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3 years ago

Solve for X: 2/5 (X - 4) = 20

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

6 0

Answer:

x=104

Step-by-step explanation:

2/5(x-4)=20

2/5x-8/5=20

2/5x=20+8/5

=21/3/5

x=21/3/5÷2/5

=104

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I don't know how to do the questions on the page can you help me

tino4ka555 [31]

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3 years ago

MANUFACTURING. The cost of manufacturing widgets is given by

nadezda [96]

Let's solve

C=2p+35

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2 years ago

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3/4x - 6y = 18<br>what's the x intercepts and y intercepts?

mash [69]

Answer:

X: (24,0)

Y(0,-3)

Step-by-step explanation:

X-intercept:

.75x-6y=18

plug in 0 for y

.75x-6(0)=18

divide by .75

x=24

(24,0)

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plug in 0 for x

.75(0)-6y=18

divide by -6

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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

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

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

5$ is the cost and the cost per ride is 50 cent , if your parents give you 20$ . Write and solve a linear equation to find how m

Lesechka [4]

ANSWER= 30 rides

STEP-BY-STEP EXPLANATION

y=mx+b

m= $0.5 (per ride)

b= $5 (the cost)

x(rides) and y(parents money)

y= 0.5 x + 5

20= 0.5 x +5

20 - 5 = 0.5 x

15=0.5x

x= $\frac{15}{0.5}$

x=30 rides

4 0

3 years ago