8 - 9 = 7
=> Ry + Sy = 2My => Ry = 2My - Sy = 2 x 7 - 5 = 9
=> R(7, 9)
Hope this helps!
:)
Answer:
(7,9)
Step-by-step explanation:
Let R:(x,y)
(x+9)/2, (y+5)/2 = 8 ,7
x + 9 = 16
x = 7
y + 5 = 14
y = 9
R: (7,9)
Teachers’ salaries in one state are very low that the educators in that state regularly complain about their compensation. The state mean is $33,600, but teachers in one district claim that the mean their district is significantly lower. They survey a simple random sample of 22 teachers in the district and calculate a mean salary of $32,400 with a standard deviation s = $ 1520. Test the teachers’ claim at the 0.05 level of significance.
Answer:
[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]
The degrees of freedom are given by:
[tex]df=n-1=22-1=21[/tex]
The p value is given by:
[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]
The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600
Step-by-step explanation:
Information given
[tex]\bar X=33400[/tex] represent the sample mean
[tex]s=1520[/tex] represent the sample standard deviation
[tex]n=22[/tex] sample size
[tex]\mu_o =33600[/tex] represent the value that we want to analyze
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to check if the true mean is lower than 33600, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 33600[/tex]
Alternative hypothesis:[tex]\mu < 33600[/tex]
The statistic is given:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the data given we got:
[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]
The degrees of freedom are given by:
[tex]df=n-1=22-1=21[/tex]
The p value is given by:
[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]
The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600
Adrian earns $16000 per month. He spends 1/4 of his income on food; 3/10 of the remainder on house rent . How much money does he have left?
Answer:
8000
Step-by-step explanation:
1/4 of 16000 is 4000
16000-4000=12000 1/3 of 12000 is also 4000
12000-4000=8000
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling or overfilling the bags. A 23 bag sample had a mean of 435 grams with a variance of 841. Assume the population is normally distributed. A level of significance of 0.05 will be used. Specify the type of hypothesis test.
Answer:
[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]
The degrees of freedom are given by:
[tex]df=n-1=23-1=22[/tex]
And the p value taking in count that we have a bilateral test we got:
[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value
Step-by-step explanation:
Information given
[tex]\bar X=435[/tex] represent the mean for the weight
[tex]s=\sqrt{841}=29[/tex] represent the sample standard deviation
[tex]n=23[/tex] sample size
[tex]\mu_o =430[/tex] represent the value that we want to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We are trying to proof if the filling machine works correctly at the 430 gram setting, so then the system of hypothesis for this case are:
Null hypothesis:[tex]\mu = 430[/tex]
Alternative hypothesis:[tex]\mu \neq 430[/tex]
In order to cehck the hypothesis the statistic for a one sample mean test is given by
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we have this:
[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]
The degrees of freedom are given by:
[tex]df=n-1=23-1=22[/tex]
And the p value taking in count that we have a bilateral test we got:
[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value
The type of hypothesis test used in this scenario is a two-tailed t-test for a single sample mean.
Step 1: State the hypotheses
- Null hypothesis H₀: The mean weight of the bags is 430 grams [tex]\mu = 430[/tex]
- Alternative hypothesis H₁: The mean weight of the bags is not 430 grams [tex]\mu \neq 430[/tex]
Step 2: Select the significance level
- The level of significance [tex]\alpha[/tex] is 0.05.
Step 3: Calculate the test statistic
- Sample mean [tex]\bar{x}\)[/tex] = 435 grams
- Population mean [tex]\mu\)[/tex] = 430 grams
- Sample size n = 23
- Sample variance s² = 841
- Sample standard deviation s = √841 = 29
The test statistic (t) is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
[tex]\[ t = \frac{435 - 430}{29 / \sqrt{23}} \][/tex]
[tex]\[ t = \frac{5}{6.048} \][/tex]
t ≈ 0.8266
Step 4: Determine the degrees of freedom and critical value
- Degrees of freedom [tex](\(df\)) = \(n - 1[/tex]
= 23 - 1
= 22
- For a two-tailed test at [tex]\(\alpha = 0.05\)[/tex] with 22 degrees of freedom, the critical t-values are approximately [tex]\(\pm 2.074[/tex].
Step 5: Make a decision
- Compare the calculated test statistic 0.8266 with the critical t-values [tex]\pm 2.074[/tex].
- If the test statistic is within the range -2.074 to 2.074, fail to reject the null hypothesis.
- If the test statistic is outside this range, reject the null hypothesis.
Since 0.8266 is within the range -2.074 to 2.074, we fail to reject the null hypothesis.
Conclusion:
- There is not enough evidence to suggest that the mean weight of the bags is different from 430 grams at the 0.05 significance level.
The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but you believe that the mean nicotine content is actually higher than advertised. To explore this, you test the hypotheses H0:μ=1.5, Ha:μ>1.5 and you obtain a P-value of 0.052. Which of the following is true? A. At the α=0.05 significance level, you have proven that H0 is true. B. This should be viewed as a pilot study and the data suggests that further investigation of the hypotheses will not be fruitful at the α=0.05 significance level. C. There is some evidence against H0, and a study using a larger sample size may be worthwhile. D. You have failed to obtain any evidence for Ha.
Answer:
Step-by-step explanation:
This is a test of a single population mean since we are dealing with mean.
From the information given,
Null hypothesis is expressed as
H0:μ=1.5
The alternative hypothesis is expressed as
Ha:μ>1.5
This is a right tailed test
The decision rule is to reject the null hypothesis if the significance level is greater than the p value and accept the null hypothesis if the significance level is less than the p value.
p value = 0.052
Significance level, α = 0.05
Since α = 0.05 < p = 0.052, the true statement would be
At the α=0.05 significance level, you have proven that H0 is true. B.
Final answer:
The P-value of 0.052 provides some evidence against the null hypothesis H0, although not enough to reject it at the α = 0.05 significance level. A study with a larger sample size may be worthwhile to investigate the true mean nicotine content further.
Explanation:
With a P-value of 0.052 for the hypothesis test, comparing it against a significance level alpha (α) = 0.05 determines whether the null hypothesis (H0) is rejected or not. Despite the P-value being slightly above the significance level, the correct interpretation is Option C: 'There is some evidence against H0, and a study using a larger sample size may be worthwhile.' This suggests that while there isn't enough evidence to reject the null hypothesis at α = 0.05, the result is close enough to warrant further investigation.
We cannot claim that the null hypothesis has been proven true as statistical hypothesis testing never proves a hypothesis, but only provides evidence against it (Option A is incorrect). The data suggests that further investigation could be useful, instead of being unfruitful (Option B is an incorrect view). Lastly, there is evidence pointing towards the alternative hypothesis (Ha), it's not that there's no evidence at all (Option D is incorrect).
α = 0.05Decision: Do not reject null hypothesis.Reason for decision: p-value > αConclusion: There is insufficient evidence to conclude that the mean nicotine content is higher than 1.5 at the 5 percent level, but further research might be beneficial.
Primary Trigonometric ratios are
used for right angled triangles
only.
True
False
f(n)= 64 + 6n complete the recursive formula of f(n)
We want to find a recursive formula for f(n) = 64 + 6n
The recursive formula is f(n) = f(n - 1) + 6
First, a recursive formula is a formula that gives the value of f(n) in relation to the value of f(n - 1) or other previous terms on the sequence.
We know that:
f(n) = 64 + 6n
f(n - 1) = 64 + 6*(n - 1) = 64 + 6*n - 6
Then we can rewrite:
f(n) = 64 + 6n + 6 - 6
f(n) = ( 64 + 6n - 6) + 6
And the thing inside the parenthesis is equal to f(n - 1)
Then we have:
f(n) = f(n - 1) + 6
This is the recursive formula we wanted to get.
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Let A= | 1 1 1,1 4 5, 1 5 6 | and D= | 7 0 0 ,0 4 0, 0 0 2 | .
Compute AD and DA.
Explain how the columns or rows of A change when A is multiplied by D on the right or on the 1 5 6 0 0 2 left.
Find a 3x3 matrix B, not the identity matrix or zero matrix, such that AB -BA.
Compute AD AD- Compute DA. DA Explain how the columns or rows of A change when A is multiplied by D on the right or on the left.
Choose the correct answer below
A. Right multiplication that is, multiplication on the right by the diagonal matrix D multiplies each row of A by the corresponding diagonal entry of D
B. Both right multiplication that is, multiplication on the right and left multiplication by the diagonal matrix D multiplies each row entry of A by the 0
C. Both right multiplication that is, multiplication on the right and left multiplication by the diagonal matrix D multiplies each column entry of A by the ?
D. Right multiplication that is, multiplication on the right by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D Left-multiplication by D multiplies each column of A by the corresponding diagonal entry of D corresponding diagonal entry of D corresponding diagonal entry of D Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D Find a 3x3 matrix B, not the identity matrix or zero matrix, such that AB BA. Choose the correct answer below There is only one unique solution. B= ?
A. Simplitfy your answers.)
B. There are infinitely many solutions. Any multiple of 13 will satisfy the expression °
C. There does not exist a matrix, B, that will satisfy the expression
Answer:
Check the explanation
Step-by-step explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
After performing the calculations, the right matrix multiplication by D changes each column of A and the left matrix multiplication by D changes each row of A. Also, there are an infinite number of 3x3 matrices that satisfy the condition AB ≠ BA
Explanation:To compute the products AD and DA, we first need to understand how matrix multiplication works. Given matrices A and D as presented, it is important to understand that when A is multiplied by D on the right side (AD), each column of A is multiplied by the corresponding diagonal entry of D. Conversely, when A is multiplied by D on the left side (DA), each row of A is multiplied by the corresponding diagonal entry of D. This is true for any two matrices you want to multiply. Therefore, the correct answer to the first part of the question is statement D.
For the second part of the question, we're seeking a non-identity, non-zero 3x3 matrix B such that AB is not equal to BA. This is generally quite difficult and requires some trial and error. There are an infinite number of matrices B that satisfy the condition AB ≠ BA. The exact form of B depends on A and can be quite complex, but B isn't necessarily unique. Therefore, the correct answer to this question is B: There are infinitely many solutions.
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The number of withdrawals a bank processes in a day follows a random variable X. The number of deposits in a day is represented by Y. X and Y are independent and have the following moment generating functions \displaystyle {M}_{X}(t)=\frac{p}{1-q{e}^{t}} \displaystyle {M}_{Y}(t)={\left(\frac{p}{1-q{e}^{t}}\right)}^{r} q = 1 - p What is the expected number of transactions in a day?
Answer:
Check the explanation
Step-by-step explanation:
Number of transactions in a day is sum of number of withdrawals and number of deposits. So,
Number of transactions in a day, Z = X + Y
Moment Generating function of Z is,
T+1
Expected number of transactions in a day = E[Z]
[tex]= \frac{\mathrm{d} }{\mathrm{d} t}_{t=0}M_Z(t) = (r+1)\left ( \frac{p}{1-qe^t} \right )^{r} * \frac{-p}{(1-qe^t)^2} * (-qe^t) for t = 0[/tex]
[tex]= (r+1)\left ( \frac{p}{1-qe^0} \right )^{r} * \frac{-p}{(1-qe^0)^2} * (-qe^0)[/tex]
[tex]= (r+1)\left ( \frac{p}{1-q} \right )^{r} * \frac{-p}{(1-q)^2} * (-q)[/tex]
[tex]= (r+1)\left ( \frac{p}{p} \right )^{r} * \frac{-p}{(p)^2} * (-q)[/tex]
[tex]= \frac{(r+1)q}{p}[/tex]
The A.C. Nielsen Company collected data on the weekly TV viewing times, in hours, of 200 people. Suppose that the sample mean is 30.25, the sample standard deviation is 12.60, and that the histogram of the viewing times is bell shaped. Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85
Answer:
By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed(bell-shaped) random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 30.25
Standard deviation = 12.60
Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85
17.65 = 30.25 - 1*12.60
So 17.65 is one standard deviation below the mean.
42.85 = 30.25 + 1*12.60
So 42.85 is one standard deviation above the mean
By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 39 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
Answer:
We conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.
Step-by-step explanation:
We are given that an article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees.
You conduct a survey of 39 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2.
Let [tex]\mu[/tex] = average time taken to finish their undergraduate degrees.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 4.5 years {means that the mean time taken to finish undergraduate degrees is equal to 4.5 years}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 4.5 years {means that the mean time taken to finish undergraduate degrees is longer than 4.5 years}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean time = 5.1 years
s = sample standard deviation = 1.2 years
n = sample of students = 39
So, the test statistics = [tex]\frac{5.1-4.5}{\frac{1.2}{\sqrt{39} } }[/tex] ~ [tex]t_3_8[/tex]
= 3.122
The value of t test statistics is 3.122.
Now, at 1% significance level the t table gives critical value of 2.429 at 38 degree of freedom for right-tailed test.
Since our test statistic is more than the critical value of t as 3.122 > 2.429, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.
Among 12 metal parts produced in a machine shop, 3 are defective.
Ok, what's the full question here
Anything helps!!! :) I need this ASAP!
Answer:
200.96 in^2
Step-by-step explanation:
We want to find the area
A = pi r^2
A = pi (8)^2
A = 3.14 (64)
A =200.96
HELP
A family’s lunch bill is $10.19 before tax and tip. Using the percents shown for sales tax and gratuity, how much money should the family pay if the gratuity is calculated after tax?
$11.80
$12.16
$12.28
$12.36
Answer:
12.36
Step-by-step explanation:
A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average weight of turkeys sold in Charlotte. According to the chain's researcher, a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds. And a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds. Use a 5% level of significance to determine whether there is a difference in the mean weight of turkeys sold in these two cities. Assume the population variances are approximately the same and use the pooled t-test
Answer:
Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom
The null hypothesis is accepted .
Assume the population variances are approximately the same
Step-by-step explanation:
Explanation:-
Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds
The first sample size 'n₁'= 20
mean of the first sample 'x₁⁻'= 17.53 pounds
standard deviation of first sample S₁ = 3.2 pounds
Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds
The second sample size n₂ = 24
mean of the second sample "x₂⁻"= 14.89 pounds
standard deviation of second sample S₂ = 2.7 pounds
Null hypothesis:-H₀: The Population Variance are approximately same
Alternatively hypothesis: H₁:The Population Variance are approximately same
Level of significance ∝ =0.05
Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42
Test statistic :-
[tex]t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }[/tex]
where [tex]S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}[/tex]
[tex]S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}[/tex]
substitute values and we get S² = 40.988
[tex]t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}[/tex]
t = 1.3622
Calculated value t = 1.3622
Tabulated value 't' = 2.081
Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom
Conclusion:-
The null hypothesis is accepted
Assume the population variances are approximately the same.
A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U.S. cities. The underlying hypothesis is that average commute times are different across cities. To test the hypothesis, the researcher randomly selects six people from each of the four cities and records their one-way commute times to work.
Refer to the below data on one-way commute times (in minutes) to work. Note that the grand mean is 36.625.
Houston Charlotte Tucson Akron
45 25 25 10
65 30 30 15
105 35 19 15
55 10 30 10
85 50 10 5
90 70 35 10
74.167 36.667 24.833 10.833
524.167 436.667 82.167 14.167
Based on the sample standard deviation, the one-way ANOVA assumption that is likely not met is _____________.
A) the populations are normally distributed
B) the population standard deviations are assumed to be equal
C) the samples are independent
D) None of these choices is correct
Answer:
B
Step-by-step explanation:
For one-way ANOVA test results to be reliable, it has to be assumed that standard deviations of populations are equal. Here standard deviations of populatinos are not mentioned.
A histogram titled Number of texts has time on the x-axis and texts on the y-axis. From 6 a m to 7:59 a m there were 15 texts, 8 to 9:59 a m: 5, 10 a m to 11:59 p m: 0, 12 p m to 1:59 p m: 0, 2 p m to 3:59 p m: 0, 4 to 5:59 p m: 29, 6 to 7:59 p m: 19, 8 to 8:59 p m, 14, 10 p m to 11:59 a m: 5 The histogram shows the number of text messages sent by two high school juniors on one Monday. Which statement most reasonably explains the hours when 0 texts were sent?
Answer: I think c,e
Step-by-step explanation:Sorry if it wrong
Answer:
C
Step-by-step explanation:
IM DORA
Solve for x in the diagram below.
Answer:
x = 12
Step-by-step explanation:
20° + (6x - 2) = 90°
18° + 6x = 90°
6x = 90° - 18°
6x = 72°
x = 72°/6
x = 12
Answer:
Hope it helps buddy! Mark as brainlist please.