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iragen [17]
3 years ago
8

What does b do in y=ax^2+bx+c. How can b being positive or negative affect a graph?

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

The positions of the vertex and the axis of symmetry change.

Step-by-step explanation:

Let’s examine three parabolas in which a and c remain constant and only b changes.

(1) y = x² -  2x + 1

(2) y = x²         + 1

(3) y = x² + 2x + 1

In the image below,

Equation (1) is the black parabola, Equation (2) is green, and Equation (3) is red.

When b is more negative, the vertex moves down, and the axis of symmetry moves to the right.

When b is more positive, the vertex moves down, and the axis of symmetry moves to the left.

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Answer:

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Hope it helps

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3 years ago
The stem-and-leaf plot shows the number of points scored by the Bears in each of their basketball games this season. Identify th
denis-greek [22]

Answer:

Mean = 30.3

Median = 30

They scored 30 or more points on 11 games.

Step-by-step explanation:

Mean = all the numbers added together divided by 21

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3 years ago
Read 2 more answers
Surface-finish defects in a small electric appliance occur at random with a mean rate of 0.3 defects per unit. find the probabil
mario62 [17]

Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

Step-by-step explanation:

Since we have given that

Mean rate defects per unit = 0.3

Since we will use "Poisson distribution":

P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}

But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.

P(X\geq 2)=1+P(X=0)+P(X=1)

So,

P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22

so, it becomes,

P(X>2)=1-(0.74+0.22)=1-0.96=0.04

Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

4 0
4 years ago
The school plans to spend $5000 to order computers for its
ioda

Answer:

$400

Step-by-step explanation:

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3 years ago
a rectangular playing field is 80 yd long and has an area of 4400 yd squared. find the width of the field
klemol [59]

The answer is 55 yd.


You can find this by dividing the area, 4400 yd^2, by the length, 80 yd, which gets you 55 yd.

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