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Kryger [21]
3 years ago
5

Help me please I’m confused.

Mathematics
1 answer:
Marianna [84]3 years ago
5 0

Answer: c

Step-by-step explanation:

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Connor gathered data about the height of customers entering a store and the time that it took them to gather their items and ent
sattari [20]

0.88. you can simply see this by using the right use of the data

7 0
2 years ago
The table below shows some inputs and outputs of the invertible function f etc ​
Snowcat [4.5K]

Answer:

(a)\ f^{-1}(-15) = -6

(b)\ f^{-1}(4) + f(9)=0

Step-by-step explanation:

Given

The attached table

(a)\ f^{-1}(-15)

This represents an inverse function.

So, we look into x row for its value.

i.e.

f^{-1}(-15) = -6

(b)\ f^{-1}(4) + f(9)

Just like (a)

f^{-1}(4) = -11 ---- by looking into the x rows

f(9) = 11

So:

f^{-1}(4) + f(9)=-11 + 11

f^{-1}(4) + f(9)=0

4 0
2 years ago
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
bekas [8.4K]

9514 1404 393

Answer:

  • (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
  • NOT linearly independent

Step-by-step explanation:

We want ...

  c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

  c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

  x(c1 +2c3) = 0

  x²(c2 -4c3) = 0

These have an infinite set of solutions:

  c1 = -2c3

  c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

  (c1, c2, c3) = (-2t, 4t, t)

__

f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0
2 years ago
A deck of playing cards has four suits, with thirteen cards in each suit consisting of the numbers 2 through 10, a jack, a queen
vekshin1
Since order does not matter, you use a combination and not a permutation, so the first one is true, which means the second one is not true.
The probability of choosing two diamonds and three hearts can be represented by (13C2 * 13C3)/52C5, which is 0.0086, not 0.089, so the third one is not true.
The probability of choosing five spades and the probability of choosing five clubs are represented by the same thing, 13C5/52C5, which is roughly 0.0005, so the fourth one is not true but the fifth one is. So the answer is the first and fifth one.

4 0
3 years ago
Read 2 more answers
The equation x2 – 1x – 90 = 0 has solutions {a, b}. What is a + b
Andrews [41]
X^2 - x - 90 =0
This equation is in standard form ax^2+ bx + c
The sum of the solutions is -b/a (a and b are the coefficients from the equation above, not the solutions to the equation)
The answer is 1/1 = 1
8 0
2 years ago
Read 2 more answers
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