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Phoenix [80]
3 years ago
6

Which of the equations is graphed below?

Mathematics
2 answers:
masya89 [10]3 years ago
8 0
D is the correct answer bc if you plug in 0 you get 1 which is a point on the graph
ra1l [238]3 years ago
4 0

Answer:

D

Step-by-step explanation:

Slope of -2 that's going through y=1

so y=-2x+1

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PLEASE HURRY!! TIMED!! WILL GIVE BRAINLIEST!!!
tensa zangetsu [6.8K]

Answer:

2/3 +9.26 = 0.6666+9.26=rational

4 0
3 years ago
A home gardener estimates that 24 apple trees will have an average yield of 104 apples per tree. But because of the size of the
d1i1m1o1n [39]

Answer:

The farmer should plant 14 additional trees, for maximum yield.

Step-by-step explanation:

Given

Trees = 24

Yield = 104

x \to additional\ trees

So, we have:

Trees = 24 + x

Yield = 104 - 2x

Required

The additional trees to be planted for maximum yield

The function is:

f(x) = Trees * Yield

f(x) = (24 + x) * (104 - 2x)

Open bracket

f(x) = 24 * 104 + 104x - 24 * 2x - x * 2x

f(x) = 2796 + 104x - 48x - 2x^2

f(x) = 2796 + 56x - 2x^2

Rewrite as:

f(x) = - 2x^2 + 56x + 2796

Differentiate

f'(x) = -4x + 56

Equate f'(x) = -4x + 56 to 0 and solve for x to get the maximum of x

-4x + 56 = 0

-4x =- 56

Divide by -4

x =14

The farmer should plant 14 additional trees, for maximum yield.

5 0
3 years ago
NEED HELP!!!!!!!!!!!!!
solniwko [45]

Answer:

5x - 4

Step-by-step explanation:

4x - 2 + x - 2

simplify

5x - 4

hope this helps <3

8 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

i)

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
Plz help !! Financial algebra
aleksklad [387]

Answer:

$8.85

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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