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Korvikt [17]
2 years ago
13

Consider the probability distribution shown below.

Mathematics
1 answer:
rusak2 [61]2 years ago
3 0

Answer:

it is b

Step-by-step explanation:

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Find the least common denominator for these two rational expressions. B/b^2 -64 -7b/b^2+7b-8
san4es73 [151]

Answer:

The least common denominator is (b-8)(b+8)(b-1)

Step-by-step explanation:

We are given expression as

\frac{b}{b^2-64}-\frac{7b}{b^2+7b-8}

Firstly, we will factor both denominators

b^2-64=b^2-8^2=(b-8)(b+8)

b^2+7b-8=(b+8)(b-1)

so, we can plug it back

\frac{b}{(b-8)(b+8)}-\frac{7b}{(b+8)(b-1)}

First term denominator is

(b-8)(b+8)

Second term denominator is

(b+8)(b-1)

So,

Least common denominator will be

(b-8)(b+8)(b-1)

So, we get

LCD=(x-8)(x+8)(x-1)


3 0
3 years ago
Jerry ate 1,070 calories total at breakfast. He ate a cup of yogurt that had 150 calories in it and some granola. Each cup of gr
Maru [420]

Answer:

4 cups that's how many cups Jerry ate

4 0
3 years ago
Assume that in a statistics class the probability of receiving a grade of A equals .30 and the probability of receiving a grade
Tom [10]

Answer:

Answer D is correct

6 0
3 years ago
What distance would be covered in 60
ddd [48]
60 x 50 = 3000 meters.......
4 0
2 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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