The improper fraction of 1 IS.... 81 0 3 2 3 S 3

Sophie [7]

Answer:

H. 40 inches

Step-by-step explanation:

On Wednesday, he is 40 inches taller. ... That would make 5 days of growth, for 100 inches. But this is only 3 days therefore he would grow 40 inches taller

5 0

2 years ago

A chord of a circle is a line segment connecting any point on the circle to the center of the circle.

andrezito [222]

Answer:

FALSE

Step-by-step explanation:

A chord of a circle is a line segment that connects two points on the circle.

5 0

3 years ago

Julie and her brother go to Atlanta to ride the SkyView Ferris wheel. It measures 200 feet in diameter.

worty [1.4K]

Answer:

Part A) Circumference

Part B) $C=\pi D$

Part C) The distance traveled in one rotation is 628.32 feet

Step-by-step explanation:

Part A) we know that

The distance around the circle is equal to the circumference.

The Ferris Wheel have a circular shape

so

To find out the distance around the Ferris Wheel you should use the circumference

Part B) What is the formula needed to solve this problem?

we know that

The circumference is equal to multiply the number π by the diameter of the circle

so

$C=\pi D$

Part C) What is the distance traveled in one rotation?

we know that

One rotation subtends a central angle of 360 degrees

The distance traveled in one rotation is the same that the circumference of the Ferris wheel

we have

$D=200\ ft$ ----> diameter of the Ferris wheel

substitute in the formula of circumference

$C=\pi (200)\\C=200\pi\ ft$

assume

$\pi=3.1416$

$C=200(3.1416)=628.32\ ft$

therefore

The distance traveled in one rotation is 628.32 feet

3 0

3 years ago

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$