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Nostrana [21]
2 years ago
6

Can someone answer these for me please. I need the answers within the next few hours.

Mathematics
1 answer:
FinnZ [79.3K]2 years ago
4 0
Answer: this was all i knew how to do, hope it helps:)

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Evaluate <br><br> 0<br> ∫ ( 3^3 - 5^2) dx <br> -2
Jobisdone [24]
3^3 = 27
5^2 = 25
27 - 25 = 2

integral of 2 = 2x

(2x) evaluated from -2 to 0
= (0) - (2(-2)) = -(-4) = 4

If you did not make any mistakes in typing your question
the answer would be 4
4 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x &ge; 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
Ramesh examined the pattern in the table.
nikdorinn [45]

Answer:

D

Step-by-step explanation:

4 0
3 years ago
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135. sale price and 25percent of sale price what was the regular price
ollegr [7]

Answer:

the regular price was 33.75

5 0
3 years ago
Solve for p: m= 8 -2(p-m)
makkiz [27]
Solve for p by simplifying both sides of the equation, then isolation the variable
p = 1 + m/2

Hope this helps! :)
4 0
3 years ago
Read 2 more answers
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