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2 years ago

C is the circle with the equation x^2+y^2=1

Mathematics

1 answer:

vodka [1.7K]2 years ago

4 0

Answer:

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suppose a parabola has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1). Write the eq

lesya [120]

Answer:

Step-by-step explanation:

f(x) = a(x - h)² + k - the vertex form of the quadratic function with vertex (h, k)

the<u> axis of symmetry</u> at<u> x = -8</u> means h = -8

the <u>maximum height of 2</u> means k = 2

So:

f(x) = a(x - (-8))² + 2

f(x) = a(x + 8)² + 2 - the vertex form of the quadratic function with vertex (-8, 2)

The parabola passing through the point (-7, -1) means that if x = -7 then f(x) = -1

so:

-1 = a(-7 + 8)² + 2

-1 -2 = a(1)² + 2 -2

-3 = a

Threfore:

The vertex form of the parabola which has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1) is:

8 0

3 years ago

Which of the following is the difference of two squares?

Blizzard [7]

Answer: C

Step-by-step explanation:

$25a^{2}=(5a)^{2}$ and $36b^{6}=(6b^{3})^{2}$

5 0

2 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

PLEASE PLEASE PLEASE I ALMOST DONE AND I NEED THIS QUESTION

sukhopar [10]

Answer:

107

Step-by-step explanation:

6 0

3 years ago

Someone help me ASAP!!! It’s a timed quiz

Aloiza [94]

Answer:

n=28

Step-by-step explanation:

4+3

7+5

12+7

19+9

the pattern adds two to the last number

4 0

3 years ago