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4 years ago

Write an expression for the perimeter of the triangle shown below:

Mathematics

2 answers:

Afina-wow [57]4 years ago

7 0

4.1x +29

Hope this helps!! :)

sergiy2304 [10]4 years ago

3 0

Answer:

The answer is A.

Step-by-step explanation:

1. Combine like terms (2.3x + 2x= 4.3x - 0.2x = 4.1x)

2. add the 14 + 15 = 2

That gives you the equation 4.1x + 29

I also took the test so I know I it's right ;)

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Debby, Ella and Unique invest $10,000 each into an oil company. Debby owns 2000 $1 common stocks, Ella owns 1000 of 5% $50 prefe

Mashutka [201]

Answer:

Ella has the greatest return in the current year.

Step-by-step explanation:

Debby would receive $0.80 for each of her 2000 common stock in the oil company,hence Debby's return on investment in the current year is $1600($0.80*2000)

Besides,Ella's return on the stock investment in the current year is computed thus:

Ella's return= 5%*1000*$50=$2,500

In addition,Unique's dollar return on the investment is computed as follows:

Unique's return on investment=4%*2000*$20=$1,600

From the above computations,Ella seems to have the highest return in the current year of $2,500 whereas the two others managed to have $1600 return each

5 0

3 years ago

Write the equivalent fractions of 11/12

skelet666 [1.2K]

Answer:

11/2, 22/4, 33/6, 44/8, 55/10, 66/12, 77/14, 88/16, 99/18, 110/20, 121/22, 132/24, 143/26, 154/28, 165/30, 176/32, 187/34, 198/36, 209/38, 220/40, and so on ...

8 0

3 years ago

7. Find the slope of the line through ( -2,6 ) & (3,14).

hammer [34]

14-6 over 3--2
8/5

.................,

5 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

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