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irga5000 [103]
3 years ago
13

Write 21/8 as a mixed number

Mathematics
2 answers:
kati45 [8]3 years ago
8 0
25/8
Answer is think so check it x
Ratling [72]3 years ago
7 0

Answer: 2 5/8

Step-by-step explanation:

you divide and get the answer

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A square has an area of 16 square millimeters. What is the length of each side of the square? [free brainliest!]
Step2247 [10]

Answer:

8mm

if it is not correct please tell me

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What is the area of the kite below.<br> 15 m<br> D<br> 3 m<br> 3 m<br> 12 m<br> С
ohaa [14]

Answer:

51 meters

Step-by-step explanation:

The area can be found by adding the areas of all the triangles. The top right triangle has an area of 7.5. The bottom right triangle has an area of 18. The bottom left, also has an area of 18. And finally, the top left which has an area of 7.5. 18 + 18 + 7.5 + 7.5 = 51.

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The area of the bottom of a shoebox can be written as a(x) = 2x2 - 4 and the height of the shoebox can be written as h(x) = 3x +
Helga [31]

Answer:V(x) = 6x³ + 4x² - 12x - 8

Step-by-step explanation:

The shoe box can either be cube shaped or cubiod shaped. The volume of the box is length × breadth × height

Since Area = Length × Breadth

Volume = Area × Height

V(x) = A(x) × H(x)

A(x) = 2x² - 4

H(x) = 3x + 2

V(x) = (2x² - 4)(3x + 2)

V(x) = 6x³ + 4x² - 12x - 8

8 0
3 years ago
Who tryna help me with this math
tatyana61 [14]

Answer:

What kind?

Step-by-step explanation:

8 0
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
3 years ago
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