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Tanya [424]
3 years ago
12

WHY does every subject i have SO LONG

Mathematics
2 answers:
Llana [10]3 years ago
7 0

Answer:

omg ikr like my history is 29 videos long

Step-by-step explanation:

slega [8]3 years ago
4 0

Answer:

probably because your in a higher grade now, and the syllabus is too much

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What is the area of 25 pi?
djverab [1.8K]
answer: the numbers
4 0
3 years ago
Timed need answer ASAP
ella [17]

Answer:

6

Step-by-step explanation:

4^2 / 8*3=

16/8*3=

2*3=

6

4 0
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Jack was the science Museum in sabotage the recycling plant had an average speed of 26KM/H. Sometime later Aaliyah love travelin
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3 0
3 years ago
What are the restrictions on the domain of g o h?
umka21 [38]

g(x) = \sqrt{x-4} and h(x)= 2x-8

g o h is a composition function

First we find g o h

g o h is g(h(x))

We plug in h(x) in g(x)

We replace x  with 2x-8 in g(x)

g(h(x)) = g (2x-8) = \sqrt{2x-8-4} = \sqrt{2x-12}

g(h(x)) = \sqrt{2x-12}

To find domain we look at the domain of h(x) first

Domain of h(x) is set of all real numbers

now we look at the domain of g(h(x))

Negative number inside the square root is imaginary. so we ignore negative number inside the square root

So to find domain we set 2x - 12 >=0   and solve for x

2x - 12 >=0

add both sides by 12

2x >= 12

divide both sides by 2

x > = 6



5 0
3 years ago
Read 2 more answers
The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control a
Dafna11 [192]

Answer:

Probability that at least 490 do not result in birth defects = 0.1076

Step-by-step explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = \sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}

                 = {^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}  + .......+ {^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}

                = 0.04541 + ......+0.0000002079

                = 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

4 0
3 years ago
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