Answer:
1:27 (D)
Step-by-step explanation:
Given:
Sphere A has a diameter of 2
Sphere A is dilated to create sphere B
Scale factor = 3
Volume of a sphere = 4/3 πr³
Radius = r = diameter/2 = 2/2
r = 1
Volume of sphere A = 4/3 ×π(1)³
Volume of sphere A = 4/3 × π
Volume of sphere B = 4/3 πR³
Since the diameter was dilated, the diameter of B = diameter of A × scale factor
diameter of B = 2×3 = 6
Radius of B = R = diameter/2 = 3
Volume of sphere B = 4/3 × π(3)³
Volume of sphere B = (4/3)(27)π
Ratio of the volume of sphere A to volume of sphere B
= [4/3 ×π]: [(4/3)(27)π]
= (4π/3)/[(4π/3)×27] = 1/27
= 1:27
Answer: 1:27
Step-by-step explanation:
The original volume * scale factor cubed = new volume.
The scale factor is 3 and 3^3 is 27, so the ratio is 1:27
Select the correct answer.
The function RX) = 2x + 3x + 5, when evaluated, gives a value of 19. What is the function's input value?
A. 1
B. -1
C. 2
D. -2
E. -3
Answer:
Correct option: C.
Step-by-step explanation:
(Assuming the correct function is R(x) = 2x^2 + 3x + 5)
To find the input value that gives the value of R(x) = 19, we just need to use this output value (R(x) = 19) in the equation and then find the value of x:
[tex]R(x) = 2x^2 + 3x + 5[/tex]
[tex]19 = 2x^2 + 3x + 5[/tex]
[tex]2x^2 + 3x -14 = 0[/tex]
Solving this quadratic function using the Bhaskara's formula (a = 2, b = 3 and c = -14), we have:
[tex]\Delta = b^2 - 4ac = 9 + 112 = 121[/tex]
[tex]x_1 = (-b + \sqrt{\Delta})/2a = (-3 + 11)/4 = 2[/tex]
[tex]x_2 = (-b - \sqrt{\Delta})/2a = (-3 - 11)/4 = -3.5[/tex]
So looking at the options, the input to the function is x = 2
Correct option: C.
The probability of rolling two dice at the same time and getting a 4 with either die or the sum of the dice is 6
Answer:
Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die. But with a second dice, there will be 24 different possibilities.
Step-by-step explanation:
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
List price is 45$ if the sales tax rate is 7% how much is the sales tax in dollars
Answer:
3.15 dollars
Step-by-step explanation:
The sales tax rate is 7% = 0.07
So, we need to multiply the listed price and the sales tax rate.
= 45 * 0.07 = 3.150 (3.15)
Hope this helps and please mark as the brainliest
Assume that the random variable X is normally distributed, with mean 60 and standard deviation 16. Compute the probability P(X < 80). Group of answer choices
Answer:
P(X < 80) = 0.89435.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 60, \sigma = 16[/tex]
P(X < 80)
This is the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 60}{16}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.89435.
So
P(X < 80) = 0.89435.
the intersection of the two legs of the right triangle and the red segment is the _________ of the triangle shown
Answer:
b median
Step-by-step explanation:
Answer:
orthocenter
Step-by-step explanation:
The red segment is an altitude of the triangle, as are the two legs. The intersection point of the altitudes is the orthocenter.
__
This is basically a vocabulary question.
altitude - the perpendicular segment from a vertex to the opposite side (or its extension)median - the segment joining a vertex with the midpoint of the opposite sidecentroid - the point where medians meetorthocenter - the point where altitudes meetA sample of 26 offshore oil workers took part in a simulated escape exercise, and their escape time (unit: second) were observed. The sample mean and sample standard deviation are 370.69 and 24.36, respectively. Suppose the investigators had believed a priori that true average escape time would be at most 6 minutes. Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using the rejection region method at a significance level of 0.05.
Answer:
Yes, it contradict this prior belief as there is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes.
Test statistic t=2.238>tc=1.708.
The null hypothesis is rejected.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the true average escape time is significantly higher than 6 minutes (360 seconds).
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=360\\\\H_a:\mu> 360[/tex]
The significance level is 0.05.
The sample has a size n=26.
The sample mean is M=370.69.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=24.36.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{24.36}{\sqrt{26}}=4.777[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{370.69-360}{4.777}=\dfrac{10.69}{4.777}=2.238[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=26-1=25[/tex]
The critical value for a right-tailed test with a significance level of 0.05 and 25 degrees of freedom is tc=1.708. If the test statistic is bigger than 1.708, it falls in the rejection region and the null hypothesis is rejected.
As the test statistic t=2.238 is bigger than the critical value t=1.708, the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes (360 seconds).
The straight line L has equation y = 1/2x+7 The straight line M is parallel to L and passes through the point (0, 3). Write down an equation for the line M.
Answer:
y = [tex]\frac{1}{2}[/tex] x + 3
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = [tex]\frac{1}{2}[/tex] x + 7 ← is in slope- intercept form
with slope m = [tex]\frac{1}{2}[/tex]
Parallel lines have equal slopes
line M crosses the y- axis at (0, 3) ⇒ c = 3
y = [tex]\frac{1}{2}[/tex] x + 3 ← equation of line M
4b • 0.5a 2ab 2a2b 2ab2 2a2b2
Answer:
(4b)•(0.5a) = (4•0.5)(a)(b) = 2ab
Step-by-step explanation:
ABCD is a kite.
B
O
y = [?]
A 40°
C
Х
Enter the number
that belongs in
the green box.
D
Answer:
50°
Step-by-step explanation:
ABCD is a kite.
Therefore, AB = BC
[tex]\therefore m\angle BCA= m\angle BAC = 40\degree \\
\because BD \perp AC.. (Diagonals \: of\: kite) \\
\therefore y + 90\degree + m\angle BCA = 180\degree \\
\therefore y + 90\degree + 40\degree = 180\degree \\
\therefore y = 180\degree - 130\degree \\
\huge\red {\boxed {y = 50\degree}} [/tex]
The U.S. Department of Agriculture guarantees dairy producers that they will receive at least $1.00 per pound of butter they supply to the market. Below is the current monthly demand and supply schedule for wholesale butter (in millions of pounds per month). Wholesale Butter Market
Price (dollars per pound) Quantity of Butter Demanded Quantity of Butter Supplied
(millions of pounds) (millions of pounds)
$0.80 107 63 0
.90 104 71
1.00 101 79
1.10 98 87
1.20 95 95
1.30 92 103
1.40 89 111
1.50 86 119
1.60 83 127
1.70 80 135
1.80 77 143
a. In the butter market, the monthly equilibrium quantity is million pounds and the equilibrium price is $ per pound.
b. What is the monthly surplus created in the wholesale butter market due to the price support (price floor) program? 22 million pounds 79 million pounds Zero 11 million pounds Suppose that a decrease in the cost of feeding cows shifts the supply schedule to the right by 40 million pounds at every price.
Answer:
a. In the butter market, the monthly equilibrium quantity is 95 million pounds and the equilibrium price is $1.2 per pound.
b. The correct option is zero.
c. See the attached excel file for the new supply schedule.
d. The monthly surplus created by the price support program is 18 million pounds given the new supply of butter.
Step-by-step explanation:
Note: This question is not complete. A complete question is therefore provided in the attached Microsoft word file.
a. In the butter market, the monthly equilibrium quantity is million pounds and the equilibrium price is $ per pound.
At equilibrium, quantity demanded must be equal with the quantity supplied.
In this question, equilibrium occurs at the price of $1.20 per pound and quantity of 95 million pounds.
Therefore, in the butter market, the monthly equilibrium quantity is 95 million pounds and the equilibrium price is $1.2 per pound.
b. What is the monthly surplus created in the wholesale butter market due to the price support (price floor) program?
Price floor refers to a government price control on the lowest price that can be charged for a commodity.
It should be noted that for a price floor to be binding, it has to be fixed above the equilibrium price.
Since the price floor of $1 per pound is lower than the equilibrium price of $1.2 per pound, the price floor will therefore not be binding. As a result, the market will still be at the equilibrium point and the monthly surplus created in the wholesale butter market due to the price support (price floor) program will be zero.
Therefore, the correct option is zero.
c. Fill in the new supply schedule given the change in the cost of feeding cows.
Since a decrease in the cost of feeding cows shifts the supply schedule to the right by 40 million pounds at every price, this implies that there will be an increase in supply by 40 million at each price.
Note: Find attached the excel file for the new supply schedule.
d. Given the new supply of butter, what is the monthly surplus of butter created by the price support program?
Since the price floor has been fixed at $1 per pound by the price support program, we can observe that the quantity demanded is 101 million pounds and quantity supplied is 119 million pounds at this price floor of $1. The surplus created is then the difference between the quantity demanded and quantity supplied as follows:
Surplus created = Quantity supplied - Quantity demanded = 119 - 101 = 18 million pounds
Therefore, the monthly surplus created by the price support program is 18 million pounds given the new supply of butter.
The arithmetic mean (average) of four numbers is 85. If the largest of these numbers is 97, find the mean of the remaining three numbers. I cannot solve this. Please help on it.
Answer:
81
Step-by-step explanation:
Let's do this systematically:
Four numbers: a, b, c, d
Whose mean is 85: [tex]\frac{a + b + c + d}{4} = 85[/tex]
Whose largest number is 97: [tex]\frac{a + b + c + 97}{4} = 85[/tex]
Lets solve for the other numbers:
a+b+c+97 = 85*4 = 340
340 - 97 = 243
a+b+c = 243
at this point it doesn't matter what the numbers are, they just need to add up to 243.
We can do 243÷3=81, which is our answer
C equals 2 pi r; Cequals62.8 (Circumference of a circle)
Answer:
about 10
Step-by-step explanation:
62.8 = 2 pi r/2
62.8/2 = pi r
31.4/pi = pi r/pi
about 10 = r
Find all the missing side lengths for the following triangles.
Answer:
Step-by-step explanation:
A) u = 4 v = 4/(sqrt)3
B) b = 5 c = 10
C) b = 2(sqrt)2 a = 4
D) m and n are both 7(sqrt)2/2
The missing side lengths for the three triangles are 10√3, 12, and 8. The first triangle is a 30-60-90 triangle, the second triangle is a 45-45-90 triangle, and the third triangle is a right triangle. The missing side lengths were found using the properties of special triangles and the Pythagorean Theorem.
Here are the missing side lengths for the following triangles:
Triangle 1:
The missing side length is 15.
The triangle is a 30-60-90 triangle, so the ratio of the side lengths is 1:√3:2. The hypotenuse of the triangle is 20, so the shorter leg is 10 and the longer leg is 10√3. The missing side length is the longer leg, so it is 10√3.
Triangle 2:
The missing side length is 12.
The triangle is a 45-45-90 triangle, so the ratio of the side lengths is 1:1:√2. The hypotenuse of the triangle is 12√2, so each of the legs is 12. The missing side length is one of the legs, so it is 12.
Triangle 3:
The missing side length is 8.
We can use the Pythagorean Theorem to find the missing side length. The Pythagorean Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, the hypotenuse is 10 and one of the other sides is 6. Let x be the missing side length.
[tex]10^{2}[/tex] = [tex]6^{2}[/tex] + [tex]x^{2}[/tex]
100 = 36 +[tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 64
x = 8
Therefore, the missing side length is 8.
Learn more about side lengths here: brainly.com/question/18725640
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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!
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]
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
A U.S. dime has a diameter of about 18 millimeters. What is the area of one side of a dime to the nearest square millimeter? Use 3.14 as an approximation for pi. The area of one side of a U.S. dime is approximately _____ square millimeters.
Area of one side of a U.S. dime is approximately 254 square millimeters.
What is Circle?A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the centre.
Given that U.S. dime has a diameter of about 18 millimeters.
We need to find the area of one side of a dime to the nearest square millimeter.
Diameter=18 millimeters
Diameter is two times of radius
D=2R
18=2R
Divide both sides by 2
Radius is 9 millimeters.
Area of dime=πr²
=3.14×(9)²
=3.14×81
=254 square millimeters.
Hence, area of one side of a U.S. dime is approximately 254 square millimeters.
To learn more on Circles click:
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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
Find the surface area of the solid shown or described. If necessary, round to the nearest tenth. A.348m^2 B.484m^2 C.180.7m^2 D.262m^2
Answer: 484m²
Step-by-step explanation: This is a question on solid shape.
The surface area of a cone is the same thing as the perimeter of the cone ie, the materials required to construct the cone.
Formula for the surface area of the cone = πrl + πr², ( the circular base )
From.the diagram,
r = 7.1m , l = 14.6m, π = 3.142
Now substitute for those values in.the formula above
SA = πrl + πr²
= 3.142 × 7.1 × 14.6 + 3.142 × 7.1²
= 325.6997 + 158.388
= 484.09
Now to the nearest tenth meter,
SA = 484m²
if a^2+b^2+c^2=169. find a, given that b=2√2, 3√c=9.
Answer:
a = ±4√5
Step-by-step explanation:
Solve for c.
3√c = 9
√c = 9/3
√c = 3
c = 3²
c = 9
Put b=2√2 and c=9, solve for a.
a² + (2√2)² + 9² = 169
a² + 8 + 81 = 169
a² = 169 - 81 - 8
a² = 80
a = ±√80
a = ±4√5
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
What does csc x cot x (1-cos^2 x) equal
Answer:
Step-by-step explanation:
(matching type) given f(x)= x+4 and g(x) = 2x + 1 match the expression to its simplication operation
choose
x+4 / 2x+1
Answer 1
Choose...
f of g
f/g
f - g
f∙g
g/f
f + g
2x+1 / x+4
Answer 2
Choose...
f of g
f/g
f - g
f∙g
g/f
f + g
3x + 5
Answer 3
Choose...
f of g
f/g
f - g
f∙g
g/f
f + g
2x + 5
Answer 4
Choose...
f of g
f/g
f - g
f∙g
g/f
f + g
-x + 3
Answer 5
Choose...
f of g
f/g
f - g
f∙g
g/f
f + g
2x2 + 9x + 12
pa help po
Answer:
1) [tex]h(x) = \frac{f(x)}{g(x)}[/tex], 2) [tex]h(x) = \frac{g(x)}{f(x)}[/tex], 3) [tex]h(x) = f(x) + g(x)[/tex], 4) [tex]h (x) = f [g (x)][/tex], 5) [tex]h(x) = f(x) - g(x)[/tex]
Step-by-step explanation:
1) Let be [tex]f(x) = x + 4[/tex] and [tex]g(x) = 2\cdot x + 1[/tex], if [tex]h (x) = \frac{x+4}{2\cdot x + 1}[/tex], then:
[tex]h(x) = \frac{f(x)}{g(x)}[/tex]
2) Let be [tex]f(x) = x + 4[/tex] and [tex]g(x) = 2\cdot x + 1[/tex], if [tex]h(x) = \frac{2\cdot x + 1}{x+4}[/tex], then:
[tex]h(x) = \frac{g(x)}{f(x)}[/tex]
3) Let be [tex]f(x) = x + 4[/tex] and [tex]g(x) = 2\cdot x + 1[/tex], if [tex]h(x) = 3\cdot x + 5[/tex], then:
[tex]h(x) = 3\cdot x + 5[/tex]
[tex]h (x) = (1 + 2)\cdot x + (4+1)[/tex]
[tex]h(x) = x + 2\cdot x + 4 +1[/tex]
[tex]h(x) = (x+4) + (2\cdot x + 1)[/tex]
[tex]h(x) = f(x) + g(x)[/tex]
4) Let be [tex]f(x) = x + 4[/tex] and [tex]g(x) = 2\cdot x + 1[/tex], if [tex]h(x) = 2\cdot x + 5[/tex], then:
[tex]h(x) = 2\cdot x + 5[/tex]
[tex]h(x) = 2\cdot x + 1 + 4[/tex]
[tex]h(x) = (2\cdot x +1)+4[/tex]
[tex]h (x) = f [g (x)][/tex]
5) Let be [tex]f(x) = x + 4[/tex] and [tex]g(x) = 2\cdot x + 1[/tex], if [tex]h(x) = -x + 3[/tex], then:
[tex]h(x) = -x + 3[/tex]
[tex]h(x) = (1 - 2)\cdot x + 4 - 1[/tex]
[tex]h(x) = x - 2\cdot x + 4 - 1[/tex]
[tex]h(x) = x + 4 - (2\cdot x + 1)[/tex]
[tex]h(x) = f(x) - g(x)[/tex]
a. dashed line, shade below
b. dashed line, shaded above
c. solid line, shade above
d. solid line, shade below
Answer:
the answer is A
Step-by-step explanation:
please assist me with the power of i(imaginary)
Let's raise i to various powers starting with 0,1,2,3...
i^0 = 1
i^1 = i
i^2 = ( sqrt(-1) )^2 = -1
i^3 = i^2*i = -1*i = -i
i^4 = (i^2)^2 = (-1)^2 = 1
i^5 = i^4*i = 1*i = i
i^6 = i^5*i = i*i = i^2 = -1
We see that the pattern repeats itself after 4 iterations. The four items to memorize are
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
It bounces back and forth between 1 and i, alternating in sign as well. This could be one way to memorize the pattern.
To figure out something like i^25, we simply divide the exponent 25 over 4 to get the remainder. In this case, the remainder of 25/4 is 1 since 24/4 = 6, and 25 is one higher than 24.
This means i^25 = i^1 = i
Likewise,
i^5689 = i^1 = i
because 5689/4 = 1422 remainder 1. The quotient doesn't play a role at all so you can ignore it entirely
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.
Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+16y=0. Separate your answers by commas. isn't the answer just ±4?
Answer: k = 4, k = -4 and k = 0.
Step-by-step explanation:
If we have y = sin(kt)
then:
y' = k*cos(kt)
y'' = -k^2*son(x).
then, if we have the relation:
y'' - y = 0
we can replace it by the things we derivated previously and get:
-k^2*sin(kt) + 16*sin(kt) = 0
we can divide by sin in both sides (for t ≠0 and k ≠0 because we can not divide by zero)
-k^2 + 16 = 0
the solutions are k = 4 and k = -4.
Now, we have another solution, but it is a trivial one that actually does not give any information, but for the diff equation:
-k^2*sin(kt) + 16*sin(kt) = 0
if we take k = 0, we have:
-0 + 0 = 0.
So the solutions are k = 4, k = -4 and k = 0.
The life in hours of a battery is known to be normally distributed, with a standard deviation of 1.25 hours. A random sample of 10 batteries has a mean life x = 40.5 hours.
a) Is there evidence to support the claim that battery life exceeds 40 hours? Use
α = 0.05.
b) What is the P-value for this test?
Answer:
a) Test statistic
Z = 1.265 < 1.96 at 0.05 level of significance
The battery life is not exceeds 40 hours
b)
p- value = 0.8962
Step-by-step explanation:
Step(i):-
Given sample size 'n' =10
Mean of the sample x⁻ = 40.5 hours
Mean of of the Population μ = 40 hours
Standard deviation of the Population = 1.25 hours
Step(ii):-
Null Hypothesis:H₀: μ = 40 hours
Alternative Hypothesis :H₁ : μ < 40 hours
step(ii):-
Test statistic
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{40.5 -40}{\frac{1.25}{\sqrt{10} } }[/tex]
Z = 1.265
Level of significance = 0.05
Z₀.₀₅ = 1.96
Z = 1.265 < 1.96 at 0.05 level of significance
The battery life is not exceeds 40 hours
Step(iii):-
P - value
P( Z < 1.265) = 0.5 + A( 1.265)
= 0.5 + 0.3962
= 0.8962
P( Z < 1.265) = 0.8962
i ) p- value = 0.8962 > 0.05
Accept H₀
There is no significant
The battery life is not exceeds 40 hours
My question is probably obvious but I don't know it. What is the z axis
Answer:
z-Axis. The axis in three-dimensional Cartesian coordinates which is usually oriented vertically. Cylindrical coordinates are defined such that the -axis is the axis about which the azimuth coordinate. is measured.
Step-by-step explanation:
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.