How many months are equal to 45 days

Need help on these math questions. please and thank u

Svetach [21]

9514 1404 393

Answer:

a) x = -3

b) y = (28/27)x -27

Step-by-step explanation:

a) College street has a slope of 0, so is a horizontal line. 2nd Ave is perpendicular, so is a vertical line, described by an equation of the form ...

x = constant

For 2nd Ave to intersect the point (-3, 1), the constant must match that x-coordinate. The equation is ...

x = -3

__

b) Since Ace Rd is perpendicular to Davidson St, its slope will be the opposite reciprocal of the slope of Davidson St. The slope of Ace Rd is ...

m = -1/(-27/28) = 28/27

Using the point-slope equation for a line, we can model Ace Rd as ...

y -y1 = m(x -x1)

y -1 = (28/27)(x -27)

y = (28/27)x -27

4 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

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

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

Mr. Diaz has 347 apple trees in his orchard. He has 162 more apple trees than peach trees in his orchard . How many peach trees

strojnjashka [21]

Answer:

Mr. Diaz has 185 peach trees

Step-by-step explanation:

x = peach trees

x + 162 = 347

solve to get x = 185

5 0

3 years ago

How to write 4.651 in expanded form with decimals?

dsp73

Answer:

$0.4651 \times {10}^{1}$

5 0

3 years ago

Find 5 consecutive whole numbers if it is known that the sum of the squares of the first 3 numbers is equal to the sum of the sq

lora16 [44]

Let the first number of the five we are looking for equal n.

Because the numbers are consecutive, the next four can be expressed as (n+1), (n+2), (n+3) and (n+4).

Our series of 5 numbers is therefore n,(n+1),(n+2),(n+3),(n+4).

We are told the sum of the squares of the first 3 is equal to the sum of the squares of the last 2 therefore:
n² + (n+1)² + (n+2)² = (n+3)² + (n+4)²

expand all the brackets to give
n² + n² + 2n + 1 + n² + 4n + 4 = n² + 6n + 9 + n² + 8n + 16
3n²+6n+5 = 2n²+14n+25
n²-8n-20=0

factorising this gives us
(n+2)(n-10) = 0 so the solutions are n=10 and n=-2

That means that n,(n+1),(n+2),(n+3),(n+4) could be [-2, -1, 0, 1 and 2] or [10, 11, 12, 13 and 14].

Since the question asks for whole numbers, and negative numbers are not classed as whole numbers, the numbers we want must be 10, 11, 12, 13 and 14.

You can check this on a calculator by doing 10² + 11² + 12² (=365) and 13² + 14² (=365).

3 0

3 years ago

How many months are equal to 45 days​

How many months are equal to 45 days