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3 years ago

15

Suppose the amounts of M&M's a machine dispenses into packages has a normal distribution with a mean of 25 M&M's and a s

tandard deviation of 1 M&M. What is the probability that there are more than 26 M&M's?

1 answer:

emmainna [20.7K]3 years ago

5 0

Answer:

16%

Step-by-step explanation:

The mean is 25, and the standard deviation is 1.

26 is 1 standard deviation above the mean.

According to the empirical rule, 68% of the population is between -1 and +1 standard deviations. That means that 32% is less than -1 or greater than +1 standard deviations. So half of that, or 16%, is greater than +1 standard deviations.

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Simplify each expression. (-3 + 3) + [8 + (-12)]

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2 years ago

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35. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

Montano1993 [528]

Answer:

The linear equation for the line which passes through the points given as (-2,8) and $(4,6), is written in the point-slope form as $y=-\frac{1}{3} x-\frac{26}{3}$ .

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-2,8) and (4,6).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-2,8) and (4,6). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute $6 \& 8$ for $y_{2}$ and $y_{1}$ respectively, and 4&-2 for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows, $m=\frac{6-8}{4-(-2)}$

$\begin{aligned}&m=\frac{-2}{6} \\&m=-\frac{1}{3}\end{aligned}$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-2 for $x_{1}$ , and 8 for $y_{1}$ in the above equation and simplify using the distributive property as follows, $y-8=-\frac{1}{3}(x-(-2))$\\ $y-8=-\frac{1}{3}(x+2)$\\ $y-8=-\frac{1}{3} x-\frac{2}{3}$$

Step 3 of 3

Simplify the equation further.

Add 8 on each side of the equation $y-8=-\frac{1}{3} x-\frac{2}{3}$ , and simplify as follows, $y-8+8=-\frac{1}{3} x-\frac{2}{3}+8$

$y=-\frac{1}{3} x-\frac{2+24}{3}$$\\ $$y=-\frac{1}{3} x-\frac{26}{3}$$

This is the required linear equation.

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1 year ago

Everbank Field, home of the Jacksonville Jaguars, is capable of seating 76,867 fans. The revenue for a particular game can be mo

k0ka [10]

The total revenue for the event would the total amount earned from the tickets sold. So if x people attended the event, then there would be $(161x). We must keep in mind though the the maximum seating is 76 876. That means that the maximum revenue that can be earned must not exceed $(161)(76 867) = . Hence we have f(x) = 161x where 0 ≤ x ≤ 76867. From this, we can see that the domain is [0, 76867] while range is [0, 12375587]<span>.</span>

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3 years ago

Point T is at (-3, 8). What are the coordinates of T' after R(y-axis) o R(x-axis)?

Whitepunk [10]

Given:

Point is T(-3,8).

To find:

The coordinates of T' after $R(y-axis)\circ R(x-axis)$ .

Solution:

We know that, $R(y-axis)\circ R(x-axis)$ means the figure reflected across the x-axis then reflected across y-axis.

If a figure reflected across x-axis, then

$(x,y)\to (x,-y)$

$T(-3,8)\to T_1(-3,-8)$

If a figure reflected across y-axis, then

$(x,y)\to (-x,y)$

$T_1(-3,-8)\to T'(-(-3),-8)$

$T_1(-3,-8)\to T'(3,-8)$

Therefore, the required point is T'(3,-8).

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2 years ago