Answer:
Step-by-step explanation:
Hello!
You need to test the hypothesis that the slope of the regression is cero.
I've run in the statistic software the given data for Y and X and estimated the regression line:
Yi= 7.82 -1.60Xi
Where
a= 7.82
b=-1.60
Sb= 3.38
The hypothesis is:
H₀: β = 0
H₁: β ≠ 0
α: 0.05
This is a two-tailed test, the null hypothesis states that the slope of the regression is cero, this means that if the null hypothesis is true, there is no linear regression between Y and X.
The statistic for this test is a Student-t
t= b - β ~t[tex]_{n-2}[/tex]
Sb
The critical values are:
Left: [tex]t_{n-2; \alpha /2} = t_{2; 0.025} = -4.303[/tex]
Right: [tex]t_{n-2; \alpha /2} = t_{2; 0.975} = 4.303[/tex]
t= -1.60 - 0 = -0.47
3.38
the p-value is also two-tailed, you can calculate it by hand:
P(t ≤ -0.47) + (1 - P(t ≤ 0.47) = 0.3423 + (1 - 0.6603) =0.6820
With the level of significance of 5%, the decision is to not reject the null hypothesis. This means that the slope of the regression is equal to cero, i.e. there is no linear regression between the two variables.
I hope this helps!
The hypothesis test, utilizing a t-statistic of 3.002 and a p-value of 0.00765, rejects the null hypothesis, demonstrating a statistically significant linear relationship. The slope coefficient (β₁ = 1.637) suggests a moderately strong relationship.
Hypothesis Test for a Linear Relationship
In this hypothesis test, we are trying to determine whether there is a statistically significant linear relationship between the independent and dependent variables. The null hypothesis (H₀) states that there is no linear relationship (β₁ = 0), while the alternative hypothesis (H₁) states that there is a linear relationship (β₁ ≠ 0).
The test statistic used in this case is the t-statistic, which is calculated as the ratio of the estimated slope (β₁) to its standard error. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic, assuming that the null hypothesis is true.
In this particular example, the t-statistic is 3.002 and the p-value is 0.00765. The level of significance, which is the threshold for rejecting the null hypothesis, is typically set at 0.05.
Since the p-value (0.00765) is less than the level of significance (0.05), we reject the null hypothesis. This means that we have statistically significant evidence to conclude that there is a linear relationship between the independent and dependent variables. In other words, the slope of the regression line is not equal to zero, indicating that there is a change in the dependent variable as the independent variable changes.
Interpretation of Results
The rejection of the null hypothesis in this case indicates that there is a statistically significant linear relationship between the independent and dependent variables. This means that we can be confident that the observed relationship is not due to chance. The magnitude of the t-statistic (3.002) suggests that the relationship is moderately strong.
The interpretation of the slope coefficient (β₁) is that for every one-unit increase in the independent variable, the dependent variable is expected to increase by 1.637 units, on average. The standard error (0.5453) indicates the variability of the slope estimate.
In conclusion, the hypothesis test provides strong evidence to support the existence of a statistically significant linear relationship between the independent and dependent variables. The magnitude of the slope coefficient indicates that the relationship is moderately strong.
Learn more about hypothesis test, here:
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The advertised claim for batteries for cell phones is set at 48 operating hours with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours.
Do these experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period (assuming proper charging procedures were followed)?
Use a hypothesis testing procedure with α = 0.01. State H0 and H1, test statistic, critical value(s), critical (rejection) region, p-value, and conclusion. Hint: Here, H1 is something against what the company claims.
Final answer:
The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
Explanation:
To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.
H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.
H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.
We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.
The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.
The critical (rejection) region is z < -2.33.
The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 6.0 percent and a standard deviation of 9.9 percent. a. What is the approximate probability that your return on these bonds will be less than −3.9 percent in a given year?
Answer:
0.1587
Step-by-step explanation:
Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 = -1. Therefore, we are looking for P(Z < -1) = 0.1587
In a multiple-choice question, a student is asked to match 3 dates withIn a multiple-choice question, a student is asked to match 3 dates with 3 events. What is the probability that sheer guessing will produce 3 correct answers? g 3 events.
Answer:
P = 1/6 or 0.1667
Step-by-step explanation:
Let A,B and C be the three events and 1,2 and 3 be the three dates. The sample space for this problem is:
{A1,B2,C3} {A1,B3,C2} {A2,B1,C3} {A2,B3,C1} {A3,B1,C2} {A3,B2,C1}
There are six possible outcomes and since there is only one correct answer, the possibility of guessing all 3 questions right is:
P = 1/6
P = 0.1667
The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be K = 9×107 kg, and k = 0.75 per year. (a) If y(0) = 2×107 kg, find the biomass a year later. (Round your answer to two decimal places.) ×107 kg (b) How long will it take for the biomass to reach 4×107? (Round your answer to two decimal places.) years
Answer:
398.411
Step-by-step explanation:
Explanation has been given in the following attachments.
Final answer:
(a) Using the differential equation[tex]\( dy/dt = ky(1 - y/K) \)[/tex] with[tex]\( y(0) = 2 \times 10^7 \)[/tex] kg, the biomass after one year is approximately [tex]\( 3.11 \times 10^7 \) kg.[/tex]
(b) Solving the equation for [tex]\( y(t) = 4 \times 10^7 \) kg yields \( t \approx 3.41 \) years.[/tex]
Explanation:
(a) To find the biomass a year later, we can use the given initial condition and the provided differential equation. Substituting[tex]\( y(0) = 2 \times 10^7 \) kg, \( k = 0.75 \), and \( K = 9 \times 10^7 \)[/tex] kg into the equation, we get:
[tex]\[ \frac{{dy}}{{dt}} = 0.75y \left(1 - \frac{{y}}{{9 \times 10^7}}\right) \][/tex]
Now, we can solve this first-order ordinary differential equation. One method is the separation of variables, then integration. Integrating both sides gives:
[tex]\[ \int \frac{{dy}}{{y(1 - \frac{{y}}{{9 \times 10^7}})}} = \int 0.75 dt \][/tex]
After solving the integrals and applying the initial condition, we find [tex]\( y(1) \approx 3.11 \times 10^7 \) kg.[/tex]
(b) To determine how long it takes for the biomass to reach [tex]\( 4 \times 10^7 \) kg,[/tex] we can use the same differential equation and solve for [tex]\( t \)[/tex] when [tex]\( y(t) = 4 \times 10^7 \) kg.[/tex]This involves solving a separable differential equation and then finding the time [tex]\( t \)[/tex] that satisfies this condition. By solving this equation, we find[tex]\( t \approx 3.41 \) years.[/tex]
In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is _________?
Answer: The 95% confidence interval estimate of the population mean is (1760, 1956) .
Step-by-step explanation:
Formula for confidence interval for population mean([tex](\mu)[/tex]) :
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= Sample size
[tex]\overline{x}[/tex] = sample mean.
[tex]z^*[/tex] = Two-tailed critical z-value
[tex]\sigma[/tex] = population standard deviation.
By considering the given information, we have
n= 81
[tex]\sigma=450 [/tex] kilowatt-hours.
[tex]\overline{x}=1858[/tex] kilowatt-hours.
By using the z-value table ,
The critical values for 95% confidence interval : [tex]z^*=\pm1.960[/tex]
Now , the 95% confidence interval estimate of the population mean will be :
[tex]1858\pm (1.960)\dfrac{450}{\sqrt{81}}\\\\=1858\pm(1.960)\dfrac{450}{9}=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)[/tex]
Hence, the 95% confidence interval estimate of the population mean is (1760, 1956) .
A boat tour company finds that if the price p , charged for an one-hour harbor tour, is $20, the average number of passengers per week, x , is 300. When the price is reduced to $18, the average number of passengers per week increases to 360.The demand and supply curves of a certain brand of running shoes are given by
p=0(x)=1 1 2-0.04x, and p-S(x)-0.06x+42, where p is the price in dollars and x is the quantity sold.
(a) Assuming that the demand curve p D(x) is linear, find its formula: P-D(x)
(b) Assuming that the equilibrium price is $22 per tour, the equilibrium demand is E
(c) The consumers' surplus at that demand is $ . (Use a fraction for the coefficient.)
Answer
The answer and procedures of the exercise are attached in a microsoft word document.
Explanation
Please consider the data provided by the exercise. If you have any question please write me back. All the exercises are solved in a single sheet with the formulas indications.
Answer:
a) p(x) = (-1/30)*x + 30
b) xe = 240
c) The consumers' surplus at that demand is $ 960.00
Step-by-step explanation:
Given info
p₁ = 20
x₁ = 300
p₂ = 18
x₂ = 360
a) We can apply the following equation
p - p₁ = m*(x - x₁)
where
m is the slope, which can be obtained as follows
m = (p₂-p₁) / (x₂-x₁)
⇒ m = (18-20) / (360-300) = -2/60 = -1/30
then
p-20 = (-1/30)*(x-300)
⇒ 30*p - 600 = -x + 300
⇒ p(x) = (-1/30)*x + 30
b) If p(x) = 22
⇒ 22 = (-1/30)*x + 30
⇒ xe = 240 passengers
c) If x = 0 ⇒ p(0) = (-1/30)*(0) + 30 = 30
then
we can get h as follows
h = p(0) - p(240) = 30 - 22 = 8
if b = x(equlibrium) = 240
we apply
A = b*h/2 ⇒ A = 240*8/2 = 960
The consumers' surplus at that demand is $ 960.00
In 1992 there was an earthquake at Little Skull Mountain, Nevada, measuring 5.5 on the Richter scale. In 1994 there was an earthquake near Double Spring Flat, Nevada, measuring 6.0 on the Richter scale. How did the power of the Double Spring Flat quake compare with that of the Little Skull Mountain quake? (Round your answer to two decimal places.)
g SupposeXis a Gaussian random variable with mean 0 and varianceσ2X. SupposeN1is a Gaussian random variable with mean 0 and varianceσ21. SupposeN2is a Gaussianrandom variable with mean 0 and varianceσ22. AssumeX,N1,N2are all independentof each other. LetR1=X+N1R2=X+N2.(a) Find the mean ofR1andR2. That is findE[R1] andE[R2].(b) Find the correlationE[R1R2] betweenR1andR2.(c) Find the variance ofR1+R2.
a. [tex]X[/tex], [tex]N_1[/tex], and [tex]N_2[/tex] each have mean 0, and by linearity of expectation we have
[tex]E[R_1]=E[X+N_1]=E[X]+E[N_1]=0[/tex]
[tex]E[R_2]=E[X+N_2]=E[X]+E[N_2]=0[/tex]
b. By definition of correlation, we have
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}[/tex]
where [tex]\mathrm{Cov}[/tex] denotes the covariance,
[tex]\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])][/tex]
[tex]=E[R_1R_2]-E[R_1]E[R_2][/tex]
[tex]=E[R_1R_2][/tex]
[tex]=E[(X+N_1)(X+N_2)][/tex]
[tex]=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2][/tex]
Because [tex]X,N_1,N_2[/tex] are mutually independent, the expectation of their products distributes over the factors:
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2][/tex]
[tex]=E[X^2][/tex]
and recall that variance is given by
[tex]\mathrm{Var}[X]=E[(X-E[X])^2][/tex]
[tex]=E[X^2]-E[X]^2[/tex]
so that in this case, the second moment [tex]E[X^2][/tex] is exactly the variance of [tex]X[/tex],
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2[/tex]
We also have
[tex]{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2[/tex]
and similarly,
[tex]{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2[/tex]
So, the correlation is
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}[/tex]
c. The variance of [tex]R_1+R_2[/tex] is
[tex]{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2][/tex]
[tex]=\mathrm{Var}[2X+N_1+N_2][/tex]
[tex]=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2][/tex]
[tex]=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2[/tex]
I need help with 3 please!
Answer:
Step-by-step explanation:
PQ=24
PS=19
PR=42
TQ=10
QR=19
SR=24
PT=21
SQ=20
m∠QRS=180-m∠PQR=180-106=74°
m∠PQS=m∠QSR=49°
m∠RPS=m∠PRQ=m∠QRS-m∠PRS=74-35=39°
m∠PSQ=m∠RQS=m∠PQR-m∠PQS=106-49=57°
m∠PQR=106°
m∠ QSR=49°
M∠PRS=35°
Find the value of x.
A. 4
B. 5
C. 6
D. 9
Answer:
C. 6
Step-by-step explanation:
You can try the answers to see which satisfies the Pythagorean theorem:
A. 4² +7² = 16+49 ≠ 117
B. 5² +8² = 25 +64 ≠ 117
C. 6² +9² = 36 +81 = 117 . . . . this (C) is the correct choice
and, for completeness, ...
D. 9² +12² = 81 +144 ≠ 117
_____
Working out
You can also actually work the problem. The Pythagorean theorem tells you ...
x² + (x+3)² = (√117)²
2x² +6x +9 = 117
2x² +6x = 108 . . . . subtract 9
x² +3x = 54 . . . . . . divide by 2
x(x +3) = 54 . . . . . . factor
Now, you can compare to your memorized times tables, where you find 6×9 = 54, so you know that x=6.
__
In case times tables are a challenge, you can continue to complete the square:
x² +3x +1.5² = 54 +1.5²
(x +1.5)² = 56.25 = 7.5² . . . . write as squares
x = 7.5 -1.5 = 6 . . . . . . . . . . . square root, subtract 1.5
The value of x is 6.
_____
The other solution to the quadratic equation is x=-9, but negative segment lengths make no sense. We acknowledge, then ignore, that extraneous solution.
The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 50 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.1 pounds, what is the probability that the sample mean will be each of the following? Appendix A Statistical Tablesa. More than 58 poundsb. More than 57 poundsc. Between 55 and 57 poundsd. Less than 53 poundse. Less than 48 pounds
Answer:
[tex]\mu = 56.8[/tex]
[tex]\sigma = 12.1[/tex]
A)what is the probability that the sample mean will be More than 58 pounds
P(x>58)
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{58-56.8}{12.1}[/tex]
[tex]Z=0.09917[/tex]
refer the z table
P(x<58)=0.5359
P(X>58)=1-P(x<58)=1-0.5359=0.4641
Hence the probability that the sample mean will be More than 58 pounds is 0.4641
B)what is the probability that the sample mean will be More than 57 pounds
P(x>57)
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{57-56.8}{12.1}[/tex]
[tex]Z=0.0165[/tex]
refer the z table
P(x<57)=0.5040
P(X>57)=1-P(x<57)=1-0.5040=0.496
Hence the probability that the sample mean will be More than 57 pounds is 0.496
C)what is the probability that the sample mean will be Between 55 and 57 pound
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{57-56.8}{12.1}[/tex]
[tex]Z=0.0165[/tex]
refer the z table
P(x<57)=0.5040
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{55-56.8}{12.1}[/tex]
[tex]Z=-0.1487[/tex]
refer the z table
P(x<55)=0.4443
P(55<x<57)=P9x<57)-P(x<55) =0.5040-0.4443=0.0597
Hence the probability that the sample mean will be Between 55 and 57 pounds is 0.0597
D)what is the probability that the sample mean will be Less than 53 pounds
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{53-56.8}{12.1}[/tex]
[tex]Z=−0.314[/tex]
refer the z table
P(x<53)=0.3783
The probability that the sample mean will be Less than 53 pounds is 0.3783
E)what is the probability that the sample mean will be Less than 48 pounds
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
Substitute the values :
[tex]Z=\frac{48-56.8}{12.1}[/tex]
[tex]Z=−0.727[/tex]
refer the z table
P(x<48)=0.2358
The probability that the sample mean will be Less than 48 pounds is 0.2358
A random sample of 300 CitiBank VISA cardholder accounts indicated a sample mean debt of 1,220 with a sample standard deviation of 840. Construct a 95 percent confidence interval estimate of the average debt of all cardholders.
Answer: (1124.5619, 1315.4381)
Step-by-step explanation:
The confidence interval for population mean[tex](\mu)[/tex] when populatin standard deviation is unknown :-
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]s[/tex] =Sample standard deviation
t* = Critical t-value.
Given : n= 300
Degree of freedom : df = n-1 = 299
[tex]\overline{x}=1220[/tex]
[tex]s=840[/tex]
Confidence interval = 95%
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Using t-distribution table ,
The critical value for 95% Confidence interval for significance level 0.05 and df = 299 : [tex]t^*=t_{\alpha/2,\ df}=t_{0.025,\ 299}=1.9679[/tex]
Then, a 95% confidence interval estimate of the average debt of all cardholders will be :-
[tex]1220\pm (1.9679)\dfrac{840}{\sqrt{300}}[/tex]
[tex]=1220\pm (1.9679)\dfrac{840}{17.3205080757}[/tex]
[tex]=1220\pm (1.9679)(48.4974226119)[/tex]
[tex]\approx1220\pm 95.4381=(1220-95.438,\ 1220+95.438)\\\\=(1124.5619,\ 1315.4381)[/tex]
Hence, a 95% confidence interval estimate of the average debt of all cardholders is (1124.5619, 1315.4381) .
Fifty people in the civilian labor force are randomly selected and the sample average age iscomputed to be 36.4.(a) Find a 90% confidence interval for the mean age, ?, of all people in the civilian laborforce. Assume that the population standard deviation for the ages of civilian labor force is12.1 years. Interpret the confidence interval.(b) It is being claimed that the mean age of the population of civilian labor force is 40. Whatdo you conclude based on the confidence interval?
Answer:
a) The 90% confidence interval would be given by (33.594;39.206)
b) Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.
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".
Part a
[tex]\bar X=36.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=12.1[/tex] represent the population standard deviation
n=50 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]36.4-1.64\frac{12.1}{\sqrt{50}}=33.594[/tex]
[tex]36.4+1.64\frac{12.1}{\sqrt{50}}=39.206[/tex]
So on this case the 90% confidence interval would be given by (33.594;39.206)
Part b
Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.
Statistics Question:
The Correlation coefficient is r = -0.992324879
Step-by-step explanation:
The figure shown is the calculation of all data
Given data :
X y
1 42.2
2 42.14
3 39.38
4 38.22
5 37.46
6 35.2
7 30.24
8 29.38
9 25.92
10 24.86
11 24.8
12 21.04
13 20.98
14 15.42
15 15.46
16 13.4
Step1: Find X mean, Y mean
Mean = [tex]\frac{\sum x }{N}[/tex]
X mean = [tex]\frac{1+2+3+4+5+...+16}{16}[/tex]
X mean = 8.5
Similarly, we get
Y mean = [tex]\frac{42.2+42.14+39.38+...+15.42+15.46+13.4}{16}[/tex]
Y mean = 28.50625
Step2: Find Standard deviation
Standard deviation = [tex]\sqrt {\frac{ \sum (Xi-X mean)^{2} }{N-1}}[/tex]
Sx = [tex]\sqrt {\frac{ (-7.5)^{2} + (-6.5)^{2} +(-5.5)^{2} ....+ (7.5)^{2} }{16-1}}[/tex]
Sx= 4.760952286
Similarly, we get
Sy= 9.765590527
Step3: Find Correlation coefficient
The formula for the Correlation coefficient is given as
[tex]r=\frac{1}{N-1} \sum(\frac{Xi-X mean }{Sx} ) (\frac{Yi- Y mean }{Sy} )[/tex]
[tex]r=\frac{1}{16-1} \sum(\frac{Xi-8.5 }{4.760952286} ) (\frac{Yi- 28.50625 }{9.765590527} )[/tex]
[tex]r= -0.992324879[/tex]
Thus, The Correlation coefficient is r = -0.992324879
A random sample of 157 recent donations at a certain blood bank reveals that 86 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of 0.01. State the appropriate null and alternative hypotheses.
Answer: Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.
Step-by-step explanation:
Since we have given n = 157
x = 86
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{86}{157}=0.55[/tex]
and we have p = 0.4
So, hypothesis would be
[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]
Since there is 1% level of significance.
So, test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.55-0.40}{\sqrt{\dfrac{0.4\times 0.6}{157}}}\\\\z=\dfrac{0.15}{0.039}\\\\z=3.846[/tex]
and the critical value at 1% level of significance , z = 2.58
Since 2.58<3.846.
So, we reject the null hypothesis.
Hence, Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.
We conducted a hypothesis test and found that the actual percentage of type A blood donations significantly differs from 40%, leading to the rejection of the null hypothesis at a 0.01 significance level.
Hypothesis Testing for Blood Type Proportion
We will perform a hypothesis test to determine whether the percentage of type A blood donations differs from 40%. The appropriate hypotheses for this test are:
Null hypothesis (H0): p = 0.40 (The true proportion of type A blood donations is 40%.)Alternative hypothesis (Ha): p ≠ 0.40 (The true proportion of type A blood donations is different from 40%.)Next, we calculate the test statistic for the sample proportion:
Sample proportion (") phat = 86/157 ≈ 0.548Standard error (SE) = √[(0.40 * 0.60) / 157] ≈ 0.039Z-score: (phat - 0.40) / SE ≈ (0.548 - 0.40) / 0.039 ≈ 3.79The critical value for a two-tailed test at a significance level of 0.01 is approximately ±2.576.
Since 3.79 > 2.576, we reject the null hypothesis (H0). Therefore, the data suggests that the actual percentage of type A donations differs significantly from 40%.
How much TV do college students watch? A survey of 361 students recorded the number of hours of television they watched per week. The sample mean was 6.504 hours with a standard deviation of 5.584. The standard error of the mean was 0.294. Find a 90% confidence interval for the population mean. Round your answer to three decimal places
Answer: 90% confidence interval would be (6.022,6.986).
Step-by-step explanation:
Since we have given that
Number of students = 361
Sample mean = 6.504
Sample standard deviation = 5.584
Standard error of the mean = 0.294
At 90% confidence level,
So, α = 0.01
So, z = 1.64
Margin of error is given by
[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]
So, Lower limit would be
[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]
Upper limit would be
[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]
So, 90% confidence interval would be (6.022,6.986).
The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 53 records of automobile driver fatalities in a certain county showed that 35 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.05.
Answer:
The p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.
Step-by-step explanation:
1) Data given and notation n
n=53 represent the random sample taken
X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver
[tex]\hat p=\frac{35}{53}=0.660[/tex] estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver
[tex]p_o=0.77[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:
Null hypothesis:[tex]p\geq 0.77[/tex]
Alternative hypothesis:[tex]p < 0.77[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is an unilateral lower test the p value would be:
[tex]p_v =P(z<-1.903)=0.0285[/tex]
So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.
Juliet runs out of gas in Barnhaven, South Carolina. She walks 7 mi west and then 4 mi south looking for a gas station. How far is she from her starting point?
Answer: she is 8.06 miles from her starting point
Step-by-step explanation:
The diagram in the attached photo describes Juliet's movement from her starting point to her current position.
A triangle ABC is formed
AB = distance that she walked towards west
BC = distance that she walked towards south
AC= x = distance that she is from her starting point
The diagram is a right angle triangle. So we can find the distance that she is from her starting point using Pythagoras theorem.
Hypotenuse^2 = opposite^2 + adjacent^2
Hypotenuse = x
Opposite = 7
Adjacent = 4
x^2 = 7^2 + 4^2
x^2 = 49 + 16 = 65
x = √65 = 8.06 miles
A newsletter publisher believes that over 64%64% of their readers own a Rolls Royce. For marketing purposes, a potential advertiser wants to confirm this claim. After performing a test at the 0.010.01 level of significance, the advertiser decides to reject the null hypothesis. What is the conclusion regarding the publisher's claim?
The conclusion is that there is sufficient evidence to reject the publisher's claim that over 64% of their readers own a Rolls Royce.
When conducting a hypothesis test, there are two competing hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1). In this scenario:
Null hypothesis (H0): The proportion of readers who own a Rolls Royce is equal to or less than 64%.
Alternative hypothesis (H1): The proportion of readers who own a Rolls Royce is greater than 64%.
The significance level, denoted by α, represents the probability of rejecting the null hypothesis when it is actually true. In this case, α = 0.01, indicating that there is a 1% chance of making a Type I error (incorrectly rejecting the null hypothesis).
After conducting the hypothesis test, the advertiser decided to reject the null hypothesis. This decision suggests that the evidence from the test was significant enough to conclude that the true proportion of readers who own a Rolls Royce is greater than 64%.
In summary, based on the test results at the 0.01 level of significance, the conclusion is that the publisher's claim that over 64% of their readers own a Rolls Royce is not supported by the evidence. The data suggest that the proportion may be higher than 64%.
Construct a confidence interval of the population proportion at the given level of confidence. x = 125, n = 250, 90 % confidence.
The 90% confidence interval is ____________?
(Use ascending order. Round to three decimal places as needed.)
Answer: 90% confidence interval would be (0.448, 0.552).
Step-by-step explanation:
Since we have given that
x = 125
n = 250
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{125}{250}=0.5[/tex]
We need to find the 90% confidence interval.
so, z = 1.64
So, interval would be
[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.5\pm 1.64\sqrt{\dfrac{0.5\times 0.5}{250}}\\\\=0.5\pm 0.0519\\\\=(0.5-0.0519,0.5+0.0519)\\\\=(0.448,0.552)[/tex]
Hence, 90% confidence interval would be (0.448, 0.552)
Final answer:
The 90% confidence interval is (0.462, 0.538). To calculate a confidence interval, use the formula p' ± z * √(p'q'/n). In this case, with x = 125, n = 250, and 90% confidence, the interval is (0.462, 0.538).
Explanation:
The 90% confidence interval is (0.462, 0.538).
To construct a confidence interval for a population proportion, you can use the formula for confidence interval: p' ± z * √(p'q'/n), where p' = x/n, q' = 1 - p', and z corresponds to the confidence level.
In this case, given x = 125, n = 250, and 90% confidence level, the confidence interval is (0.462, 0.538).
An appliance dealer sells three different models of upright freezers having 14.5, 16.9, and 19.1 cubic feet of storage space. Let x = the amount of storage space purchased by the next customer to buy a freezer. Suppose that x has the following probability distribution.x p(x)14.5 .216.9 .519.1 .3(a) Calculate the mean and standard deviation of x.(b) If the price of the freezer depends on the size of the storage space, x, such that Price = 25x - 8.5, what is the mean value of the variable Price paid by the next customer?(c) What is the standard deviation of the price paid?
The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.
Explanation:The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.
To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}
The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).
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If WZ is the perpendicular bisector of VY , what conclusion can you make?
VWZ=YWZ
VWZ=WVZ
VYW=YWV
WYZ=ZWV
Answer:
[tex]\triangle VWZ = \triangle YWZ[/tex]. Justification is given below.
Step-by-step explanation:
Given:
[tex]\angle WVZ = \angle WYZ[/tex]
WZ is the perpendicular bisector of VY.
∴[tex]\angle WZV = \angle WZY = 90\°\\and\\VZ = YZ[/tex]
In [tex]\triangle VWZ and \triangle YWZ[/tex]
[tex]\angle WVZ = \angle WYZ[/tex] Given
[tex]VZ = YZ[/tex]
[tex]\angle WZV = \angle WZY = 90\°[/tex]
∴ [tex]\triangle VWZ \cong \triangle YWZ[/tex] by ASA test
[tex]\triangle VWZ = \triangle YWZ[/tex]
Here the correspondence of the vertices of a triangle should be match hence the option is first one that is.
[tex]\triangle VWZ = \triangle YWZ[/tex]
Answer:
Step-by-step explanation:
VWZ=YWZ
. In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an abnormality. Is this strong evidence that the risk has increased?
a. Define the parameter and state the hypotheses.
b. Define the sampling distribution (mean and standard deviation).
c. Perform the test and calculate P-value
d. State your conclusion.
e. Explain what the p-value means in this context.
Answer:
Reject null hypothesis
Step-by-step explanation:
a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]
(Right tailed test at 5% level)
p = risk proportion
Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]
p difference = [tex]0.1198-0.05=0.0698[/tex]
b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698
c) Z = test statistic = p diff/std error
= 6.29
p value <0.05
d) Since p < alpha we reject null hypothesis.
e) The probability that null hypothesis is rejected when true is negligible =p value
Answer:
Step-by-step explanation:
Reject null hypothesis
Consider a triangle with vertices at (0,0) and (x1, 71) and (x2, 92). a. Use the interpretation of a determinant as area of a parallelogram to explain why the following formula calculates the area of the triangle. (Hint: A diagonal of a parallelogram cuts the parallelogram into two triangles of equal size A =\det (i 2) b. Set (x1, y.) = (1,2) and (x2,Y2) = (3,-1). Use the formula to calculate the area of the triangle.
I can't really read the formula so I'll give a lecture.
With two vectors we can make a parallelogram. If the vectors are u and v, and one vertex is the origin O, the others are O+u, O+v, and O+v+u. If we draw any diagonal of the parallelogram, that divides it into two congruent triangles. So each triangle is half the area of parallelogram.
The signed area of the parallelogram is given by the two D cross product of the vectors, aka the determinant. So the area of a triangle with vertices (0,0), (a,b) and (c,d) is
[tex]A = \frac 1 2 |ad - bc| = \left| \frac 1 2 \begin{vmatrix} a & b \\ c & d\end{vmatrix} \right|[/tex]
Applying that to vertices
[tex](0,0), (x_1, 71), (x_2, 92)[/tex]
we get area
[tex]A = \frac 1 2 |\ 92x_1 - 71x_2|[/tex]
That formula is accurate for any values of x₁ and x₂
b. (1,2),(3,-1)
[tex]A=\frac 1 2|1(-1)-2(3)| = \frac 1 2 |-7| = \frac 7 2[/tex]
Answer: 7/2
A group of 24 people have found 7.2 kg of gold. Assuming the gold is divided evenly, how much gold will each one get in grams?
Answer:
Step-by-step explanation:
Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300
The total resistance R of two resistors connected in parallel circuit is given by 1/R = 1/R_1 + 1/R_2. Approximate the change in R as R_1 is decreased from 12 ohms to 11 ohms and R_2 is increased from 10 ohms to 11 ohms. Compute the actual change.
Answer:
a) Approximate the change in R is 0.5 ohm.
b) The actual change in R is 0.04 ohm.
Step-by-step explanation:
Given : The total resistance R of two resistors connected in parallel circuit is given by [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
To find :
a) Approximate the change in R ?
b) Compute the actual change.
Solution :
a) Approximate the change in R
[tex]R_1=12\ ohm[/tex] and [tex]R_2=10\ ohm[/tex]
[tex]R_1[/tex] is decreased from 12 ohms to 11 ohms.
i.e. [tex]\triangle R_1=21-11=1\ ohm[/tex]
[tex]R_2[/tex] is increased from 10 ohms to 11 ohms.
i.e. [tex]\triangle R_2=11-10=1\ ohm[/tex]
The change in R is given by,
[tex]\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}[/tex]
[tex]\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}[/tex]
[tex]\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}[/tex]
[tex]\triangle R=\frac{(1)(1)}{1+1}[/tex]
[tex]\triangle R=\frac{1}{2}[/tex]
[tex]\triangle R=0.5\ ohm[/tex]
b) The actual change in Resistance
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R}=\frac{1}{12}+\frac{1}{10}[/tex]
[tex]R=\frac{10\times 12}{10+12}[/tex]
[tex]R=\frac{120}{22}[/tex]
[tex]R=5.46\ ohm[/tex]
When resistances are charged, [tex]R_1=R_2=11[/tex]
[tex]\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}[/tex]
[tex]R'=\frac{11}{2}[/tex]
[tex]R'=5.5\ ohm[/tex]
Change in resistance is given by,
[tex]C=R'-R[/tex]
[tex]C=5.5-5.46[/tex]
[tex]C=0.04\ ohm[/tex]
Final answer:
The actual change in total resistance R of a parallel circuit as R_1 is decreased from 12 ohms to 11 ohms and R_2 is increased from 10 ohms to 11 ohms is calculated using the formula 1/R = 1/R_1 + 1/R_2.
Explanation:
The question is asking for the change in total resistance R of a parallel circuit when the resistances R_1 and R_2 are changed from 12 ohms to 11 ohms and 10 ohms to 11 ohms, respectively. To calculate the actual change, we can use the formula:
1/R = 1/R_1 + 1/R_2
Before the change, the total resistance is:
1/R_initial = 1/12 + 1/10
After the change, it becomes:
1/R_final = 1/11 + 1/11
By calculating both 1/R_initial and 1/R_final, we determine R_initial and R_final separately and find the difference between them to get the actual change in resistance.
According to the Census Bureau, 3.36 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.71 residents. The standard deviation of this sample was 1.10 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.36 persons? State the null hypothesis and the alternate hypothesis
Answer:
We conclude that the mean number of residents in the retirement community household is less than 3.36 persons.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 3.36
Sample mean, [tex]\bar{x}[/tex] = 2.71
Sample size, n = 25
Alpha, α = 0.10
Sample standard deviation, s = 1.10
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 3.36\text{ residents per household}\\H_A: \mu < 3.36\text{ residents per household}[/tex]
We use One-tailed(left) t test to perform this hypothesis.
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{2.71 - 3.36}{\frac{1.10}{\sqrt{25}} } = -2.95[/tex]
Now, [tex]t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } =-1.31[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We reject the null hypothesis and fail to accept it.
Thus, we conclude that the mean number of residents in the retirement community household is less than 3.36 persons.
The null hypothesis for testing if the mean number of residents in Arizona retirement community households is less than the national average is that the mean is equal to or greater than 3.36 residents. The alternate hypothesis is that it is less than 3.36 residents. A one-sample t-test would typically be used to test these hypotheses.
Conducting a hypothesis test to determine if the mean number of residents in retirement community households in Arizona is significantly less than the national average reported by the Census Bureau, which is 3.36 residents per household. To address this,
The null hypothesis (H0) is that the mean number of residents per household in the Arizona retirement communities is equal to or greater than the national average, = 3.36 residents.
The alternate hypothesis (Ha) is that the mean number of residents per household in the Arizona retirement communities is less than the national average, < 3.36 residents.
To test this hypothesis at a 0.10 significance level, we would typically use a one-sample t-test to compare the sample mean (2.71 residents) against the national average (3.36 residents) considering the sample standard deviation (1.10 residents) and sample size (25 households).
The hypothesis test would determine if the observed difference is statistically significant, implying that the mean number of residents in retirement community households is indeed less than 3.36 persons.
Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1,2,-5). Also, deternine if the force is conservative.
Find the work required to move an object in the fo
Answer:
Work = e+24
F is not conservative.
Step-by-step explanation:
To find the work required to move an object in the force field
[tex]\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})[/tex]
along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.
Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with
r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1
so
r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1
is a parameterization of the segment.
the work W required to move an object in the force field F along the straight line from A to B is the line integral
[tex]\large W=\int_{C}Fdr [/tex]
where C is the segment that goes from A to B.
[tex]\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt[/tex]
Integrating by parts the last integral:
[tex]\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1[/tex]
and
[tex]\large \boxed{W=\int_{C}Fdr=e+24}[/tex]
To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24
Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.
The segment that goes from A to (1,0,0) can be parameterized as
r(t) = (t,0,0) with 0≤ t ≤ 1
so the work required to move the particle from A to (1,0,0) is
[tex]\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1 [/tex]
The segment that goes from (1,0,0) to B can be parameterized as
r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1
so the work required to move the particle from (1,0,0) to B is
[tex]\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}[/tex]
Hence, the work required to move the particle from A to B along D is
e - 1 + (25e)/2 = (27e)/2 -1
since this result differs from e+24, the force field F is not conservative.
Suppose that 250 students take a test, of which 100 are sophomores, 140 are juniors and 110 are seniors. Let X be the random variable representing the score of each student on the test. Suppose that the sophomores have an average of 60 juniors have an average of 67 and the seniors have an average of 77. Find E(X) using conditioning. Use the appropriate notation to explain your answer.
Answer:
95.4
Step-by-step explanation:
The overall average score would be the total score divided by the total number of students. The total score is the sum of the product of the average of each student body and their average score
[tex]E(x) = \frac{E_1n_1 + E_2n_2+E_3n_3}{n_1+n_2+n_3}[/tex]
[tex]E(x) = \frac{100*60+140*67+110*77}{250} = \frac{23850}{250} = 95.4[/tex]
In a competition between players X and Y, the first player to win three games in a row or a total of four games wins. How many ways can the competition be played if X wins the first game and Y wins the second and third games? (Hint: Draw a tree.)
Answer:
The competition can be played in 7 different ways
Step-by-step explanation:
First game : X
Second game : Y
Third game : Y
After that either X or Y can win
¹Case - 1: Y wins the fourth game:The competition ends with Y winning as he/she won 3 games in a row
Case - 2: X wins the fourth game:more games are required to decide the winner
Sub-case - a: X wins the fifth game:more games are required to decide the winner
²Sub-sub-case - i: X wins the sixth game:The competition ends with X winning as he/she won 3 games in a row
³⁻⁴Sub-sub-case - ii: Y wins the sixth game:The competition ends after the seventh game. Whoever wins seventh game wins the competition as he/she would've won 4 games in total by then.
Sub-case - b: Y wins the fifth game:more games are required to decide the winner
⁵Sub-sub-case - i: Y wins the sixth game:The competition ends with Y winning as he/she won 4 games in total
⁶⁻⁷Sub-sub-case - ii: X wins the sixth game:The competition ends after the seventh game. Whoever wins seventh game wins the competition as he/she would've won 4 games in total by then.
∴The competition can be played in 7 different ways
The problem requires combinatorial analysis to find all potential sequences of wins leading to player X or Y's victory. There are different scenarios of victory based on the number of games player X and Y wins and the sequence of their wins.
Explanation:This problem showcases combinatorial analysis where we consider the different sequences of wins that leads either team to victory. In the given scenario X has already won the first game and Y has won the second and third games.
From this point, some possible winning combinations could be:
X winning next three games in a rowY winning next game, X winning three in a rowY winning next two games, X winning next three gamesY winning next game, X winning the next one, then Y wins, followed by another X winX wins next game, Y wins the two after, and X wins the next twoThese are representative of the different sequences of wins that could occur. To calculate the total number of possible sequences, you would consider the different ways the remaining games can unfold until one player satisfies the win conditions.
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