All categories

4 years ago

What is -3/4 + 1/5 + 2/9 fraction or decimal form

Mathematics

1 answer:

aivan3 [116]4 years ago

7 0

Answer:

-59/180

Step-by-step explanation:

if you add these three fractions you get -59/180.

You might be interested in

I need help please!!!!

givi [52]

Answer:

1/3( x-5)= -2/3

Multiply both sides by 3

x = 3

Step-by-step explanation:

1/3( x-5)= -2/3

Multiply both sides by 3

3*1/3( x-5)= -2/3*3

x-5 = -2

Add 5 to each side

x-5+5 = -2+5

x = 3

5 0

3 years ago

Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that

vivado [14]

The Profit = Revenue - CostP ( x ) = ( 40 x - 0.5 x² ) - ( 5 x + 15 ) = = 40 x - 0.5 x² - 5 x - 15 = - 0.5 x² + 35 x - 15The maximum profit is when: P`( x ) = 0P ` ( x ) = ( - 0.5 x² + 35 x - 15 ) ` = - x + 35- x + 35 = 0x = 35 ( the number of units )P max = - 0.5 · 35² + 35 · 35 - 15 = - 612.5 + 1225 - 15 = 597.5Answer: The maximum profit is 597.5 and the number of units is 35.

7 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

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