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3 years ago

HELP ME ASAP! Will give BRAINLIEST! Please read the question THEN answer correctly! No guessing. Check ALL that apply

Mathematics

1 answer:

sammy [17]3 years ago

4 0

Answer:

C. 65

Step-by-step explanation:

Compare to the standard form ...

ax² +bx +c

to find the values of a, b, c. They are ...

a=2, b=3, c=-7

The discriminant is ...

d = b² -4ac

d = (3)² -4(2)(-7) = 9 +56

d = 65 . . . . . . matches choice C

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How many $0.21 pencils can be purchased for $7.00? Show your work.

IgorLugansk [536]

N*(21/100)=(700/100)

n=(700/100)/(21/100)

n=(700/100)*(100/21)

n=700/21

n=33+1/3

So thirty three $0.21 pencils can be purchased for $7.00.

8 0

2 years ago

Read 2 more answers

A parabola and its directrix are shown on the graph. On a coordinate plane, a parabola opens upward. It has a vertex at (0, 0) a

sukhopar [10]

Answer:

The focus of the parabola is at the point (0, 2)

Step-by-step explanation:

Recall that the focus of a parabola resides at the same distance from the parabola's vertex, as the distance from the parabola's vertex to the directrix, and on the side of the curve's concavity. In fact this is a nice geometrical property of the parabola and the way it can be constructed base of its definition: "All those points on the lane whose distance to the focus equal the distance to the directrix."

Then, the focus must be at a distance of two units from the vertex, (0,0), on in line with the parabola's axis of symmetry (x=0), and on the positive side of the y-axis (notice the directrix is on the negative side of the y-axis. So that puts the focus of this parabola at the point (0, 2)

5 0

3 years ago

Read 2 more answers

Can you help me with number 6? Confused abit Please

Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.

To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.

First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is

$\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}$

Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is

$\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}$

Third question:

To get exactly one six, it can either be the first, second or third roll.

In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.

In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus

$\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}$

And since there are three of these combinations, The answer is

$3\frac{5^2}{6^3}$

Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is

$1 - \frac{5^3}{6^3}$