PLEASE HELP! The AHS football team did a weigh-in at the start of training camp. The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg. What percentage of players are under 86kg?

=============================================================

Explanation:

Notice how 98-2*6 = 98-12 = 86. So we're 2 standard deviations below the mean.

Start at the center, which is where the highest point is located. The value 98 is at the center. Then move 2 spots to the left, which has us move 2 standard deviations lower than the mean, and we arrive at 86. Refer to the diagram below.

From here, we add up the percentages in that diagram that are below the 86 mark. Those percentages are the 0.15% and 2.35%

So we get (0.15%) + (2.35%) = 2.5% which is the answer

Side note: This diagram is using the Empirical Rule. Some textbooks call it the "68-95-99.7 rule", but that's more clunky of a name in my opinion. The three values 68, 95 and 99.7 refer to the approximate percentage values within 1, 2 and 3 standard deviations of the mean.

step-by-step explanation:

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form