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2 years ago

What are three consecutive integers that add up to 4,590

1 answer:

7 0

1529+1530+1531=4590

All you have to do is divide 4590 by three and you get 1530. After that add 1 and you get 1531. Then go back to 1530 and subtract 1 and you get 1529 and there are your three consecutive integers that equal 2590.(1529, 1530,1531)

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

Use lagrange multipliers to find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).

horsena [70]

Lagrangian:

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-5)^2+\lambda(x-2y+3z-6)$

where the function we want to minimize is actually $\sqrt{x^2+(y-2)^2+(z-5)^2}$ , but it's easy to see that $\sqrt{f(\mathbf x)}$ and $f(\mathbf x)$ have critical points at the same vector $\mathbf x$ .

Derivatives of the Lagrangian set equal to zero:

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-5)+3\lambda=0\implies z=5-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0$

Substituting the first three equations into the fourth gives

$-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6$
$11-7\lambda=6\implies \lambda=\dfrac57$

Solving for $x,y,z$ , we get a single critical point at $\left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right)$ , which in turn gives the least distance between the plane and (0, 2, 5) of $\dfrac5{\sqrt{14}}$ .

7 0

3 years ago

Ignore the stuff underneath I did it in pen and messed up.

Vinil7 [7]

U didn't mess up....
V = pi * r^2 * h....divide both sides by (pi)r^2
V / (pi)r^2 = h

7 0

3 years ago

Someone help plz correct answers only and explain how you got the answer

Harrizon [31]

Δ YX Z And Δ VUW are SIMILAR because all their respective angles are congruent:

∠ y = 90°; ∠ v = 90°
∠ x = 67°; ∠ u = 67°
∠ z = 23°; ∠ w = 23°

Two triangles are similar if their related angles are congruent (answer B)

7 0

3 years ago

Which equation relates the number of triangles in the figure (n) to the perimeter of the figure (p)

WITCHER [35]

Answer:

Option d is correct.

Equation : P=7n+20.

Explanation;

Given the perimeter of each figure is;

Perimeter of triangle is equal to the sum of all the sides of a triangle.

Perimeter of 1 triangle is 21

Perimeter of 2 triangle is 34 and

Perimeter of 3 triangle is 41

Let n be the number of figure and P be the perimeter of the figure;

the only equation which satisfy the given perimeter is;

$P=7n+20$

Check:

for n =1 which means 1 triangle then;

$P =7n+20 = 7 \cdot 1 +20 = 7+20 = 27$

for n = 2 , [ i.e 2 triangle]

$P =7n+20 = 7 \cdot 2 +20 = 14+20 = 34$

and for n =3 [i.e, 3 triangles]

$P =7n+20 = 7 \cdot 3 +20 = 21+20 = 41$

Therefore, the equation P =7n+20 relates the number of triangles in the figure(n) to the perimeter of the figure(P).

3 0

3 years ago