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sp2606 [1]
2 years ago
15

By running the same software on a computerthat is 40% faster, how many more frames per second can be displayedthan on the slower

, current computer?
I can’t show the picture because I’m not on computer. It’s on itc pitsco on rca 4 video production. If you know how to do this I’ll give you a reward ;)
Mathematics
1 answer:
labwork [276]2 years ago
8 0

Answer:

Step-by-step explanation:

Let the current display rate be f (in frames per second).  If the new computer is 40% faster, then the new display rate will be 1.40f (frames per second).

You might be interested in
Find the answer to this problem 15/16 - -3 / 16​
posledela

Answer:

The answer would be 18/16 or 1 1/8.

4 0
3 years ago
A standard roulette wheel with slots labeled 0, 00, 1, 2, 3, … , 36 is spun 60 times. Of these spins, the number 7 is spun 7 tim
WARRIOR [948]

Since the p-value of the test is of 0.00001 < 0.01, these results have statistical significance.

<h3>What is the relation between the p-value and the conclusion of the test hypothesis?</h3>

Depends on if the p-value is less or more than the significance level:

  • If it is more, the null hypothesis is not rejected, which means that the results do not have statistical significance.
  • If it is less, it is rejected, which means that the results have statistical significance.

In this problem, the probability is the p-value, hence since the p-value of the test is of 0.00001 < 0.01, these results have statistical significance.

More can be learned about p-values at brainly.com/question/13873630

#SPJ1

6 0
2 years ago
Solve for a. 7a + 18 = 102
Oliga [24]

Answer:

a=12

Step-by-step explanation:

7a+18=102

7a=102-18

7a=84

a=84/7

a=12 <----- answer

6 0
3 years ago
Read 2 more answers
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
Functions.
never [62]

Answer:

2,5 and 6

Step-by-step explanation:

The answer is going to be option 2, 5 and 6 because the line touches -4,2 and 4 on the x axis. And these are the x values so that's why this is going to be the answer.

Hope this information helps you!

4 0
3 years ago
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