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2 years ago

Can someone awsner this for me honestly I’m sorry but I really need these awsner but I need the answers for these 3 to

Mathematics

2 answers:

zalisa [80]2 years ago

5 0

1. 50 + 20x = 240

2. 85% so more than 75%

3. 25.9% so about 26%

Hope this helps and please consider marking branliest if this helped you

allsm [11]2 years ago

5 0

The second question is more than 75%

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A company that makes nails finds that it would sell 42 boxes of nails per month if they are priced at $20 per box. If the compan

denpristay [2]

The answer is D.

If x = number of boxes sold and y = price of nails box, then it is best represented by the equation y = 62 - x.
Checking:
If y = 20 and x = 42,
20 = 62 - 42
20 = 20

If y = 10 and x = 52.
10 = 62 - 52
10 = 10

Thank you for posting your question. I hope you found you were after. Please feel free to ask me another.

8 0

3 years ago

8m^2+ 40m ------------------- 8m

dsp73

Answer:

$(m+5)$

Step-by-step explanation:

Take out 8m as a factor from the top side of the equation. You are left with $\frac{8m(m+5)}{8m} = m+5$

6 0

2 years ago

Find the length of the tangent segment AB to the circles centered at O and O' whose radii are a and b respectively when the circ

bixtya [17]

Answer:

$AB=2\sqrt{ab}$

Step-by-step explanation:

From figure,

$OA=a, \quad \quad O'B=b\\\Rightarrow OO'= (a+b) \quad \quad \text{and}\quad OD=(a-b)$

In triangle $OO'D$

$(OO')^2=(OD)^2+(O' D)^2$

$\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}]$

Hence, $AB=2\sqrt{ab}$

6 0

3 years ago

Write the six step in division

joja [24]

Answer:

Divide the tens column dividend by the divisor. Multiply the divisor by the quotient in the tens place column. Subtract the product from the divisor

Step-by-step explanation:

5 0

2 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

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

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

2 years ago