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likoan [24]
2 years ago
5

Let v⃗ 1=⎡⎣⎢⎢⎢0.50.50.50.5⎤⎦⎥⎥⎥, v⃗ 2=⎡⎣⎢⎢⎢0.50.5−0.5−0.5⎤⎦⎥⎥⎥, v⃗ 3=⎡⎣⎢⎢⎢0.5−0.5−0.50.5⎤⎦⎥⎥⎥. v→1=[0.50.50.50.5], v→2=[0.50.5−0

.5−0.5], v→3=[0.5−0.5−0.50.5]. find a vector v⃗ 4v→4 in r4r4 such that the vectors v⃗ 1v→1, v⃗ 2v→2, v⃗ 3v→3, and v⃗ 4v→4 are orthonormal.
Mathematics
1 answer:
Morgarella [4.7K]2 years ago
6 0

Answer:

can you post a question that actually has the right graphics

Step-by-step explanation:

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The radius of the circular lens of a magnifying glass is 4 centimeters.
Nadusha1986 [10]

Answer: 50.24 square centimeters.

Step-by-step explanation:

Given: The radius of the circular lens of a magnifying glass = 4 centimeters

The area of a circle is given by :-

\text{Area}=\pi r^2, where r is the radius of the circle.

Then the area of the circular lens of a magnifying glass is given by :-

\text{Area}=\pi (4)^2=(3.14)(16)=50.24\text{ square centimeters }

Hence, the area of the circular lens of a magnifying glass = 50.24 square centimeters.

5 0
2 years ago
Which point is located at (4, –2)? On a coordinate plane, point A is 2 units to the left and 4 units up. Point B is 4 units to t
ycow [4]

Answer:

Point B is 4 units to the right and 2 units down

Step-by-step explanation:

Since our <em>x</em> is positive, we are going to the right of the x-axis (where positive values lay).

Since our <em>y</em> is negative, we are going down in the y-axis (where negative values lay).

7 0
2 years ago
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7. Classical linear model assumptions for time series Consider the following stochastic process {(x1, x2, x3, . . . , xk, yt): t
iogann1982 [59]

Answer:

Options are missing.

The options for the above question are:

TS.1: Linear in parameters.

TS.2:No perfect collinearity

TS.3: Zero conditional mean.

TS.4: Homoskedasticity.

TS.5: No serial correlation

TS.6: Normality.

Hence the correct answer is TS1 to TS 5

Step-by-step explanation:

Assumptions TS 1 to TS 5 are the minimum set of assumptions needed to for the OLS estimates to be the best linear unbiased estimators conditional on explanatory variables for all time periods.

The assumptions of Normality is not needed for the estimators to show the BLUE property

3 0
2 years ago
Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou
monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )

                   = P ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) = P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )

                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = P (\frac{6.2-8}{1.5} <  \frac{x - u}{ std} < \frac{7-8}{1.5} )

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0
2 years ago
a flower garden is in the shape of a right triangle. The longest side of the triangle measures 13 meters. One of the shorter sid
netineya [11]
The answer is 10.4 Hope this helps
7 0
2 years ago
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