What is the difference between drawing and constructing something

kondor19780726 [428]

9/4 is the answer :)

6 0

3 years ago

n the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In the

aniked [119]

Answer:

In the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In the second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half. Which expression represents the total number of points the player scored in the game?

a) 2x + 3x + 9

b) 2x + 3 + 9

c) 2x + 3x + 9x

d) 2 + 3x + 9

Answer:

a) 2x + 3x + 9

Step-by-step explanation:

3 0

3 years ago

A game show awards 30 points for each correct answer and deducts 50 points for each in correct answer. A contestant answers 2 qu

mina [271]

Answer:

-10 points

Step-by-step explanation:

A game show awards 30 points for each correct answer and deducts 50 points for each in correct answer. A contestant answers 2 questions incorrectly and 3 questions correctly. What is his final score?

Let us say Initial number of points = 0

Hence,

He answers 2 incorrectly = 0 - 50 - 50

= -100

He answers 3 correctly =-100 + 30 + 30 + 30

His final score = -100 + 30 + 30 + 30

= -10 points

8 0

3 years ago

Calcula el área de la siguiente figura

andreyandreev [35.5K]

Answer:

272.5 m²

Step-by-step explanation:

1/2x4x9=18

(9+18)/2=13.5. 13.5x11=148.5

(18+14)/2=16. 16x7=112

12+148.5+112= 272.5

5 0

3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0

3 years ago