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3 years ago

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A researcher is studying a group of field mice. the distribution of the weight of field mice is approximately normal with mean 2

5 grams and standard deviation 4 grams. which of the following is closest to the proportion of field mice with a weight greater than 33 grams?
a)0.046
b)0.977
c)0.954
d)0.023

2 answers:

Svetllana [295]3 years ago

4 0

Answer:

b

Step-by-step explanation:

morpeh [17]3 years ago

3 0

Answer:

B

Step-by-step explanation:

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Which equation represents the line that is perpendicular to graph of 4x+3y=9 and passes through (-2,3)

goldfiish [28.3K]

4y = 3x + 18

Step-by-step explanation:

NOTE THAT A line that is perpendicular to another has a negative inverse of the slope of the other line. The products of their slopes, that is, is always -1

Therefore we can begin by finding the slope of this line defined by the function 4x+3y=9

3y = -4x + 9

y = -4/3 x + 9/3

y = -4/3 x + 3

The slope of the perpendicular line is, therefore;

¾ - this is the negative inverse of -4/3

Now that we know the slope, we need to find the y-intercept. This is where x = 0 and the line meets the y-axis;

i.e (0, y)

The other given point, where the line crosses is (-2, 3). Remember that to get the gradient we use the formula;

Gradient = Δ y / Δ x

¾ = (3 – y) / (-2 – 0)

¾ = (3 –y) / -2

¾ * -2 = 3 – y

-3/2 = 3 – y

-3/2 – 3 = -y

9/2 = y ←– This is the y-intercept

Remember the function of a straight line is given;

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

y = 3/4 x + 9/2

4y = 3x + 18

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Answer:

(a) Interest = Principal x Rate x Time

Step-by-step explanation:

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Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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