We select a sample size n from the confidence interval with the mean $\mu$ and 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

Factorise 49x^2-121 what’s the answer

asambeis [7]

Answer:

( 7x + 11 )( 7x- 11 )

Step-by-step explanation:

( 7x + 11 )( 7x- 11 )

Because 7*7 is 49

-11*11= -121

7*11=77

-7*11= -77

77-77=0

so the answer is 49x^2-121

8 0

2 years ago

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