simplify ( 7^5) ^3
A (1/7) ^15
B 7^15
C 7^8
D 35^3​

Answer:

wait for another one

I am doing.

The correct option is option a) mutually exclusive.

Mutually exclusive events are events that cannot occur simultaneously. They are also referred to as disjoint or incompatible events. In probability theory and statistics, mutually exclusive events have a probability of zero occurring together.

For example, in a coin flip, getting "heads" and getting "tails" are mutually exclusive events because a coin cannot land on both "heads" and "tails" at the same time.

Another example would be the events "rolling a six on a dice" and "rolling an even number on a dice", these events are mutually exclusive because a dice can't land on a six and an even number at the same time.

In general, mutually exclusive events have no intersection, meaning that they don't share any outcomes.

Therefore, The correct option is option a) mutually exclusive.

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the correct choice is {0, 18}. These elements are unique to set C and do not appear in set A.

To find the set difference C - A, we need to remove all elements from A that are also present in C. Let's examine the sets:

C = {0, 6, 12, 18}

A = {2, 4, 6, 8, 10, 12}

We compare each element of A with the elements of C. If an element from A is found in C, we exclude it from the result. After the comparison, we find that the elements 2, 4, 8, 10 are not present in C.

Thus, the set difference C - A is {0, 18}, as these are the elements that remain in C after removing the common elements with A.

Therefore, the correct choice is {0, 18}. These elements are unique to set C and do not appear in set A.

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Answer:

y = 3x - 7

Step-by-step explanation:

We can use the two-point form formula to find the equation of a line that passes through two given points (x1, y1) and (x2, y2):

y - y1 = (y2 - y1) / (x2 - x1) * (x - x1)

where (x1, y1) and (x2, y2) are the two points and (x, y) represents any point on the line.

Using the given points, we have:

x1 = 1, y1 = 10

x2 = -4, y2 = -5

Substituting these values into the formula, we get:

y - 10 = (-5 - 10) / (-4 - 1) * (x - 1)

Simplifying the equation, we have:

y - 10 = 3(x - 1)

Expanding and rearranging, we get:

y = 3x - 7

Therefore, the equation of the line that passes through (1, 10) and (-4, -5) is y = 3x - 7

chatgpt

Answer:

a= 6

Step-by-step explanation:

(7a-4)= 38

7×6= 42

42-4= 38

The value of x is 65. Please note that this solution is based on the assumption that the angles QRP and PSQ are supplementary. If this assumption doesn't hold, feel free to let me know.

We need to find the value of x in the equation ZQRP = (2x + 20)° given that ZPSQ = 30°. Since the question doesn't provide enough information about the relationship between angles QRP and PSQ, I'll assume that they are supplementary angles (angles that add up to 180°). This assumption is based on the possibility that the angles form a straight line or a linear pair.

If angles QRP and PSQ are supplementary, their sum is 180°:

(2x + 20)° + 30° = 180°

Now, we can solve for x:

2x + 50 = 180

Subtract 50 from both sides:

2x = 130

Divide by 2:

x = 65

Answer:

rectangle

Step-by-step explanation:

thank u

(a) The statement "if a and b are diagonalizable, so is ab" is true. If both matrices a and b are diagonalizable, it is written as a product of diagonal matrix D and invertible matrix P: a = PDP^(-1) and b = PEQ^(-1).

To show that ab is also diagonalizable, we can calculate ab as ab = (PDP^(-1))(PEQ^(-1)) = PDEQ^(-1).

Since both D and E are diagonal matrices, their product DE is also a diagonal matrix.

Therefore, we can write ab as ab = PFQ^(-1), where F = DE is a diagonal matrix.

This demonstrates that ab can be expressed as a product of a diagonal matrix F and an invertible matrix P, indicating that ab is diagonalizable.(b) The statement "if a and b are diagonalizable, so is ab" is false. A counterexample to this statement can be given by considering the matrices:

a = [[1, 0], [0, 1]]

b = [[0, 1], [0, 0]]

Both matrices a and b are diagonalizable since they are already diagonal matrices. However, their product ab = [[0, 1], [0, 0]] is not diagonalizable because it has a non-diagonal Jordan form. Hence, this counterexample disproves the statement.(c) The statement "if a^2 is diagonalizable, then a is diagonalizable" is true. If a^2 is diagonalizable, it means that it can be written as a product of diagonal matrix D and invertible matrix

P: a^2 = PDP^(-1).

To show that a is also diagonalizable, we can take square root of both sides of equation: a = (PDP^(-1))^(1/2). Since D is a diagonal matrix, taking the square root of D means taking the square root of each diagonal element. This operation yields a new diagonal matrix D^(1/2).Thus, we can express a as a = PD^(1/2)P^(-1), which shows that a can be written as a product of a diagonal matrix D^(1/2) and an invertible matrix P. Therefore, a is diagonalizable.

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Based on the number of members and the ratio in which they chose the types of film, the number who chose Action in the second week more than the first week is 6 people.

Assuming that the number of members is 99 members, the number who chose Action on the second week were:

= (7 / (5 + 7 + 6)) x 99

= 39 people

The number who chose Action in the first week:

= (5 / (2 + 5 + 8)) x 99

= 33

The difference is:

= 39 - 33

= 6 people

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There were 16 ounces of oil in the lantern at noon.

Let's start by defining the variables we know. We'll call the amount of oil in the lantern at noon "x," the rate at which the oil burns "r," and the time elapsed from noon to 2 pm "t." We know that the amount of oil in the lantern at 2 pm is 12 ounces, and at 4 pm, it's 8 ounces.

We can use the rate of oil burning to create an equation relating the amount of oil in the lantern to the time elapsed. The equation is:

x - rt = y

where "y" is the amount of oil in the lantern at any given time after noon. We can solve for "x" by plugging in the values we know at 2 pm:

x - 2r = 12

And at 4 pm:

x - 4r = 8

Now we have two equations with two variables. We can solve for "r" by subtracting the second equation from the first:

2r = 4

r = 2

Now we can plug in "r" to one of the equations to solve for "x." Let's use the first equation:

x - 2(2) = 12

x - 4 = 12

x = 16

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The main difference between an area variance and a use variance is that area variance allows for an exception to zoning regulations related to the physical characteristics of a property, while use variance allows for an exception to regulations related to the intended use of a property.

An area variance is typically granted when a property owner is unable to comply with zoning regulations related to setbacks, building height, lot coverage, or other physical characteristics of a property.

In contrast, a use variance allows a property owner to use their property for a purpose that is not permitted under the current zoning regulations. This may include using a residential property for commercial purposes or using a commercial property for residential purposes.

Use variances are generally more difficult to obtain than area variances, as they require a showing of a unique hardship or practical difficulty that cannot be addressed through other means.

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Therefore, the transformation of the double integral can be expressed as follows: ∫∫g(x', y') (x' + 1)(y' - 1) dx' dy' = ∫∫g(x', y') (x' + 1)(y' - 1) |J| dA'

the alteration of variables A mathematical trick known as a formula is used to switch the variables in an integral from one set of variables to another. The integral is frequently simplified or made easier to solve by doing this. Given a function f(x) and two sets of variables x and y, the integral of f(x) with respect to x may be represented as the integral of a transformed function g(y) with respect to y, where g(y) is connected to f(x) by a functional connection between x and y. For instance, the integral of f(x) with respect to x can be expressed as the integral of f(sqrt(y)) with respect to y if x and y are connected by the equation y = x2.

How to solve?

x' = x - 1

y' = y + 1

∫∫g(x, y) xy dA = ∫∫g(x', y') (x' + 1)(y' - 1) dx' dy'

∫∫f(x, y) dA = ∫∫f(x', y') |J| dA'

|J| = | ∂x'/∂x ∂x'/∂y |

| ∂y'/∂x ∂y'/∂y |

|J| = | 1 0 |

| 1 1 |

Therefore, the transformed double integral can be expressed as follows:

∫∫g(x', y') (x' + 1)(y' - 1) dx' dy' = ∫∫g(x', y') (x' + 1)(y' - 1) |J| dA'

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Answer:

10, 60, 70, 120

Step-by-step explanation:

Correct on edge 2020

Answer:

10<s<60

70<p<120

Step-by-step explanation:

35-25=10

35+25=60

Answer:

Because we know the height of the mailbox is 4 feet and the length is 6 feet, and we also know that it's located 54 feet from the base of the flagpole, we can use proportions to solve the problem.

\(\frac{4}{6\\}\)=  \(\frac{x}{54}\)  

Cross-multiply 4 with 54 and 6 with x to result with 216=6x. Divide 6 to both sides to get 36. So, the height of the flagpole would be 36 feet.

In this scenario, the dependent variable is the time the pencil takes to become dull and the type of pencil.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

It is given that, there are three different pencils that Monique possesses. She's interested in finding out what kind of pencil maintains its sharpness the longest. Each pencil she uses leaves an identical mark on a piece of paper.

There are often two different sorts of variables in analytics. We obtained that independent variables will have an impact on dependent variables. What occurs as a result of the independent variable is referred to as a dependent variable.

Thus, in this scenario, the dependent variable is the time the pencil takes to become dull and the type of pencil.

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Answer:

6.4 because subtract 9.6 from 15

(a) The median lifetime of an engine is approximately 2.772 months, which is the value that divides the distribution into two equal halves.

b) The probability of having at least 2 successful engines in a sample of 10 engines is approximately 0.98168, or about 98.17%.

(a) To determine the median lifetime of an engine, we need to find the value of the random variable that divides the distribution into two equal halves.

The Gamma distribution is defined by two parameters: shape (k) and scale (θ). The mean and variance of a Gamma distribution are given by:

Mean = k * θ

Variance = k * θ^2

Given that the mean is 2 months and the variance is 4 months squared, we can equate these values to the corresponding formulas:

2 = k * θ

4 = k * θ^2

We can solve these equations simultaneously to find the values of k and θ:

Dividing the second equation by the first equation, we get:

4/2 = θ/θ^2

2 = 1/θ

θ = 1/2

Substituting this value back into the first equation, we have:

2 = k * (1/2)

k = 4

Therefore, the parameters of the Gamma distribution are k = 4 and θ = 1/2.

To find the median lifetime, we need to solve for the value of x such that the cumulative distribution function (CDF) of the Gamma distribution equals 0.5:

CDF(x) = 0.5

Using a statistical software or tables for the Gamma distribution, we can find that the median lifetime of an engine is approximately 2.772 months.

(b) The probability of at least 2 successful engines in a sample of 10 engines can be calculated using the cumulative distribution function (CDF) of the Gamma distribution. We need to find the probability of getting 2 or more engines with a lifetime exceeding 10 months.

Let X be the number of successful engines in a sample of 10. Since the lifetime of an engine is a continuous random variable, we need to use the cumulative distribution function for X.

P(X ≥ 2) = 1 - P(X < 2)

Using the properties of the Gamma distribution, we can calculate this probability:

P(X < 2) = P(X = 0) + P(X = 1)

The probability mass function (PMF) of the Gamma distribution is given by:

P(X = x) = (k^x * exp(-k)) / (x!)

For x = 0:

P(X = 0) = (4^0 * exp(-4)) / (0!) = exp(-4)

For x = 1:P(X = 1) = (4^1 * exp(-4)) / (1!) = 4 * exp(-4)

Therefore:

P(X < 2) = exp(-4) + 4 * exp(-4) ≈ 0.01832

Finally, we can calculate the probability of at least 2 successful engines:

P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.01832 ≈ 0.98168

Therefore, the probability of at least 2 successful engines in a sample of 10 engines is approximately 0.98168, or about 98.17%.

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Answer:

  24 km

Step-by-step explanation:

Multiply the given ratio by 3:

  1 cm : 8 km

  3 cm : 24 km . . . . . . ratio values multiplied by 3

The real distance to the shop is 24 km.

Answer: 16

Step-by-step explanation:

Answer:

above sea level

Step-by-step explanation:

HOPE THIS HELPS YOU.

GOOD NIGHT

In triangle GHI, we can see that GH and GI are the legs, and HI is the hypotenuse. So,   GH^2 + GI^2 = HI^2

To complete the table, we need to use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the legs (the two shorter sides) is equal to the square of the hypotenuse (the longest side).

In triangle ABC, we can see that AB and AC are the legs, and BC is the hypotenuse. So, we have:

AB^2 + AC^2 = BC^2

Similarly, in triangle DEF, we can see that DE and DF are the legs, and EF is the hypotenuse. So, we have:

DE^2 + DF^2 = EF^2

Finally, in triangle GHI, we can see that GH and GI are the legs, and HI is the hypotenuse. So, we have:

GH^2 + GI^2 = HI^2

By using the Pythagorean theorem, we can confirm that the relationship holds for all the triangles above.

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The two-period moving average model and the two-period weighted moving average model are both common forecasting methods used to predict future demand. and we understand that the model with the lower MAD and MSE values will have the most accurate forecast.

To determine which model is better for this particular data set, we need to compare the accuracy of each model. To do this, we will calculate the Mean Absolute Deviation (MAD) and the Mean Squared Error (MSE) for each model.
For the two-period moving average model, we can calculate the forecast for period 7 by taking the average of p5 and 6:

Period 7 forecast = (Gallons in Period 5 + Gallons in Period 6)/2

For the two-period weighted moving average model, we can calculate the forecast for period 7 by using the weights of 0.6 and 0.4:

Period 7 forecast = (0.6 x Gallons in Period 5) + (0.4 x Gallons in Period 6)

We can then compare the accuracy of each model by calculating the MAD and MSE. To calculate MAD, we need to subtract the actual demand in each period from the forecasted demand and take the absolute value:

MAD = |Actual demand – Forecasted demand|

To calculate MSE, we need to square the differences between the actual demand and the forecasted demand:

MSE = (Actual demand – Forecasted demand)^2

After calculating the MAD and MSE for each model, we can compare the results to determine which model is better for this data set. The model with the lower MAD and MSE values will have the most accurate forecast.

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The answer is -9

Step-By-Step explanation:

The number of weet to buy is 84.

We have amount given to one friend and we need to find number of weet given to all of his friends. We use unitary methods to calculate the number of weets.

Let total number of weet be X

Number of weet given to 1 friend = 3

Number of weet given to 28 friend = 28 X 3

                                                      = 84

Therefore X = 84

84 weet will be distributed among 28 friend.

Wasim wants 10 weet extra

So, total number of wee he will purchase =

Total weet given to his 28 friends + total extra weet

= 84 + 10

= 94

Hence, Waim will buy 94 weets.

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Answer: 40

Step-by-step explanation:

38+ 52=90

230-90=40

Answer:

a)2 * 2 = 4

b)2 * 3/2 = 3

c) 20 * 3/2 = 30

d) 100 * 8/5 = 160

e)473*1/100=4.73

f)111*1/100=1.11

Answer:

answers are written

Step-by-step explanation:

Answer:

(i) From 0 to 50.

(ii) From 3 to 5.

(iii) From 3R to 5R, where R=3,4,5,...

Step-by-step explanation:

Let N be the number of players in a team.

As each team can have numbers of player ranging from 3 to 5, so

\(3\leqN\leq5\cdots(i)\)

Let P be the points scored by one player in one round of the game.

As each player can score up to 10 points in each round of the game, so

\(0\leq P \leq 10\cdots(ii)\)

(1) There are 5 members in the team formed by Elena.

So, points earned by 5 players in 1 round = 5P

From equation (ii), the range is

\(0\leq 5P \leq 50\)

Hence, the points earned by Elena's team is ranging from 0 to 50.

(2) If each played misses 1 point, then,

Points earned by 1 player in 1 round = 10-1=9

Hence, points earned by 5 players in 1 round = 9x5=45.

(3) Let R (an integer) denoted the number of teams.

As there are at least 3 teams in the game, so

\(R\geq3\cdots(iii)\)

As the number of players in one team is N, so

the number of players in R team is NR, where R=3,4,5,...

Now, from equation (i)

\(3R\leqNR\leq5R\)

Hence, if there are at least 3 teams, then the number of players is ranging from 3R to 5R.

For R=3, the number of players in the game is ranging from 3x3=9 to 5x3=15.

The expressions when evaluated are 3/5 and 1/2√5

From the question, we have the following parameters that can be used in our computation:

√9/√25

Evaluate the roots

So, we have

3/5

For the second expression, we have

√15/√12

This gives

√(15/12)

Divide

√(5/4)

So, we have

1/2√5

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False. The given sequence X; where (e ; t = 0,1,... is an i.d sequence with zero mean and constant variance of σ^2, does not necessarily imply that the process is asymptotically uncorrelated.

The term "asymptotically uncorrelated" refers to the property where the autocovariance between observations of the time series tends to zero as the lag between the observations increases. In the given sequence, since the random variables e; are independent, the cross-covariance between different observations will indeed tend to zero as the lag increases. However, the process may still have non-zero autocovariance for individual observations, depending on the properties of the underlying random variables.

In order for the process to be asymptotically uncorrelated, not only should the cross-covariance tend to zero, but the autocovariance should also tend to zero. This would require additional assumptions about the distribution of the random variables e; beyond just being i.d with zero mean and constant variance.

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