Answer:
a) D_u . ∀ f ( 2 , 1 ) = 3 , Q = 72 degrees
b) D_u . ∀ f ( 2 , 1 ) = 4*sqrt(2) , Q = 80 degrees
c) D_u . ∀ f ( 2 , 1 ) = - sqrt(2) , Q = -54.74 degrees
d) u = 1 / sqrt(34) *(-3i +5j) , D_u . ∀ f ( 2 , 1 ) = +/- sqrt (34) , Q = 80.27 degrees
Step-by-step explanation:
Given:
- The terrain is modeled as a surface:
f(x , y) = x*y + y^3 - x^2
At point P( 2 , 1 , -1 ) is the current position:
Find:
(a) Determine the slope you would encounter if you headed due West from your position. What angle of inclination does this correspond to?
(b) Determine the slope you would encounter if you headed due North-West from your position. What angle of inclination does this correspond to?
(c) Determine the slope you would encounter if you headed due South-West from your position. What angle of inclination does this correspond to?
(d) Determine the steepest slope you could encounter from your position and the direction of that slope (as a unit vector).
Solution:
- We will compute the gradient of the vector ∀ f @ point P( 2 , 1 , -1 ) as follows:
∀ f ( x , y ) = (y - 2x ) i + ( x + 3y^2) j
- Evaluate at point P ( 2 , 1 ):
∀ f ( 2 , 1 ) = (1 - 2*2 ) i + ( 2 + 3*1^2) j
∀ f ( 2 , 1 ) = -3 i + 5 j
- We will use the result of ∀ f @ point P( 2 , 1 , -1 ) for the all the parts.
a)
- The direction due west can be written as a unit vector u_w = - i .
- Now compute the directional derivative in the direction of u_w = - i
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . - i
D_u . ∀ f ( 2 , 1 ) = 3
- Now compute the angle of inclination Q for the following direction:
Q = arctan(3) = 71.57 = 72 degrees
Hence, along the direction due west from position we ascend with an inclination of approximately 72 degrees.
b)
- The direction due north-west can be written as a unit vector u_w = - i + j .
- Now compute the directional derivative in the direction of u_w = - i + j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i + j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = 8 / sqrt(2) = 4*sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(4*sqrt(2)) = 79.98 = 80 degrees
Hence, along the direction due north-west from position we ascend with an inclination of approximately 80 degrees.
c)
- The direction due south-west can be written as a unit vector u_w = - i - j .
- Now compute the directional derivative in the direction of u_w = - i - j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i - j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = -2 / sqrt(2) = - sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(-sqrt(2)) = -54.74 degrees
Hence, along the direction due south-west from position we descend with an inclination of approximately 55 degrees.
d)
- We know that ∀ f ( 2 , 1 ) gradient points in the direction greatest increase, hence, So from P the direction of greatest increase is ∇f(P) = -3i +5j. The unit vector pointing in this direction is:
u = 1 / sqrt(34) *(-3i +5j)
- so we have:
D_u . ∀ f ( 2 , 1 ) = u . ∀ f ( 2 , 1 ) = (-3i +5j) . (-3i +5j) / sqrt(34)
= +/- sqrt (34)
- Hence, the steepest ascent or decent is of :
Q = arctan ( sqrt(34)) = 80.27 degrees
Trevor decides to start saving money for a new car. He knows he can invest money into an account which will earn 6.8% APR, compounded weekly, and would like to have saved $12,000 after 4 years. How much money will he need to invest into the account now so that he has $12,000 after 4 years.
Answer:
PV=$9,143.88
Step-by-step explanation:
Compound Interest
When a principal amount (also called present value PV) is saved at some rate of interest r for some time t, the future value FV that includes the original investment plus the interests is computed as
[tex]FV=PV\left(1+r\right)^t[/tex]
If the investment is compounded other than annually, then r and t must be scaled to the proper time units.
If we already know the future value, then the present value is computed by solving the above equation
[tex]PV=FV\left(1+r\right)^{-t}[/tex]
The Annual Percentage Rate (APR) is 6.8% compounded weekly, so the value of r is (assuming 52 weeks per year)
[tex]\displaystyle r=\frac{6.8}{100\times 52}=0.00131[/tex]
And the time is computed in weeks
t=4*52=208
The present value of the investment Trevor needs to save now is
[tex]PV=12,000\left(1+0.00131\right)^{-208}[/tex]
[tex]\boxed{PV=\$9,143.88}[/tex]
Trevor will need to invest $9,143.88 into the account to have $12,000 after 4 years
A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms. Find the current i(t) if i(0) = 0. i(t) = Determine the current as t → [infinity]. lim t→[infinity] i(t) =
Answer:
attached below
Step-by-step explanation:
The current as t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms
What is ohms law?It is defined as the relationship between current and voltage according to the ohms law the voltage is directly proportional to the current.
It is given that:
A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms.
Using the equation:
LdI/dt + RI = v(t)
I(0) = 0
LdI/dt = v(t)
V = 20 volts
After integration:
I = 2/7 + c
c = e⁻⁷⁰⁰ⁿ (n = t) = 0
t → ∞
I = 2/7 A
Thus, the current as t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms
Learn more about the ohms law here:
brainly.com/question/14874072
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A closed box with a square bottom is three times high as it is wide. a) Express the surface area of the box in terms of its width. b) Express the volume of the box in terms of its width. c) Express the surface area in terms of the volume. d) If the box has a volume of 24 m³, what is its surface area?
Answer:
a) [tex]S(s) = 14s^2[/tex]
b) [tex]V(s) = 3s^3[/tex]
c) [tex]S(s) = \dfrac{14V(s)}{3s}[/tex]
d) 56 square meter
Step-by-step explanation:
We are given the following in the question:
A closed box with a square bottom is three times high as it is wide.
Let s be the side of square and h be the height.
[tex]h = 3s[/tex]
a) Surface area of box
[tex]2(lb + bh + hl)[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]S = 2(s^2 + sh +sh)\\S = 2(s^2 + 3s^2 + 3s^2)\\S(s) = 14s^2[/tex]
b) Volume of box
[tex]l\times b \times h[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]V = s\times s\times h\\V= s\times s\times 3s\\V(s) = 3s^3[/tex]
c) Surface area in terms of volume
[tex]S(s) = 14s^2 = \dfrac{14V(s)}{3s}[/tex]
d) Surface area
Volume = 24 m³
[tex]V(s) = 24\\3s^3 = 24\\s^3 = 3\\s = 2[/tex]
[tex]S(2) = 14(2)^2 = 56\text{ square meter}[/tex]
Nine cars (3 Pontiacs [labeled 1-3], 4 Fords [labeled 4-7], and 2 Chevrolet's [labeled 8-9]) are divided into 3 groups for car racing. each group consistes of 3 cars, and they are allocated to 3 tracks (1-3), respectively.
1. How many different ways can you arrange the 9 cars?
Answer:
The number of different ways to arrange the 9 cars is 362,880.
Step-by-step explanation:
There are a total of 9 cars.
These 9 cars are to divided among 3 racing groups.
The condition applied is that there should be 3 cars in each group.
Use permutation to determine the total number of arrangements of the cars.
For group 1:There are 9 cars and 3 to be allotted to group 1.
This can happen in [tex]^9P_{3}[/tex] ways.
That is, [tex]^9P_{3}=\frac{9!}{(9-3)!} =504[/tex] ways.
For group 2:There are remaining 6 cars and 3 to be allotted to group 2.
This can happen in [tex]^6P_{3}[/tex] ways.
That is, [tex]^6P_{3}=\frac{6!}{(6-3)!} =120[/tex] ways.
For group 3:There are remaining 3 cars and 3 to be allotted to group 3.
This can happen in [tex]^3P_{3}[/tex] ways.
That is, [tex]^3P_{3}=\frac{3!}{(3-3)!} =6[/tex] ways.
The total number of ways to arrange the 9 cars is: [tex]^9P_{3}\times ^6P_{3}\times ^3P_{3}=504\times120\times6=362880[/tex]
Thus, the number of different ways to arrange the 9 cars is 362,880.
A baker blends 600 raisins and 400 chocolate chips into a dough mix and, from this, makes 500 cookies. (a) Find the probability that a randomly picked cookie will have no raisins. (b) Find the probability that a randomly picked cookie will have exactly two chocolate chips. (c) Find the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
Answer:
a) P(X = 0) for raisins = 0.33287
b) P(X = 2) for chocolate chips = 0.14379
c) P(X ≥ 2) for both bits = 0.59399
Step-by-step explanation:
The average amount of raisin per cookie is 600/500 = 1.2
The average amount of chocolate chips per cookie = 400/500 = 0.8
a) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For raisin, Mean = λ = 1.1
x = 0
P(X = 0) = (e⁻¹•¹)(1.1⁰)/(0!) = 0.33287
b) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For chocolate chips, Mean = λ = 0.8
x = 2
P(X = 2) = (e⁻⁰•⁸)(0.8²)/(2!) = 0.14379
c) the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
For this, the average number of bit in a raisin = (600+400)/500 = 2 bits per raisin.
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
For P(X = 0)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 0
P(X = 0) = (e⁻²)(2⁰)/(0!) = 0.13534
For P(X = 1)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 1
P(X = 1) = (e⁻²)(2¹)/(1!) = 0.27067
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
P(X ≥ 2) = 1 - (0.13534 + 0.27067) = 1 - 0.40601 = 0.59399
Using the Poisson distribution, it is found that there is a:
a) 0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
b) 0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
c) 0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes e = 2.71828 is the Euler number [tex]\mu[/tex] is the mean in the given interval.Item a:
600 raisins in 500 cookies, hence, the mean is:
[tex]\mu = \frac{600}{500} = 1.2[/tex]
The probability is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.2}1.2^{0}}{(0)!} = 0.3012[/tex]
0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
Item b:
400 chips in 500 cookies, hence, the mean is:
[tex]\mu = \frac{400}{500} = 0.8[/tex]
The probability is P(X = 2), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-0.8}0.8^{2}}{(2)!} = 0.1438[/tex]
0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
Item c:
1000 bits, as 600 + 400 = 1000, in 500 cookies, hence, the mean is:
[tex]\mu = \frac{1000}{500} = 2[/tex]
The probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}2^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}2^{1}}{(1)!} = 0.2707[/tex]
Hence:
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.4060[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594[/tex]
0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
A similar problem is given at https://brainly.com/question/16912674
Find the equation for the circle with center (4,3) and passing through (2,-4)
Answer:
[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]
Step-by-step explanation:
The general equation of a circle is as follows:
[tex](x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}[/tex]
In which the center is [tex](x_{c}, y_{c})[/tex], and r is the radius.
In this problem, we have that:
[tex]x_{c} = 4, y_{c} = 3[/tex]
So
[tex](x - 4)^{2} + (y - 3)^{2} = r^{2}[/tex]
Passing through (2,-4)
We replace into the equation to find the radius.
[tex](2 - 4)^{2} + (-4 - 3)^{2} = r^{2}[/tex]
[tex]4 + 49 = r^{2}[/tex]
[tex]r^{2} = 53[/tex]
The equation of the circle is:
[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]
In square inches, the area of the square is 4x^2 - 2x - 6 and the area of the triangle 2x^2 + 4x - 5. What polynomial represents the area of the shaded region?
Answer: the polynomial representing the shaded region is
2x² - 6x - 1
Step-by-step explanation:
The area of the square is expressed as 4x² - 2x - 6
The area of the triangle is expressed as 2x² + 4x - 5. The area of the shaded region would be the area of the square - the area of the area of the triangle.
The polynomial that represents the area of the shaded region would be
4x² - 2x - 6 - (2x² + 4x - 5)
Collecting like terms, it becomes
= 4x² - 2x² - 2x - 4x - 6 + 5
= 2x² - 6x - 1
Final answer:
The area of the shaded region is represented by the polynomial [tex]2x^2 - 6x - 1[/tex] square inches, which is the difference between the area of the square and the area of the triangle.
Explanation:
The student asked for the polynomial that represents the area of the shaded region given the area of the square and the area of the triangle within it. To find this area, we must subtract the area of the triangle from the area of the square. The given area of the square is [tex]4x^2 - 2x - 6[/tex] square inches, and the area of the triangle is [tex]2x^2 + 4x - 5[/tex] square inches. Therefore, the polynomial representing the area of the shaded region is found by the following subtraction:
Area of the square - Area of the triangle = [tex](4x^2 - 2x - 6) - (2x^2 + 4x - 5) = 2x^2 - 6x - 1[/tex]
This simplifies to 2x^2 (from the square's area) minus 6x (the change in linear dimensions caused by the triangle) minus 1 (the constant term from completing the square). Thus, the polynomial representing the shaded region is [tex]2x^2 - 6x - 1[/tex] square inches.
What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Answer:
Given that an article suggests
that a Poisson process can be used to represent the occurrence of
structural loads over time. Suppose the mean time between occurrences of
loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a
4-year period? c). How long must a time period be so that the probability of no loads
occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a
4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads
occurring during that period is at most 0.3 is given by: 3.3 years
Step-by-step explanation:
(3 pts) Given sample statistics for two data sets: _ Set A: x 23 Med 22 S 1.2 _ Set B: x 24 Med 29 S 3.1 a) Calculate the Pearson’s Index of Skewness for both sets. 3 b) Based on your findings, what type of distribution each set has (circle correct answer): Set A: symmetric skewed left skewed right uniform Set B: symmetric skewed left skewed right uniform c) Which set, A or B, can be considered and analyzed as symmetric _______
Answer:
Step-by-step explanation:
Given Data:
Set A : x = 23, Med = 22, S = 1.2
Set B : x = 24, Med = 29, S = 3.1
The Formula for Pearson's Index of Skewness (for Median in given data) is:
[tex]Sk_{2} = 3(\frac{x - Med}{S} )[/tex]
where,
[tex]Sk_{2} =[/tex] Pearson's Coefficient of Skewness
[tex]Med =[/tex] Median of Distribution
[tex]x=[/tex] Mean of Distribution
[tex]S=[/tex] Standard Deviation of Distribution
a) Finding Skewness:
For Set A:
[tex]Sk_{2_{A}} = 3(\frac{23 - 22}{1.2} )\\\\Sk_{2_{A}} = (\frac{3}{1.2} )\\\\Sk_{2_{A}} = 2.5[/tex]
For Set B:
[tex]Sk_{2_{B}} = 3(\frac{24 - 29}{3.1} )\\\\Sk_{2_{B}} = 3(\frac{-5}{3.1} )\\\\Sk_{2_{B}} = -4.84[/tex]
b) Type of Distribution:
For Set A:
As the value of skewness is a positive value (i.e. 2.5). Hence, Set A is right (positively) skewed.
For Set B:
As the value of skewness is a negative value (i.e. -4.84). Hence, Set B is skewed left (or negatively skewed).
c) Which Set can be considered as symmetric?
As the Pearson's Coefficient of skewness for Set A (2.5) is closer to 0 as compared to that of Set B (-4.84). Set A is more closer to that of a symmetric distribution and therefore can be considered as one.
Answer:
Is there a picture?
Step-by-step explanation:
It would be better if i could see a image!
According to a recent poll 53% of Americans would vote for the incumbent president. If a random sample of 100 people results in 40% who would vote for the incumbent, test whether the claim that the actual percentage is different from 53% is supported or not supported.
(1) State the null hypothesis.
(2) State the alternative hypothesis.
(3) What is the test statistic used for the test (z or t)?
(4) State the significance or alpha (α) level?
Answer:
1) Null hypothesis:[tex]p=0.53[/tex]
2)Alternative hypothesis:[tex]p \neq 0.53[/tex]
3) [tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]
4) We assume that [tex]\alpha=0.05[/tex]
[tex]p_v =2*P(z<-2.605)=0.0092[/tex]
So the p value obtained was a very 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 people who would vote for the incumbent is different from 0.53.
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
[tex]\hat p=0.4[/tex] estimated proportion of people who would vote for the incumbent
[tex]p_o=0.53[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level (assumed)
Confidence=95% or 0.95 (Assumed)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is 0.53 or not.:
1) Null hypothesis:[tex]p=0.53[/tex]
2)Alternative hypothesis:[tex]p \neq 0.53[/tex]
When we conduct a proportion test we need to use the z statistic, 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{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method 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 assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-2.605)=0.0092[/tex]
So the p value obtained was a very 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 people who would vote for the incumbent is different from 0.53 .
A boxer takes 3 drinks of water between each round for the first four rounds of a championship fight. After the fourth round he starts to take his three drinks plus one additional drink between each of the remaining rounds. If he continues to increase his drinks by 1 after each round, how many drinks will he take between the 14th and 15th round
Answer:
56 drinks
Step-by-step explanation:
Given
n = 15-4 = 11 rounds (with one additional drink)
a₀ = 3*4 = 12 drinks (between the first and the fourth round)
r = 4 (number of drinks per round)
We can use the formula
aₙ = a₀ + n*r
⇒ a₁₁ = 12 + 11*4 = 12 + 44
⇒ a₁₁ = 56 drinks
Kareem bought 5 feet of fabric. how much is this in yards?
Answer:
1.66667
Step-by-step explanation:
You would do 5 divided by 3.
5 divided by 3 = 1.66667
You can verify this answer by multiplying 1.66667 * 3
1.66667 * 3 = 5
Hope this helps!!!
Sarah opens a savings account that has a 2.75% annual interest rate,
compounded monthly. She deposits $500 into the account. How much will
be in the account after 15 years?
$500.00
$754.94
$1255.27
$255.27
Answer: $754.94
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 500
r = 2.75% = 2.75/100 = 0.0275
n = 12 because it was compounded monthly which means 12 times in a year.
t = 15 years
Therefore,.
A = 500(1+0.0275/12)^12 × 15
A = 500(1+0.0023)^180
A = 500(1.0023)^180
A = $754.94
Paul consumes only books and DVDs. At his current consumption bundle, his marginal utility from DVDs is 1212 and from books is 44. Each DVD costs $66, and each book costs $33. Is he maximizing his utility? Explain. Let MU Subscript Upper BMUB be the marginal utility of books, MU Subscript Upper DMUD be the marginal utility from DVDs, Upper P Subscript Upper BPB be the price of books, Upper P Subscript Upper DPD be the price of DVDs, and MRS be the marginal rate of substitution. Paul is
Answer: MUb / Pb ∠ MUd / Pd
Step-by-step explanation:
First let us define the parameters given;
we have that :
Marginal utility of from book MUb = 44
Marginal utility of from DVD MUd = 1212
Cost of DVD Pd = $ 66
Cost of Book Pb = $ 33
Using,
⇒ MUb / Pb = 44 / 33 = 1.33
⇒ MUd / Pd = 1212 / 66 = 18.36
From this we can see that;
MUb / Pb ∠ MUd / Pd
The question asked is if he is maximizing his utility?
From the law of equi-marginal utility, a consumer will maximize utility only when the utility derived from every unit of spending on goods and services is the same. That is to say that for this to happen,
MUb / Pb = MUd /Pd.
But from the question solved, this is not true, so from the answer gotten The consumer is not maximizing his utility because;
MUb / Pb ∠ MUd / Pd
cheers i hope this helps.
List and explain the elements that make up the structure of a fugue, then list and explain at least four techniques Bach utilized when writing fugues to create musical development.
Final answer:
The structure of a fugue includes elements such as the subject, counter-subject, expositions, and episodes. Bach used techniques like inversion, retrograde, augmentation, and diminution to develop his fugues.
Explanation:
The structure of a fugue traditionally includes a subject, which is the main theme introduced at the beginning, a counter-subject, which complements the subject, expositions where the subject is introduced in different voices, and episodes, which are free sections providing contrast. Johann Sebastian Bach used various techniques to create musical development within his fugues, some of which include:
Inversion, where the subject is mirrored vertically.
Retrograde, where the subject is played backwards.
Augmentation, where the subject is presented with longer note values, thus slower.
Diminution, where the subject is presented with shorter note values, making it faster.
These techniques helped to create complexity and interest, ensuring that the fugue evolves musically throughout the piece.
All human blood can be "ABO-typed" as one of O,A, B, or AB, but the distribution of thetypes varies a bit among groups of people. Here is the distributionof blood types for a randomly chosen person in the United States.Blood type O A B ABU.S. probability 0.46 0.41 0.12 0.01Choose a married couple at random. It is reasonable to assume thatthe blood types of husband and wife are independent and follow thisdistribution.a. What is the probability that the wife has type A and the husbandhas type B?b. What is the probability that one of the couple has type A blood andthe other has type B?
Answer:
(a) 0.0492
(b) 0.0984
Step-by-step explanation:
The probability distribution of blood type in the US is:
[tex]\begin{array}{cc}Type&Probability\\O&0.46\\A&0.41\\B&0.12\\AB&0.01\end{array}[/tex]
(a) The probability that the wife has type A and the husband has type B is:
[tex]P(w=A\ and\ h=B) =0.41*0.12\\P(w=A\ and\ h=B) =0.0492[/tex]
(b) The probability that one of the couple has type A blood and the other has type B is given by the probability that the wife has type A and the husband has type B added to the probability that the wife has type B and the husband has type A.
[tex]P= P(w=A\ and\ h=B) + P(w=B\ and\ h=A)\\P=0.41*0.12+0.12*0.41=0.0984[/tex]
Final answer:
The probability that the wife has type A blood and the husband has type B blood is 4.92%. The probability that one of the couple has type A blood and the other has type B blood is 9.84%.
Explanation:
The student is being asked to calculate probabilities related to the blood types of a randomly chosen married couple in the United States using the given distribution of ABO blood types. The probabilities for each blood type are: type O at 0.46, type A at 0.41, type B at 0.12, and type AB at 0.01. To solve these questions, we use the principle that the probability of independent events occurring together is the product of their respective probabilities.
Probability wife has type A and husband has type B: This is a straightforward calculation using the probabilities of each event happening independently. Since the blood type of the wife and the husband are independent events, you multiply their individual probabilities:P(wife has type A and husband has type B) = P(wife has type A) × P(husband has type B) = 0.41 × 0.12 = 0.0492, or 4.92%.Probability one has type A and the other has type B: Here we need to calculate the probability for two scenarios - Scenario 1: Wife has type A and husband has type B, and Scenario 2: Wife has type B and husband has type A. We add the probabilities of these two independent scenarios together.P(one has type A and the other has type B) = P(wife has type A and husband has type B) + P(wife has type B and husband has type A) = 0.0492 + 0.0492 = 0.0984, or 9.84%.The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation
5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.
(a) what is the probability that a part is defective?
(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in
such a sample? Assume that each part is independent of the others.
Answer:
a) There is a 15.87% probability that a part is defective.
b) The expected number of parts defective in such a sample is 0.7935.
Step-by-step explanation:
To solve this question, we use concepts of the normal probability distribution and the binomial probability distribution.
Normal probability distribution
Problems of normally distributed samples can be solved using 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.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
(a) what is the probability that a part is defective?
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation
5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.
Here we have [tex]\mu = 40, \sigma = 5[/tex]
This probability is the pvalue of Z when X = 35.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 40}{5}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
There is a 15.87% probability that a part is defective.
(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in such a sample? Assume that each part is independent of the others.
This is the expected value of a binomial distribution when [tex]n = 5, p = 0.1587[/tex].
So
[tex]E(X) = np = 5*0.1587 = 0.7935[/tex]
The expected number of parts defective in such a sample is 0.7935.
Of the 44 students in a class, 38 are taking the class because it is a major requirement, and the other 6 are taking it as an elective. If two students are selected at random from this class, what is the probability that the first student is taking the class as an elective and the second is taking it because it is a major requirement?
Answer:
The probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.
Step-by-step explanation:
The number of students taking the class because it is a major requirement is, n (M) = 38.
The number of students taking the class because it as an elective is, n (E) = 6.
The total number of students is, N = 44.
Two students are selected at random.
Assume that the selection is without replacement.
Compute the probability that the 1st student selected is taking the class as an elective and the 2nd is taking it because it is a major requirement as follows:
P (1st Elective ∩ 2nd Major) = P (1st Elective) × P (2nd Major)
[tex]=\frac{6}{44}\times \frac{38}{43}\\ =0.120507\\\approx0.121[/tex]
Thus, the probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.
Complete an amortization schedule for a $27,000 loan to be repaid in equal installments at the end of each of the next three years. The interest rate is 12% compounded annually. If an amount is zero, enter "0". Do not round intermediate calculations. Round your answers to the nearest cent.
Answer:
Amount paid annually = $11241.42
Step-by-step explanation:
The steps are as shown in the attached file
It is proposed to build a raft of pine logs for carrying a cargo on a river. The cargo will weigh 500 kg, and it must be kept entirely above the water level. How many kilograms of pine logs must we use to make the raft, if the logs may be entirely submerged, and they have SG
Answer:
They will need 200kg of pine logs.
Gauge pressure in the tank = 24045.29 Pa
Absolute pressure= 125370.29Pa
Step-by-step explanation:
Full question:sp.gr=0.8.The fluid in the nanometer is ethyl iodide with sp.gr =1.93. The manometric fluid height is 50 inches. What is the gauge pressure and absolute pressure in the tank?
Converting 50inches to metres. Multiply by 0.0254
50×0.0254=1.27m
Gauge pressure in tank is given by:
P = pgh
Where p= density,g= acceleration due to gravity, h= height
P= 1.93× 1000× 9.81× 1.27= 24045.29Pa
Absolute pressure=Ptotal= Pliquid + Patm
Absolute pressure =24045.29 + 101325=125370.29pa
Eighty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. (Enter your answers to three decimal places.)(a)P(all of the next three vehicles inspected pass)(b)P(at least one of the next three inspected fails)(c)P(exactly one of the next three inspected passes)(d)P(at most one of the next three vehicles inspected passes)(e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)? (Round your answer to three decimal places.)
Answer:
(a) P(all of the next three vehicles inspected pass) = 0.512 .
(b) P(at least one of the next three inspected fails) = 0.488 .
(c) P(exactly one of the next three inspected passes) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
Step-by-step explanation:
We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.
So, Probability that the next vehicle examined fails the inspection is 20%.
Also, it is given that successive vehicles pass or fail independently of one another.
(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection
= 0.8 * 0.8 * 0.8 = 0.512
(b) P(at least one of the next three inspected fails) =
1 - P(none of the next three inspected fails) = 1 - P(all next three passes)
= 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .
(c) P(exactly one of the next three inspected passes) is given by ;
First vehicle pass the inspection, second and third vehicle doesn't passSecond vehicle pass the inspection, first and third vehicle doesn't passThird vehicle pass the inspection, first and second vehicle doesn't passHence, P(exactly one of the next three inspected passes) = Add all above cases ;
(0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = P(that none
of the next three vehicle passes) + P(Only one of the next three passes)
We have calculated the P(Only one of the next three passes) in the above part of this question;
Hence, P(at most one of the next three vehicles inspected passes) =
(0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)
= 0.008 + 0.096 = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;
P(All three passes/At least one of the next three vehicles passes inspection)
= P( All next three passes [tex]\bigcap[/tex] At least one of next three passes) /
P(At least one of next three passes)
= P( All next three passes ) / P(At least one of next three passes)
= P( All next three passes ) / 1 - P(none of the next three passes)
= [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .
Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
a) P(all of the next three vehicles pass) = 0.512
(b) P(at least one of the next three inspected fails) = 0.488
(c) P(exactly one of the next three inspected passes) = 0.096
(d) P(at most one of the next three vehicles inspected passes) = 0.104
(e) Given that at least one of the next three vehicles passes inspection, the probability that all three pass is approximately 1.049 (rounded to three decimal places).
Let's calculate the probabilities step by step:
(a) To find the probability that all of the next three vehicles pass, we can use the probability of a single vehicle passing (0.80) raised to the power of 3 (because the events are independent).
P(all pass) = (0.80)^3 = 0.512
(b) To find the probability that at least one of the next three vehicles fails, we can use the complement rule. It's easier to find the probability that none of them fail and then subtract that from 1.
P(at least one fails) = 1 - P(all pass) = 1 - 0.512 = 0.488
(c) To find the probability that exactly one of the next three vehicles passes, we need to consider three cases: Pass-Fail-Fail, Fail-Pass-Fail, and Fail-Fail-Pass. Each of these cases has a probability of (0.80) * (0.20) * (0.20).
P(exactly one passes) = 3 * (0.80) * (0.20) * (0.20) = 0.096
(d) To find the probability that at most one of the next three vehicles passes, we sum the probabilities of exactly one passing and none passing.
P(at most one passes) = P(none pass) + P(exactly one passes) = (0.20)^3 + 0.096 = 0.104
(e) To calculate the conditional probability that all three pass given that at least one passes, we use the formula for conditional probability:
P(all three pass | at least one passes) = P(all three pass and at least one passes) / P(at least one passes)
P(all three pass and at least one passes) = P(all pass) = 0.512 (from part a)
So, P(all three pass | at least one passes) = 0.512 / 0.488 ≈ 1.049 (rounded to three decimal places).
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Sammy read the line plot and said that there were 8 little league players who are 3 years old.
(View the picture.)
Is Sammy correct? If not, construct a viable argument and critique his response.
Answer:
where is the age three she is incorrect
Step-by-step explanation:
Answer:
she is correct its 8 people who are 3 years old and it says it
Step-by-step explanation:
Solve 5x − 6y = −38
3x + 4y = 0
(4, 3)
(−4, 3)
(4, −3)
(−4, −3)
Answer:
The answer is :- ( -4 , 3 )
List more than two events (i.e., categorical events) that might describe the outcome of each situation given below (a) A student applies for admission to Oxnard University. (You may select more than one answer. Click the box with a check mark for the correct answer and double click to empty the box for the wrong answer.) a) Admitted unconditionally b) Awarded a degree c) Not admitted d) Granted a visa e) Admitted conditionally
Answer:
Step-by-step explanation: From the above question, the possible outcome are:
a) Admitted unconditionally √
b) Awarded a degree √
c) Not admitted √
d) Granted a visa ×
e) Admitted conditionally √
Are G and F mutually exclusive events? Explain. G and F are mutually exclusive because if you are not a finalist, then you cannot receive a Green Card. G and F are not mutually exclusive because you could be a finalist and also win a Green Card. G and F are mutually exclusive because becoming a finalist will not allow a person to receive a Green Card. G and F are mutually exclusive because becoming a finalist and receiving a Green Card cannot occur together.
Answer: G and F are mutually exclusive because they cannot occur together
Step-by-step explanation:
According to the definition of mutually exclusive events,
The events which can not occur together and probability of them occurring together is 0 are known as mutually exclusive events.
The first statement gives an implication that if one happens then other happens meaning they could both still happen so it is not true.
The second statement contradict the question about being mutually exclusive events.
The third statement also is a implication that if one event occurs then other does or does not occur.
The last statement is correct one that conforms with the question and obeys the definition of mutually exclusive events.
Assume the random variable X is normally distributed, with mean 46 and standard deviation 8. Find the 7 th percentile.
Answer:
The 7th percentile is 34.2
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 problem, we have that:
[tex]\mu = 46, \sigma = 8[/tex]
Find the 7 th percentile.
The value of X when Z has a pvalue of 0.07. So it is X when Z = -1.475.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.475 = \frac{X - 46}{8}[/tex]
[tex]X - 46 = -1.475*8[/tex]
[tex]X = 34.2[/tex]
The 7th percentile is 34.2
A levee was designed to protect against floods with an annual exceedance probability of 0.02. A larger flood would cause the levee to fail. What is the risk that the levee will NEVER fail in the next 20 years?
Answer:
66.76% probability that the levee will NEVER fail in the next 20 years.
Step-by-step explanation:
For each year, there are only two possible outcomes. Either a levee fails during the year, or no levees fail. In each year, the probabilities of levees failing are independent from each other. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
A levee was designed to protect against floods with an annual exceedance probability of 0.02. This means that [tex]p = 0.02[/tex]
What is the risk that the levee will NEVER fail in the next 20 years?
This is [tex]P(X = 0)[/tex] when [tex]n = 20[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}*(0.02)^{0}*.(0.98)^{20} = 0.6676[/tex]
66.76% probability that the levee will NEVER fail in the next 20 years.
The risk that a levee, designed to withstand floods with an annual exceedance probability of 0.02, will never fail in the next 20 years is approximately 67%, assuming each year's flood risk is independent and identical.
Explanation:The question asks about a levee's failure probability over the next 20 years. If a levee is designed to protect against floods with an annual exceedance probability of 0.02, this means that there's a 2% chance each year that the levee will fail due to a flood.
To determine the risk that the levee will never fail in 20 years, we must first understand the probability that it won't fail in any given year, which is 1 - 0.02 = 0.98 (or 98%). The probability that the levee will not fail in 20 consecutive years is therefore 0.98^(20), or approximately 0.67 (or 67%), assuming each year's flood risk is independent and identical.
This calculation, while seemingly straightforward, doesn't account for variables like changes in climate, improvements in the levee's design and maintenance, or other unforeseen factors that might affect the levee's performance.
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There are 20 chemistry students to be scheduled for labs this term. 6 will be assigned to the Adams Hall lab, 11 to the Baker Hall lab, and the rest to the Craig Hall lab. How many possible assignments are there of students to labs?
Given the specific assignment of students to labs, there is only one possible way to assign the 20 students to the three labs.
Explanation:To solve this problem, we are essentially counting the number of different ways to distribute 20 students among three labs: Adams Hall, Baker Hall, and Craig Hall. Given that 6 students will be assigned to Adams Hall and 11 to Baker Hall, we already know where 17 of the 20 students will be placed. The only students left to place are the remaining 3 students, who must go to the Craig Hall lab. Therefore these students could be assigned in only one way given the constraints of the problem. Thus, only one assignment is possible.
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The diameter of a brand of ping-pong balls is approximately normally distributed, with a mean of 1.32 inches and a standard deviation of 0.08 inch. A random sample of 4 ping-pong balls is selected. Complete parts (a) through (d). a. What is the sampling distribution of the mean? A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. B. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 can not be found. C. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will not be approximately normal. D. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will be the uniform distribution.
Answer:
a) A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
b) [tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
c) [tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Part a
Let X the random variable that represent the diameter of a brand of ping pong of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1.32,0.08)[/tex]
Where [tex]\mu=1.32[/tex] and [tex]\sigma=0.08[/tex]
And we select a sample of size n =4 and we want to find the distribution for [tex]\bar X[/tex]
Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
So the best option for this case would be:
A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
Part b
[tex] P(\bar X<1.28)[/tex]
We can use the z score given by:
[tex]z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
Part c
[tex] P(1.28< \bar X<1.34)[/tex]
[tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
The sampling distribution of the mean diameter of the ping-pong balls, given that the population diameter is normally distributed, will also be normally distributed. The expected mean is 1.32 inches and the standard deviation will be 0.04 inches.
Explanation:The correct answer to this question should be A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. This is grounded on the Central Limit Theorem which states that the distribution of sample means tends to be normal regardless of the shape of the population distribution, especially when the sample size is large.
In this case, though our sample size (4) is fairly small due, because the distribution of the population is specified as normal, the sampling distribution will remain normal. The important numbers to know would be the expected value (mean which is the same as population mean i.e., 1.32 inches) and the standard deviation (which is the population standard deviation divided by square root of sample size i.e., 0.08/sqrt(4)).
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Find the volume of a cube with side length of 7 in.
A.) 147 in3
B.) 343 in3
C.) 49 in3
D.) 215 in3
Answer:
343 in³ - B
Step-by-step explanation:
Volume of cube = l³ = 7 in x 7 in x 7 in = 343 in³