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serious [3.7K]
4 years ago
9

Copy and complete the table of value for the equation y =2x square - 7x - 9 for - < x < 6

Mathematics
1 answer:
Radda [10]4 years ago
5 0
Vo tdusri cheesa mucha confuse kar ria hai haina
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506

Step-by-step explanation:

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3 years ago
Which one of the following is true when ds/st =3t^2/2s
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Middle school math!  I actually know how to teach differential calculus to middle schoolers, but I wouldn't start with this example, where we're gonna have to integrate eventually.

We have

\dfrac{ds}{dt} = \dfrac{3t^2}{2s}

We have an odd mix of s and t on the right, which means this probably was an implicit derivative to begin with.  In other words, instead of s as a function of t, we have some equation of s and t that is satisfied.

We move the variables to their respective sides 

2s \ ds = 3t^2 dt

Integrating,

s^2 =t^3 + C

We only need one constant of integration.

I won't bother to write the square root. This is choice D.

7 0
3 years ago
What is the approximate percent increase when the population of a town increases from 12,089 to 15,025
MAXImum [283]

Answer:

✨The percent increase is 25%.✨

✨ Round the population amounts to the nearest thousand to find an estimate of the percent increase✨

✨~GOD LOVES YOU~✨

8 0
3 years ago
Solve the equation:<br> (-11) + [9] =
Sidana [21]

Answer:

-2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Fifty draws are made at random with replacement from the box [ 0 0 1 1 1]. There are 33 ticket 1’s among the draws. The expected
Aleks04 [339]

Answer:

- The Expected value for the sum is 30.

- The Observed sum of 50 draws is 33.

- The Chance Error on the 50 draws is 3.

- The Standard Error on 50 draws is 2.191.

Step-by-step explanation:

The box contains [0, 0, 1, 1, 1]

Using probability to predict the expected outcome.

On one draw, the probability of drawing a 0 is (2/5).

And the probability of drawing a 1 is (3/5).

Probability mass function would look like

X | P(X)

0 | 0.40

1 | 0.60

So, expected value on one draw would be

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (0×0.40) + (1×0.60) = 0.60.

Standard error on one draw = √[Σ(xᵢ - μ)²/N]

μ = E(X) = 0.60

Σ(xᵢ - μ)² = (0 - 0.60)² + (0 - 0.60)² + (1 - 0.6)² + (1 - 0.6)² + (1 - 0.6)² = 1.20

SE = √(1.2/5) = 0.490

So, for 50 draws (with replacement),

E(50X) = 50E(X) = 50 × 0.60 = 30.

For 50 draws, standard error = √50 × 0.490 = 2.191

The expected value for the sum = 30

The observed valued for the sum = (33×1) + (17×0) = 33

Chance Error = (Observed Outcome) - (Expected Outcome) = 33 - 30 = 3

Standard error gives an idea of how large the chance error would be.

Standard error on 50 draws = 2.191

Hope this Helps!!!

5 0
3 years ago
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