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2 years ago

Ayudaaaaaaaaaaaaaaaaaa aaa aaa aaa Doy corona

Mathematics

1 answer:

zvonat [6]2 years ago

6 0

Answer:

The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

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On a mountain, the temperature decreases by 18°F every 5000 feet. What integer represents the change in temperature at 20,000 ft

attashe74 [19]

To find this answer, use multiples. For every 5000 feet ascended, deduct 18 degrees. For 20,000 feet, which is (4 * 5000), use (4 * -18). In this case, the answer is -72 degrees F, so the integer described would be -72.

7 0

3 years ago

Write -6.1 as a mixed number in simplest form.

butalik [34]

-6.1=-(6+1/10)

-6 1/10

8 0

2 years ago

Please help, I I beg you please

Tatiana [17]

Answer:

The volume of the cube is $\mathit{\frac{27}{z^3x^{9}}}$ cu in.

Step-by-step explanation:

The Volume of a Cube

Let's have a cube of side length a. The volume of the cube is:

$V=a^3$

The cube of the image has a side length of

$\displaystyle a=\frac{3x^{-3}}{z}\ inches$

Simplifying the expression of the base by converting the negative exponent in the numerator to the denominator:

$\displaystyle a=\frac{3}{zx^{3}}\ inches$

Now find the volume:

$\displaystyle V=\left(\frac{3}{zx^{3}}\ inches\right)^3$

Applying the exponents:

$\displaystyle V=\frac{3^3}{z^3x^{9}}\ inches^3$

$\displaystyle V=\frac{27}{z^3x^{9}}\ inches^3$

The volume of the cube is $\mathbf{\frac{27}{z^3x^{9}}}$ cu in.

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3 years ago

Does a ti 83 plus calculator work the same as a ti 84?

Roman55 [17]

Answer:

Yes. Both are graphing calculators. TI 83 has a black only screen while the TI 84 has a colored screen

5 0

2 years ago

Read 2 more answers

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

2 years ago