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aivan3 [116]
2 years ago
8

33=10∙q+r cual es el valor de q y r

Mathematics
2 answers:
Svetach [21]2 years ago
8 0

Answer:

Q=33/10-r/10 R=33−10q

Step-by-step explanation:

Espero que sea correcto para ti también puedes hacerme más inteligente si es correcto

Sophie [7]2 years ago
6 0
Can you someone help me understand
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If you begin with the basic equation of a vertical parabola:  y-k=a(x-h)^2, where (h,k) is the vertex, then that equation, when the vertex is (-3,-2), is 

y + 2 = a (x + 3)^2.  If we solve this for y, we get

y        = a(x+3)^2 - 2.  Thus, eliminate answers A and D.  That leaves B, since B correctly shows (x+3)^2.
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Savvas Texas Algebra II
Elan Coil [88]

Answer:

The number of small balloons = 46

The number of large balloons = 159

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

Step-by-step explanation:

The volume of each small balloon = 300 in.³

The volume of each large balloon = 1000 in.³

The cost of one package of 50 small balloons = $2.59

The cost of one package of 24 large balloons = $1.99

The volume of one small tank of pressurized helium = 20 ft³

The volume cost of one small tank of pressurized helium = $10.49

By conversion

20 ft³ = 34,560 in.³

The volume of 30 small balloons = 30 × 300 in.³ = 9,000 in.³

The volume of 30 large balloons = 30 × 1000 in.³ = 30,000 in.³

The volume taken by 24 large balloons = 24,000 in.³,

24 large balloons cost = $1.99

The volume taken by 50 small balloons = 15,000 in.³

50 small balloons cost = $2.59

Therefore, we buy more large balloons

We have;

A×300 + B×1000 = C×34,560

Given that all the helium will be used, the volume of helium will be a multiple of 300, therefore, when 5 bottles of helium is used, we have;

A×300 + B×1000 = 5×34,560 = 172,800 in.³

Given that there will be at least 30 of each balloon, gives;

We remove the minimum possible quantity (more than 30) of small balloons from 172,800 in.³ leaving multiples of 1,000s as follows;

Using Excel, the quantities of 36 and 46 small balloons give the same cost

For 46 small balloons

The volume of 46 small balloons = 13,800 in.³

Therefore, 172,800 in.³ - 13,800 in.³ = 159,000  in.³ = The volume of the large balloons

The quantity of the large balloons = 159,000  in.³/(1,000  in.³/balloon) = 159 balloons

Therefore, we have 46 small balloons, and 159 large balloons, with 5 small bottles of helium gas

The number of package of large balloons is 159/24 = 6.625 ≈ 7 packages

The cost of the 7 packages of large balloons = 7 × $1.99 = $13.93

The number of package of small balloons is 46/50= 0.92 ≈ 1 package

The cost of the 1 packages of small balloons = 1 × $2.59 = $2.59

The cost of the five bottles of helium = 5 × $10.49 = $52.45

$68.97

The number of small balloons = 46

The number of large balloons = 159

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

At the same cost, we can also have;

The number of small balloons = 36

The number of large balloons = 162

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

Therefore, we can have 36 or 46 small balloons and 162 or 159 large balloons along with 5 bottles of helium gas with a total cost of $13.93 + $2.59 + $52.45 = $68.97.

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3 years ago
Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g
Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also,  $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0
2 years ago
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