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3 years ago

Solve: x^4-34x^2=-225

1 answer:

8 0

X
4
−34x
2
+225=0

2 Factor
x
4
−
34
x
2
+
225
x
4
−34x
2
+225.
(
x
2
−
25
)
(
x
2
−
9
)
=
0
(x
2
−25)(x
2
−9)=0

3 Solve for
x
x.
x
=
±
5
,
±
3
x=±5,±3

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A woman in a highland village in the Andes knits sweaters and sells them for export. She also takes care of her family and helps

Mnenie [13.5K]

Answer:

Expected number of sweaters per month can be given as follows:

E(X) = Σ x P(X = x)

Now,

E(X) = [2 * 0.1 + 3* 0.1+ 4* 0.2 + 5* 0.3 + 6* 0.2 + 7 * 0.1]

E(X) = 4.7.

Var(X) = E(X^2) – [E(X)]^2

We have E(X) = 4.7. Thus, [E(X)]2= 4.7*4.7 = 22.09.

Now E(X^2) = [2*2 * 0.1 + 3*3* 0.1+ 4*4* 0.2 + 5*5* 0.3 + 6*6* 0.2 + 7*7*0.1]

E(X^2) = 24.1

Thus by formula, Var(X) = E(X2) – [E(X)]2

Var(X) = 24.1-22.09

Var(X) = 2.01

Given that exporter pays the $12 for each sweater. The woman pays $2 per sweater. The cost of shipment is $3 irrespective of the number of sweaters. Now, let m is the number of sweaters she made. Thus, the total cost she would have to pay would be

Total cost by woman = 2m+3

The total cost paid by the exporter would be = 12m.

Now the profit of woman would be given by,

= The total cost exporter pay – cost paid by the woman

= 12m – (2m +3)

= 12m – 2m -3

= 10m – 3.

Now expected profit made by the woman is given in the following table below:

E(Profit) = Σ profit* P(X = x)

In a similar way, as we have done in part (a).

E(Profit) = [17 * 0.1 + 27* 0.1+ 37* 0.2 + 47* 0.3 + 57* 0.2 + 67 * 0.1]

E(Profit) = 44.

Now, we calculate the variance:

Var(profit) = E(profit^2) – [E(profit)]^2

Var(profit) =

E(profit^2) = [17*17 * 0.1 + 27*27* 0.1+ 37*37* 0.2 + 47*47* 0.3 + 57*57* 0.2 + 67*67*0.1]

E(profit^2) = 2137.

[E(profit)]^2 = 44*44 = 1936.

Thus, the variance can be given as =

Var(profit)= 2137 – 1936

Var(profit) = 201.

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3 years ago

Marco's mom gave him d dollars for his allowance one week. He also earned $14.55 for his newspaper route that week.

myrzilka [38]

Answer:

The answer is d+14.55

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4 years ago

Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y minus 5

My name is Ann [436]

Answer:

Vertices at (-7, 5) and (-1, 5).

Foci at (-9, 5) and (1,5).

Step-by-step explanation:

(x + 4)²/9 - (y - 5)²/16 = 1

The standard form for the equation of a hyperbola with centre (h, k) is

(x - h²)/a² - (y - k)²/b² = 1

Your hyperbola opens left/right, because it is of the form x - y.

Comparing terms, we find that

h = -4, k = 5, a = 3, y = 4

In the general equation, the coordinates of the vertices are at (h ± a, k).

Thus, the vertices of your parabola are at (-7, 5) and (-1, 5).

The foci are at a distance c from the centre, with coordinates (h ± c, k), where c² = a² + b².

c² = 9 + 16 = 25, so c = 5.

The coordinates of the foci are (-9, 5) and (1, 5).

The Figure below shows the graph of the hyperbola with its vertices and foci.

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