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erik [133]
2 years ago
13

Never mind lol I did it wrong

Mathematics
1 answer:
True [87]2 years ago
6 0

Answer:

what did u do wrong

Step-by-step explanation:

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10/22 in simplest form math
hram777 [196]
10/22 in simplest form is 5/11
7 0
3 years ago
Read 2 more answers
If SDE ~ SWT, find WT.
faust18 [17]

Formula used: 40+x = 4x-3 x is used for <em>ET</em>

Step-by-step explanation:

40+x = 4x-3

<u>+</u><u>3</u> <u>+</u><u>3</u>

43+x = 4x

<u>-</u><u>x</u> <u>-</u><u>x</u>

<u>43</u> = <u>3x</u>

3 3

14.3 = x

then substitude this on 5x+3 on <em>W</em><em>T</em> and you arr good

5 0
3 years ago
If u(x) = x5 – x4 + x2 and v(x) = –x2, which expression is equivalent to (U/V) (x)?
lions [1.4K]
Hello,

Answer C if x≠0


(x^5-x^4+x²)/(-x²)=-x²(x^3-x²+1)/x²=-(x^3-x²+1)=-x^3+x²-1

3 0
3 years ago
Read 2 more answers
(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure
Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0
2 years ago
Can anyone help me on this?
alina1380 [7]
That should be right
Answer x=-299

3 0
3 years ago
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