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Setler [38]
3 years ago
10

Find the solution set of this inequality. Select the correct graph. |-8x+4|>12

Mathematics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

The first option is correct

Step-by-step explanation:

Hope it is helpful....

You might be interested in
On an interval of [0, 2π), can the sine and cosine values of a radian measure ever be equal? If so, enter the radian measure(s)
Lisa [10]

Answer:

Yes, they are equal in the values (in radians):

π/4, 5π/4

If cos(x) and sin(x) are defined to you as nonnegative functions (in terms of lengths), then 3π/4 and 7π/4 are also included

Step-by-step explanation:

Remember that odd multiples of 45° are special angles, with the same sine and cosine values (you can prove this, for example, by considering a right triangle with an angle of 45° and hypotenuse with length 1, and finding the trigonometric ratios).

The radian measure of 45° corresponds to π/4, hence the odd multiples on the interval [0, 2π) are π/4, 3π/4, 5π/4, 7π/4.

If you define sin(x) and cos(x) using the cartesian coordinate system (via unit circle), then cos(3π/4)=-sin(3π/4) and cos(7π/4)=-sin(7π/4). In this case, only π/4 and 5π/4 are valid choices.  

6 0
3 years ago
Alguien me ayuda a resolver esto por favor es urgente 1/2+1/4=​
natulia [17]

Answer:

48

Step-by-step explanation:

Solo quiero ganar puntos

5 0
3 years ago
A player of the National Basketball Association’s Portland Trail Blazers is the best free-throw shooter on the team, making 94%
g100num [7]

Answer:

The data for the probabilities are shown in the table below.

- A represents the probability of making the two shots for each of the best and worst shooter on the Portland Trail Blazers' team

- B represents the probability of making at least one shot for each of the best and worst shooter on the Portland Trail Blazers' team.

- C represents the probability of not making any of the two shots for each of the best and worst shooter on the Portland Trail Blazers' team.

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Step-by-step explanation:

Part 1

Probability of the best shooter of the National Basketball Association’s Portland Trail Blazers making a shot = P(B) = 94% = 0.94

Probability that he doesn't make a shot = P(B') = 1 - 0.94 = 0.06

a) Probability that the best shooter on the team makes the two shots awarded = P(B) × P(B) = 0.94 × 0.94 = 0.8836

b) Probability that the best shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(B) × P(B') + P(B') + P(B)

= (0.94 × 0.06) + (0.06 × 0.94) = 0.1128

Probability that he makes two shots = 0.8836 (already calculated in part a)

Probability that he makes at least one shot = 0.1128 + 0.8836 = 0.9964

c) Probability that the best shooter on the team makes none of the two shots = P(B') × P(B') = 0.06 × 0.06 = 0.0036

d) If the worst shooter on the team, whose success rate is 56% is now fouled to take the two shots.

Probability of the worst shooter on the team making a shot = P(W) = 56% = 0.56

Probability that the worst shooter on the team misses a shot = P(W') = 1 - 0.56 = 0.44

Part 2

a) Probability that the worst shooter on the team makes the two shots = P(W) × P(W)

= 0.56 × 0.56 = 0.3136

b) Probability that the worst shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(W) × P(W') + P(W') + P(W)

= (0.56 × 0.44) + (0.44 × 0.56) = 0.4928

Probability that he makes two shots = 0.3136 (already calculated in part a)

Probability that he makes at least one shot = 0.4928 + 0.3136 = 0.8064

c) Probability that the worst shooter makes none of the two shots = P(W') × P(W') = 0.06 × 0.06 = 0.1936

From the probabilities obtained

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
Find the value of x in the triangle
AURORKA [14]
Cos38 = 7,8/x
X= 7,8/cos38
= 9,90
8 0
3 years ago
Read 2 more answers
3<br><img src="https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B7%7D%20%20%3D%20" id="TexFormula1" title=" {3}^{7} = " alt=" {3}^{7} =
DedPeter [7]

Answer:2187


Step-by-step explanation:

3*3*3*3*3*3*3=2187

3 0
3 years ago
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