Answer:
I guess that the geometric sequence is:
An = 2*5^(n - 2)
we usually have that that a geometric sequence is something like:
A*r^n.
so i will write the first two terms of our sequence separated, and i will calculate the summation of the first seven terms after n = 2.
A0 = 2*5^(-2) = 2/25
A1 = 2*5^-1 = 2/5
Then, the sum for a geometric sequence is:
SN = A(1 - r^N)/(1 - r)
in this case, A = 2 and r = 5, and i will use N = 7 for the sequence 2*5^n
S7 = 2*(1 - 5^7)/(1 - 5) = 39,062.
now we add the first terms that we obtained before and get:
S9 = 2/25 + 2/5 + 39,062 = 39,062.48
6. Create a probability distribution for a coin flipping game. That is, toss a coin at least 25 times and keep up with the number of heads and the number of tails. (8 points for each part) a. Compile your data into a probability distribution. Be sure to show that your distribution meets the properties for a probability distribution. Trial 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Res. T H H H T H H H H T H H T H T H T T H T T H H T H H 15/25=3/5 T 10/25=2/5 3/5+2/5=1 Can anyone help with part a im lost!
Answer:
Event Probability
Heads [tex]\dfrac{3}{5}[/tex]
Tails [tex]\dfrac{2}{5}[/tex]
Step-by-step explanation:
If a coin is tossed 25 times, then from the given table it is clear that
Number of heads = 15
Number of tails = 10
Total number of tosses = 15+10 = 25
We know that,
[tex]Probability=\dfrac{\text{Favorable outcomes}}{\text{Total number of outcomes}}[/tex]
[tex]P(H)=\dfrac{\text{Number of heads}}{\text{Total number of tosses}}=\dfrac{15}{25}=\dfrac{3}{5}[/tex]
[tex]P(T)=\dfrac{\text{Number of tails}}{\text{Total number of tosses}}=\dfrac{10}{25}=\dfrac{2}{5}[/tex]
So, probability distribution table is
Event Probability
Heads [tex]\dfrac{3}{5}[/tex]
Tails [tex]\dfrac{2}{5}[/tex]
According to the properties for a probability distribution, the sum of probability of all events is 1.
Since,
[tex]\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{3+2}{5}=\dfrac{5}{5}=1[/tex]
Hence, the distribution meets the properties for a probability distribution.
A certain group of test subjects had pulse rates with a mean of 80.9 beats per minute and a standard deviation of 10.7 beats per minute. Use the range rule of thumb to identify the limits separating values that are significantly low or significantly high. Is a pulse rate of 142.3 beats per minute significantly low or significantly high?
Answer:
We want to find the usual limits using the rule of thumb and for this case rule states that we have most of the values within 2 deviations from the mean so then we can find the usual limits like this:
[tex] \mu -2*\sigma = 80.9 -2*10.7= 59.5[/tex]
[tex] \mu +2*\sigma = 80.9 -2*10.7= 102.3[/tex]
And for this case a value of 142.3 is higher than the upper limti so then we can conclude that would be significantly high with the rule of thumb criteria
Step-by-step explanation:
For this case we have the follwing info given:
[tex] \mu = 80.9[/tex] represent the mean
[tex]\sigma = 10.7[/tex] represent the deviation
We want to find the usual limits using the rule of thumb and for this case rule states that we have most of the values within 2 deviations from the mean so then we can find the usual limits like this:
[tex] \mu -2*\sigma = 80.9 -2*10.7= 59.5[/tex]
[tex] \mu +2*\sigma = 80.9 -2*10.7= 102.3[/tex]
And for this case a value of 142.3 is higher than the upper limti so then we can conclude that would be significantly high with the rule of thumb criteria
Q3. Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500 Americans is obtained. In a random sample of 500 Americans, what is the probability that more than 20% do not have health insurance?
Answer:
[tex]P(X > 100.5) = 0.0062 \\\\[/tex]
Therefore, there is 0.0062 probability that more than 20% of Americans do not have health insurance.
Step-by-step explanation:
Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500.
From the above information,
p = 16% = 0.16
n = 500
The mean is given by
[tex]\mu = n \times p \\\\\mu = 500 \times 0.16 \\\\\mu = 80[/tex]
The standard deviation is given by
[tex]\sigma = \sqrt{n \times p(1-p)} \\\\\sigma = \sqrt{500 \times 0.16(1-0.16)} \\\\\sigma = 8.197[/tex]
What is the probability that more than 20% do not have health insurance?
We can use the Normal distribution as an approximation to the Binomial distribution since the following conditions are satisfied.
n×p ≥ 5 (satisfied)
n×(1 - p) ≥ 5 (satisfied)
So the probability is given by
500×0.20 = 100
[tex]P(X > 100) = 1 - P(X < 100)[/tex]
We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).
[tex]P(X > 100.5) = 1 - P(X < 100.5)\\\\P(X > 100.5) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 100.5) = 1 - P(Z < \frac{100.5 - 80}{8.197} )\\\\P(X > 100.5) = 1 - P(Z < \frac{20.5}{8.197} )\\\\P(X > 100.5) = 1 - P(Z < 2.50)\\\\[/tex]
The z-score corresponding to 2.50 is 0.9938
[tex]P(X > 100.5) = 1 - 0.9938\\\\P(X > 100.5) = 0.0062 \\\\P(X > 100.5) = 0.62\% \\\\[/tex]
Therefore, there is 0.0062 probability that more than 20% of Americans do not have health insurance.
PLS HELP. The question is in the photo :)
Answer:
So each strawberry is 4 calories
Step-by-step explanation:
First find the slope of the line
Two points on the line are
(0,0) and (3,12)
m = (y2-y1)/(x2-x1)
= (12-0)/(3-0)
= 12/3
= 4
So each strawberry is 4 calories
C(n)=-6(-1/3) n-1 what’s the 2nd term in the sequence
Answer:
The second term is 3
Step-by-step explanation:
C(n)=-6(-1/3) n-1
The first term of the sequence
C(1) = -6(-1/3)1-1
C(1) = (2)-1
C(1) = 1
C(n)=-6(-1/3) n-1
The second term of the sequence
C(2) = -6(-1/3)2-1
C(2) = -6(-2/3)-1
C(2) = 4-1
C(2) = 3
Answer:
2
Step-by-step explanation:
C(2)= -6(-1/3)^2-1
=2
Simplify.
In e =
In e 2x=
In 1 =
Answer:
ln e = 1
ln e 2x = 2x
ln 1 = 0
Step-by-step explanation:
ln e
ln(2.718282) = 1
In e 2x
ln(2.718282)(2)x = 2x
ln 1 = 0
If 5ex = 300, x = A) ln (300) = 5.703 . . . B) 5 ln (60) = 20.471 . . . C) ln (300)∕5 = 1.140 . . . D) ln (60) = 4.094 . . .
Answer:
D) ln (60) = 4.094...
Step-by-step explanation:
Let's solve for x
We'll be careful because it's exponential.
5ex = 300
ex= 300/5
ex= 60
x = e^-1(60)
Let's note that this expression e^-1 is equal to In
x= In (60)
x = 4.0943
So the value of x is equal to 4.0943
Thank you.
convert 3days to minutes
Answer:
4320 minutes
Step-by-step explanation:
Recall,
1 day ---> 24 hours
but each hour has 60 minutes, hence 1 day can also be expressed:
1 day -----> 24 x 60 = 1440 minutes
3 days -----> 1440 min/day x 3 days = 4320 minutes
Answer: 4,320 minutes
Step-by-step explanation: 1 day = 1440 days. 1440 * 3 = 4,320 minutes
Please answer this correctly
Answer:
50%
Step-by-step explanation:
The numbers that are not odd are 2, 4, and 6 on a dice.
3 numbers out of 6.
3/6 = 1/2 = 0.5
P(not odd)= 50%
Mary invested $200 for 3 years at 5% per annum. John invested $300 at the same rate. If they both received the same amount of interest how much did john invest???
Answer:
Step-by-step explanation:
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 95% of the lead to decay?
Answer:
[tex] N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}[/tex]
Where [tex]t_{1/2}= 3.3 hr[/tex] represent the half life and the intial amount would be [tex] N_o = 1[/tex]
And we want to find the time in order to have a 95% of decay so we can set up the following equation:
[tex] 0.05 = 1 (0.5)^{t/3.3}[/tex]
If we apply natural log on both sides we got:
[tex] ln(0.05) = \frac{t}{3.3} ln (0.5)[/tex]
And solving for t we got:
[tex] t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26[/tex]
So then would takes about 14.26 hours in order to have 95% of the lead to decay
Step-by-step explanation:
For this case we can define the variable of interest amount of Pb209 and for the half life would be given:
[tex] N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}[/tex]
Where [tex]t_{1/2}= 3.3 hr[/tex] represent the half life and the intial amount would be [tex] N_o = 1[/tex]
And we want to find the time in order to have a 95% of decay so we can set up the following equation:
[tex] 0.05 = 1 (0.5)^{t/3.3}[/tex]
If we apply natural log on both sides we got:
[tex] ln(0.05) = \frac{t}{3.3} ln (0.5)[/tex]
And solving for t we got:
[tex] t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26[/tex]
So then would takes about 14.26 hours in order to have 95% of the lead to decay
Given the equation y = 7 sec(6x– 30)
The period is:
The horizontal shift is:
Answer:
The period is of [tex]\frac{\pi}{3}[/tex] units.
The horizontal shift is of 30 units to the left.
Step-by-step explanation:
The secant function has the following general format:
[tex]y = A\sec{(Bx + C)}[/tex]
A represents the vertical shift.
C represents the horizontal shift. If C is positive, the shift is to the right. If it is negative, it is to the left.
The period is [tex]P = \frac{2\pi}{B}[/tex]
In this question:
[tex]y = 7\sec{6x - 30}[/tex]
So [tex]B = 6, C = -30[/tex]
Then [tex]P = \frac{2\pi}{6} = \frac{\pi}{3}[/tex]
The period is of [tex]\frac{\pi}{3}[/tex] units.
The horizontal shift is of 30 units to the left.
a. What is a residual? b. In what sense is the regression line the straight line that "best" fits the points in a scatterplot? a. What is a residual?
Answer:
a. A residual is how far off a point is from the expected value. For example, if I were to estimate the weight of my Southeastern Lubber Grasshopper, I would say it's maybe 5 ounces. But, in reality, it might be 4 ounces. So, the residual would be the reality minus the prediction, or 4 - 5, or -1 ounce.
b. The regression line is the line of predicted values for the points in the scatterplot. It tries to predict the points and make all the points be on the line.
Hope this helps!
To test H0: μ=100 versus H1:≠100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d).(a) If x =104.2 and s=9.6, compute the test statistic.t= _ (Round to three decimal places as needed.)(b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical values.The critical values are __ .(c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in thet-distribution?(d) Will the researcher reject the null hypothesis?
Answer:
a) Test statistic = 1.960
b) The critical values include -2.50 and 2.50.
The critical regions of rejection are thus
t < -2.50 or t > 2.50
c) The sketch of the curve is presented in the attached image to this solution. The shaded parents indicate the rejection regions.
d) The t-statistic obtained (1.96), lies within the acceptance region (-2.50 ≤ x ≤ 2.50), we fail to reject the null hypothesis.
Step-by-step explanation:
a) Test statistic is computed using the expression
t = (x - μ₀)/σₓ
x = Sample mean = 104.2
μ₀ = the standard we are comparing Against
σₓ = standard error of the mean = (σ/√n)
σ = 9.6
n = Sample size = 24
σₓ = (9.6/√24) =
t = (0.425 - 0.35) ÷ 0.07816
t = 1.9595917942 = 1.960
b) To obtain these critical values, we first find the degree of freedom
Degree of freedom = n - 1 = 24 - 1 = 23
The critical values for significance level of 0.01 and degree of freedom of 23 is given as
t(0.01, 23) = 2.50
So, since the test is two-tailled (we are testing in both directions; greater than or less than), the regions of rejection include
t < -2.50 and t > 2.50
c) since the test is two-tailled (we are testing in both directions; greater than or less than), the regions of rejection include
t < -2.50 and t > 2.50
The t-distribution curve is very similar to the normal distribution curve. The t-distribution curve is also a bell shaped curve, but it is heavier at the limits indicating that the t-distribution favours outliers more than the normal distribution.
The sketch of the curve is presented in the attached image with the shaded regions indicating the rejection region.
d) Since the t-statistic obtained (1.96), lies within the acceptance region (-2.50 ≤ x ≤ 2.50), we fail to reject the null hypothesis.
Hope this Helps!!!
Use the distributive property to find an expression equivalent to 6(s -7) -5 (4 - t)
Answer:
6s-62+5t
Step-by-step explanation:
6s-42-20+5t=6s-62+5t
Which expression is the simplest form of -(x + 5) - 3(x + 2)?
Answer:
-4x -11
Step-by-step explanation:
-(x + 5) - 3(x + 2)
Distribute
-x -5 -3x -6
Combine like terms
-x-3x -5-6
-4x -11
Answer:
[tex] = - (4x + 11)[/tex]
Step-by-step explanation:
[tex]-(x + 5) - 3(x + 2) \\ -x - 5 - 3x - 6 \\ -x - 3x -5 - 6 \\ - 4x - 11 \\ = -(4x + 11)[/tex]
AHH!! PLEASE HELP ME IM STUCK :(
Answer:
x = 6
Step-by-step explanation:
Compare the given formula to the equations shown in the attachment. First of all, you see that all of the numbers are scaled by a factor of 4. Removing that gives ...
[tex]r=\dfrac{6.6}{1+1.1\cos{\theta}}=\dfrac{1.1\cdot6}{1+1.1\cos{\theta}}[/tex]
This matches the formula for the hyperbola with e=1.1 and d=6, for a directrix of x = 6.
the distance around the edge of a circular pond is 88m. the radius in meters is ?
(a)88π
(b)176π
(c)88/π
(d)88/2π
Answer: (d) 88/ 2π
Step-by-step explanation:
Perimeter = 88m
Perimeter of a circle = 2πr
88 = 2π x r
r = 88 / 2π
Answer:
88/2π = r
Step-by-step explanation:
The circumference is 88 m
The circumference is given by
C = 2*pi*r
88 = 2 * pi *r
Divide each side by 2 pi
88 / 2pi = 2 * pi *r / 2 * pi
88 / 2 pi = r
What is cos A?
15
С
36
B
Enter your answer as a simplified fraction, in
the box
COS A
Answer:
[tex] cos A = \frac{5}{13} [/tex]
Step-by-step explanation:
The given triangle, ∆ABC is a right triangle.
To find cos A, we'd need to apply the trigonometric ratio formula, which is cos A = adjacent length/hypotenuse length
From the ∆ given,
AC = adjacent = 15,
BC = opposite = 36
We are not given the hypotenuse length AB.
==>Find the hypotenuse length AB, using the Pythagorean theorem formula:
c² = a² + b²
AB² = 15² + 36² = 225 + 1296 = 1521
AB = √1521
AB = 39
==>Find cos A:
cos A = adjacent/hypotenuse
[tex] cos A = \frac{adjacent}{hypotenuse} [/tex]
Adjacent = AC = 15
Hypotenuse = AB = 39
[tex] cos A = \frac{15}{39} [/tex]
[tex] cos A = \frac{5}{13} [/tex]
Determine whether the sequence converges or diverges. If it converges, find the limit (if an answer does not exist, enter DNE.)
{lnn/ln3}
limn→[infinity]{lnn/ln3n}=________
Answer:
The sequence converges. The limit DNE.
Step-by-step explanation:
Find the limit of n as n tends to infinity (in other words, positive infinity) in {Ln(n)/ Ln(3n)}
Positive infinity values for n start from 1,2,3,4,5,6,7,8,9,10,11,12,13,14,...,infinity
So I solved for values of n, up to n=20. All values are rounded up to 3 decimal places; for better accuracy.
When n is 1, the function is equal to 0.000
When n is 2, the function is = 0.387
When n is 3, the function = 0.500
When n is 4, the function = 0.558
When n is 5, the function = 0.594
When n is 6, the function = 0.619
When n is 7, the function = 0.639
When n is 8, the function = 0.654
When n is 9, the function = 0.667
When n is 10, the function = 0.677
When n is 11, the function = 0.686
When n is 12, the function = 0.693
When n is 13, the function = 0.700
When n is 14, the function = 0.706
When n is 15, the function = 0.711
When n is 16, the function = 0.716
When n is 17, the function = 0.721
When n is 18, the function = 0.725
When n is 19, the function = 0.728
When n is 20, the function = 0.732
We say there is a convergence because the space between the values of n gets smaller and smaller as n tends to infinity and there is no definite limit. Limit DNE.
I want to fence in a rectangular vegetable patch. The fencing for the east and west sides costs $2 per foot, and the fencing for the north and south sides costs only $1 per foot. I have a budget of $40 for the project. What is the largest area I can enclose
Answer:
largets area is 32 feet cubed
Step-by-step explanation:
8=4 foot 2 for each side w and e and 32feet n and s 16 each side
The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 41 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 38 sales representatives reveals that the mean number of calls made last week was 42. The standard deviation of the sample is 3.9 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 41?H0 : µ = 40
H1 : µ > 401. Compute the value of the test statistic. 2. What is your decision regarding H0?
Answer:
1. Test statistic t=1.581.
2. The null hypothesis H0 failed to be rejected.
There is not enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 41.
NOTE: if the null hypothesis is µ = 40, there is enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 40 (test statistic t=3.161).
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the mean number of calls per salesperson per week is significantly more than 41.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=41\\\\H_a:\mu> 41[/tex]
The significance level is 0.025.
The sample has a size n=38.
The sample mean is M=42.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.9.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.9}{\sqrt{38}}=0.633[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{42-41}{0.633}=\dfrac{1}{0.633}=1.581[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=38-1=37[/tex]
This test is a right-tailed test, with 37 degrees of freedom and t=1.581, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>1.581)=0.061[/tex]
As the P-value (0.061) is bigger than the significance level (0.025), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 41.
For µ = 40:
This is a hypothesis test for the population mean.
The claim is that the mean number of calls per salesperson per week is significantly more than 40.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=40\\\\H_a:\mu> 40[/tex]
The significance level is 0.025.
The sample has a size n=38.
The sample mean is M=42.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.9.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.9}{\sqrt{38}}=0.633[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{42-40}{0.633}=\dfrac{2}{0.633}=3.161[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=38-1=37[/tex]
This test is a right-tailed test, with 37 degrees of freedom and t=3.161, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>3.161)=0.002[/tex]
As the P-value (0.002) is smaller than the significance level (0.025), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the mean number of calls per salesperson per week is significantly more than 40.
calculating the five number summary
Answer:
2) 43
4) 65
Step-by-step explanation:
The first and third quartile of the data can be found by calculating the median of the first and second halves of the data. For example, the first quartile of the data can be calculated thus:
40,41,43,50,56
41,43,50
43
and the third quartile thus:
62,63,65,78,97
63,65,78
65
Hope it helps <3
Answer:
A) 43
B) 65
Step-by-step explanation:
A) First Quartile = [tex](N+1)\frac{1}{4}[/tex]
Where N is the number of observations
=> 1st Quartile = (11+1)(1/4)
=> 1st Quartile = (12)(1/4)
=> 1st Quartile = 3rd number
=> 1st Quartile = 43B) Third Quartile = [tex](N+1)\frac{3}{4}[/tex]
=> 3rd Quartile = (11+1)(3/4)
=> 3rd Quartile = (12)(3/4)
=> 3rd Quartile = 3*3
=> 3rd Quatile = 9th number
=> 3rd Quartile = 65The hourly rate for a staff nurse is £13.75.
A staff nurse works 30 hours a week and 6 hours overtime at time-and-a-half.
What is her total pay for the week?
Answer:
P = £536.25
Her total pay for the week is £536.25
Step-by-step explanation:
The total pay can be written as;
P = t1(r1) + t2(r2)
Where;
t1 = normal time
r1 = normal time pay rate
t2 = overtime
r2 = overtime pay rate
Given;
t1 = 30 hours
t2 = 6 hours
r1 = £13.75
r2 = 1.5(r1) = 1.5(£13.75)
Substituting the given values into the equation 1;
P = 30(£13.75) + 6(1.5(£13.75))
P = £536.25
Her total pay for the week is £536.25
In the figure angle G measures 102° and angle B measures 30° what is the measurement of angle B
Answer:
I hope this helps
Step-by-step explanation:
The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with a potential function
V(x,y)equals=c ln (ro/√ x²+y²)
where c > 0 is a constant and r0 is a reference distance at which the potential is assumed to be 0.
Required:
Find the components of the electric field in the x- and y-directions, where E(x,y) =∇V (x,y )
Answer:
c(x,y)/(x²+y²)
Step-by-step explanation:
Since E(x,y) = -∇V (x,y ) and V(x,y) =c ln (ro/√ x²+y²)
Let ro/√(x²+y²) = u and √(x²+y²) = v
dV/du = c/u = c√(x²+y²)/ro,
du/dv = -ro/(x²+y²) = -ro/² and dv/dx = x/√(x²+y²)
Using the chain rule,
So dV/dx = dV/du × du/dv × dv/dx
= c√(x²+y²)/ro × -ro/(x²+y²) × x/√(x²+y²)
dV/dx = -cx/(x²+y²)
- dV/dx = -(-cx/(x²+y²)) = cx/(x²+y²)
Also, dv/dy = y/√(x²+y²)
Using the chain rule
dV/dy = dV/du × du/dv × dv/dy
= c√(x²+y²)/ro × -ro/(x²+y²) × y/√(x²+y²)
dV/dy = -cy/(x²+y²)
- dV/dy = -(-cy/(x²+y²)) = cy/(x²+y²)
E(x,y) = -∇V (x,y )
= -(dV/dx)i + [-(dV/dy)]j
= [cx/(x²+y²)]i +[ cy/(x²+y²)]j
= c(x,y)/(x²+y²)
Assume that adults have IQ scores that are normally distributed with a mean of 94 and a standard deviation of 14. Find the probability that a randomly selected adult has an IQ greater than 107.1. (Hint: Draw a graph.) To help visualize the area of interest, draw a standard normal curve. Label the given values for x and mu . x 94107.1
Answer:
0.6517
Step-by-step explanation:
z = (x - μ)/σ
Z= standard score
x= observed value
μ= mean of the sample
σ= standard deviation of the sample
z = (x - μ)/σ = (107.1 - 94 )/ 14 = 0.9357
probability that a randomly selected adult has an IQ greater than 107.1. = P(Z > 0.935) = 0.6517
NB: the value is 0.6517 is pulled from the z table which can be found at the back of most math text.
Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oil-filled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than 0.05 mm. What hypotheses should be tested, and why?The appropriate hypotheses areH0: σ0.05 mmversusHa: σ0.05 mm.With this formulation, the burden of proof is on the data to show that the requirementbeen met.In this context, what are the type I and type II errors?In this context, the type I error occurs if wea shipment that should have been . A type II error occurs if we a shipment that should have been.Need Help? Read It Talk to a Tutor
Answer:
Null and alternative hypothesis:
[tex]H_0: \mu=0.05\\\\H_a:\mu< 0.05[/tex]
The alternative hypothesis is the one that needs evidence to be supported, while the null hypothesis is the one that can be nullified (reject).
Only if there is enough evidence that thickness is less than 0.05 the null hypothesis will be rejected and the alternative hypothesis claim supported.
A Type I error happens when a true null hypothesis is rejected. In this case we will be purchase a order that is not fulfilling the thickness required.
A Type II error happens when a false null hypothesis is failed to be rejected. In this case, the order has a thickness significantly smaller than 0.05, but the sample gives no enough evidence and the order will not be purchased.
Step-by-step explanation:
A hypothesis test to see conclusive evidence that the true standard deviation of sheath thickness is less than 0.05 mm will have the following hypothesis:
[tex]H_0: \mu=0.05\\\\H_a:\mu< 0.05[/tex]
The alternative hypothesis Ha will state that the true mean is significantly smaller than 0.05, while the null hypothesis H0 will state the opposite: that the true mean is not significantly smaller than 0.05.
The alternative hypothesis is the one that needs evidence to be supported, while the null hypothesis is the one that can be nullified (reject).
Only if there is enough evidence that thickness is less than 0.05 the null hypothesis will be rejected and the alternative hypothesis claim supported.
Need Assistance With This Problem
Answer:
not sure how to really answer this question.
Answer:
4.56, 4.65, 5.46, 5.64, 6.45, 6.54
Step-by-step explanation:
First we have to compare the first digits in each number as less is this digit as less is the number. So the least off all are
4.56 and 4.65
which of these two numbers is least ? Now we have to look to the 2-nd digits of these numbers:
they are 5 and 6 . 5<6 so 4.56<4.65
Lets select next numbers whicj first digit is 5. They are:
5.46 and 5.64. However the second digit of the number 5.64 -6 is bigger than the second digit of number 5.46 -4. That is why 5.46<5.64
Similarly 6.45< 6.54
Solve the two-step equation.-0.45x + 0.33 = -0.66What is the solution?x = -2.2x = -1.4x = 1.4x = 2.2f
Answer:
x = 2.2
Step-by-step explanation:
-0.45x + 0.33 = -0.66
Subtract .33 from each side
-0.45x + 0.33-.33 = -0.66-.33
-.45x = -.99
Divide each side by -.45
-.45x./-.45 = -.99/-.45
x = 11/5
x = 2.2