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Rus_ich [418]
3 years ago
14

Where would \sqrt{130}be on the the number line?

Mathematics
1 answer:
Marrrta [24]3 years ago
3 0
12 x 12 = 144
11 x 11 = 121

Sqrt130 between the two,
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Step-by-step explanation:

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Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro
saw5 [17]

Answer:

(a) P(0 <= Z <= 2.24)  = 0.4927

(b) P(0 <= Z <= 2)  = 0.4773

(c) P(-2.60 <= Z <= 0) = 0.4953

(d) P(-2.60 <= Z <= 2.60) = 0.9906

(e) P(Z <= 1.64)  = 0.9495

(f) P(-1.75 <= Z) = 0.0047

(g) P(-1.60 <= Z <= 2.00) =  0.9425

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(j) P(|Z| <= 2.50) = 0.9876

Step-by-step explanation:

(a) P(0 <= Z <= 2.24) = P(Z <= 2.24)- P(Z <= 0)

using the STANDARD NORMAL DISTRIBUTION TABLE

P(0 <= Z <= 2.24) = 0.9927 -  0.5  = 0.4927

(b) P(0 <= Z <= 2) = P(Z <= 2)- P(Z <= 0)

= 0.9773 -  0.5  = 0.4773

(c) P(-2.60 <= Z <= 0) = P(Z <= 0)- P(-2.60)

=   0.5 - 0.0047 = 0.4953

(d) P(-2.60 <= Z <= 2.60) = P(Z <= 2.6)- P(-2.60)

=   0.9953 - 0.0047 = 0.9906

(e) P(Z <= 1.64)  = 0.9495

(f) P(-1.75 <= Z) =1 - P(Z < 2.6) = 1 -  0.9953  = 0.0047

(g) P(-1.60 <= Z <= 2.00) = P(Z <= 2.0)- P(-1.60)

= 0.9773- 0.0548  = 0.9425

(i) P(1.60 <= Z)=1 - P(Z < 1.6) = 1 -  0.9452  = 0.0548

(j) P(|Z| <= 2.50) = P(-2.5 < Z <= 2.50)= P(Z <= 2.5)- P(-2.5)

=0.9938 - 0.0062 = 0.9876

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The answer for this is 4.
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Answer:

Step-by-step explanation:

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