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3 years ago

This table represents a quadratic function.

Mathematics

1 answer:

AVprozaik [17]3 years ago

7 0

Answer:

B. 1/2

Step-by-step explanation:

y = ax^2 + bx + c

14 = a(0)^2 + b(0) + c

c = 14

10.5 = a(1)^2 + b(1) + 14

10.5 = a + b + 14 ____(i)

8 = a(2)^2 + b(2) + 14

8 = 4a + 2b + 14

4 = 2a + b + 7 ___ (ii)

i - ii

10.5 - 4 = -a + 7

6.5 = -a + 7

a = 7- 6.5

a = 0.5

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Select other ways to express 10². Mark all that apply.

Elanso [62]

F and D is the answer
hope this helps
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7 0

3 years ago

Read 2 more answers

3 + 4 + 7 = 14 multiplication

SSSSS [86.1K]

Multiplication:
7×2=14.

8 0

3 years ago

Read 2 more answers

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago

Expressions is equal to 2(40 - p)?

Bumek [7]

80 - 2p
You use the distributive property

8 0

2 years ago

What positive value of x makes the following equation true: 125 = x^3? Explain

alina1380 [7]

Answer: The correct answer is x=5

Step-by-step explanation:

x=2−5−−75=2−5−5i3=−2.5000−4.3301i

x=2−5+−75=2−5+5i3=−2.5000+4.3301i

x=5

Hoped I helped

can you mark me as brainliest

4 0

3 years ago