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3 years ago

Consider the inequality 5x<30

1 answer:

4 0

If any x value less than 25 makes the inequality true then 5 x 7 should not be greater than 30 therefore saying any value of x less than 25 could be greater or equal to 30.

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ANSWER QUICKLY PLZ :)

AfilCa [17]

The answer is (-1,1)
The solution is the point where the two lines cross each other
Hope this helps

5 0

3 years ago

The number by which the dividend is being divided? Means?

Ann [662]

That would be called the quotient.

7 0

3 years ago

Read 2 more answers

The Japanese 5 yen coin has a hole in the middle. What are the possible ways to calculate the difference between the distances f

ipn [44]

Answer:a) Subtract the two diameters and multiply by 3.14

d) Subtract the circumference of the hole from the circumference of the coin

Step-by-step explanation:

i just did the assignment

9 0

3 years ago

Read 2 more answers

The base of a rectangular prism has an area of 26.25, and the height of the prism is 4.5. What is the volume? Round to the neare

kramer

The right answer is 118.

Look at the attached picture

Hope it will help you....

Good luck on your assignment.

4 0

3 years ago

Read 2 more answers

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

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

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

8 0

3 years ago