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AysviL [449]
3 years ago
13

Please help asap!!! I’ll give you a thank you

Mathematics
1 answer:
stich3 [128]3 years ago
7 0

Answer:

V≈536.38

Step-by-step explanation:

V=a3

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Simplify the expression p(p^-7*q^3)^-2*q^-3<br> A.p/q^3<br> B.p^15/q^9<br> C.1/(p^9*q^2<br> D.p^14*q
Alenkinab [10]

Answer:

B.p^15/q^9

Step-by-step explanation:

p(p^-7•q^3)^-2•q^-3

answer: B. p^15/q^9

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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. what is the pro
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2/3*2/3*1/3 + the chance that the fourth is the 5 or 6. 
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Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

4 0
2 years ago
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