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3 years ago

Last week's and this week's low temperatures are shown in the table below.

Mathematics

1 answer:

kicyunya [14]3 years ago

8 0

Answer:

the mean of this week’s temperatures

the mean of last week’s temperatures

the range of this week’s temperatures

Step-by-step explanation:

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A company’s stock earned $151,000,000. If these earnings represent $1.14 per share. How many shares of stock are there?

miskamm [114]

$\dfrac{151 \textrm{ \$M}}{1.14 \textrm{ \$/share}} = 132.45... \textrm{ M share}$

Answer: About 132 million shares

3 0

2 years ago

Pleas help me!!!

tigry1 [53]

The answer is about 900.

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

Find the mean and the standard deviation for each set of values. Show your work. 15,17,19,20,14,23,12

vekshin1

1. Mean.

$\overline{x}=\dfrac{15+17+19+20+14+23+12}{7}=\dfrac{120}{7}\approx\boxed{17.14}$

2. <span>Standard deviation.

First we have to square each value and calculate the mean of that squares:

$\overline{x^2}=\dfrac{15^2+17^2+19^2+20^2+14^2+23^2+12^2}{7}=\boxed{\dfrac{2144}{7}}$

so the variance:

$s^2=\overline{x^2}-\overline{x}^2=\dfrac{2144}{7}-\left(\dfrac{120}{7}\right)^2\approx\boxed{12.41}$

and the standard deviation:

$s=\sqrt{s^2}=\sqrt{12.41}\approx\boxed{3.52}$
</span>

3 0

3 years ago

Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height ov

Semmy [17]

Answer: p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

3 0

3 years ago