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3 years ago

Chromebook - $249

Mathematics

1 answer:

STALIN [3.7K]3 years ago

3 0

Answer:

Discount- $62.25

Price after discount-$186.75

Step-by-step explanation:

25% of 249 is 62.25.

Therefore, 249-62.25=186.75

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GIVING 100 POINTS!!!!

IrinaK [193]

Answer:

As per dot plots we see the distribution of prices is close but majority of prices are concentrated in different zones. So MAD would be more similar by the look.

<u>Let's verify</u>

<h3>Neighborhood 1</h3>

55, 55, 60, 60, 70, 80, 80, 80, 90, 120

(55*2+ 60*2+ 70+ 80*3 + 90+ 120)/10 = 75

(20*2+15*2+5+5*3+15+45)/10 = 15

<h3>Neighborhood 2</h3>

100, 110, 110, 110, 120, 120, 120, 140, 150, 160

(100 + 110*3+ 120*3+ 140 + 150+ 160)/10 = 124

(24+14*3+4*3+16*3+16+26+36)/10 = 20.4

As we see the means are too different (75 vs 124) than MADs (15 vs 20.4).

7 0

3 years ago

Read 2 more answers

Find the length of the arc

worty [1.4K]

Answer:

33 Km

Step-by-step explanation:

The length of the arc is calculated as

arc = circumference of circle × fraction of circle

= 2πr × $\frac{135}{360}$

= 2π × 14 × $\frac{3}{8}$

= 28π × $\frac{3}{8}$ ≈ 33 Km ( to nearest Km )

5 0

3 years ago

The graph of an absolute value function opens down and has a vertex of (0, -3). The domain of the function is . The range of the

Firdavs [7]

Answer:

Domain: All real numbers

Range: $y\leq -3$

Step-by-step explanation:

The domain is the set of all x values. The quadratic equation continue to operate over all x-values in both directions. The domain is all real numbers.

The range is the set of all y-values. Here the quadratic begins at -3 and continues in a downward direction forever. Then range is $y\leq -3$

6 0

4 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .