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3 years ago

Solve the system of equations {3x-y=10, -x y=12} for Y

1 answer:

5 0

3x - y = 10 . . . (1)
-x + y = 12 . . . (2)

(1) + (2) => 2x = 22 => x = 22/2 = 11

From 2, -11 + y = 12 => y = 12 + 11 = 23

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Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in 1994. suppose a sample of 100

seropon [69]

Answer: The probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.

Explanation:

Step 1: Estimate the standard error. Standard error can be calcualted by dividing the standard deviation by the square root of the sample size:

$SE=\frac{0.8}{\sqrt{100}}=\frac{0.8}{10} = 0.08$

So, Standard Error is 0.08 million or $80,000.

Step 2: Next, estimate the mean is how many standard errors below the population mean $1 million.

$\frac{1 - 1.5}{0.08}$

$=-6.250$

-6.250 means that $1 million is siz standard errors away from the mean. Since, the value is too far from the bell-shaped normal distribution curve that nearly 100% of the values are greater than it.

Therefore, we can say that because 100% values are greater than it, probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.

5 0

3 years ago

GIVING BRAINLIEST Use the inequality 186< -3 (4x - 2).

a_sh-v [17]

Answer:

x < -15

Step-by-step explanation:

I solved the inequality but i don't know if that was what you were asking for lol, if not sorry.

3 0

2 years ago

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Vikentia [17]

$\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}$

6 0

3 years ago

24 points scored in 4 games; 48 points scored in 10 games... are they equivalent?

ch4aika [34]

Hello.

24 points scored in 4 games = 24 * 4.

24 * 4 = 96

48 points scored in 10 games = 48 * 10

48 * 10 = 480

<em>96≠480, they are not equivalent.</em>

5 0

3 years ago

Write the common rule to describe the translation using coordinate notation.

UNO [17]

The common rule is (x - 1, y + 3) can be used to describe the translation.

Step-by-step explanation:

Step 1:

The point J (-3, -4) becomes $J^{1}$ (-4, -1).

In order to write the rule for translation from J to $J^{1}$ , we subtract the x coordinate of J from $J^{1}$ and subtract the y coordinate of J from $J^{1}$ .

The x coordinate $= -4-(-3) = -4+3 = -1.$

The y coordinate $= -1-(-4) = -1+4 = 3.$

Step 2:

So from the calculations, we get that the x coordinate is subtracted by 1 i.e. $x-1$ and the y coordinate is increased by 3 i.e. $y+3$ .

So the common rule is (x - 1, y + 3).

4 0

3 years ago