Help asap they are in order form top to bottom <br><br> A<br> B<br> C<br> D

Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 22 days and

Alika [10]

Answer:

25.5 days

Step-by-step explanation:

Mean number of days (μ) = 22 days

Standard deviation (σ) = 6 days

Z-score for the 72nd percentile (according to tabulated values) = 0.583

The z-score for any number of days, X, is determined by:

$z=\frac{X-\mu}{\sigma}$

The value of X that is greater than 72% of the trial times is:

$0.583=\frac{X-22}{6}\\ X=25.5\ days$

Therefore, 72% of all of these types of trials are completed within 25.5 days.

6 0

3 years ago

Given the geometric sequence where a1=3 and the common ratio is -1, what is the domain for n?

zimovet [89]

Answer:

all integers where n ≥ 1

Step-by-step explanation:

n is a counting number, since it counts the terms of the sequence. It is in the set {1, 2, 3, ...}, any of the integers 1 or greater.

5 0

3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$