Agovernment agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. in standard english text, a particular letter is used at a rate of 5.7%. a. find the mean and standard deviation for the number of times this letter will be found on a typical page of 1600 characters. muequals 91.2 (do not round.) sigmaequals 9.3 (round to one decimal place as needed.) b. in an intercepted message, a page of 1600 characters is found to have the letter occurring 102 times. is this unusual? \

A) μ = 91.2; σ = 9.3

B) No

Step-by-step explanation:

To find the mean, μ, we use the formula

μ = np, where n is the sample size and p is the probability of success (percentage of times the letter is used).

This gives us

μ = 0.057(1600) = 91.2

To find the standard deviation, we use the formula

σ = √(np(1-p)

This gives us

σ = √(1600×0.057×(1-0.057))

= √(1600×0.057×0.943) = √86.0016 = 9.2737 ≈ 9.3≈

Any value that is more than two standard deviations from the mean is considered unusual.  The value 102 is

(102-91.2)/9.3 = 10.8/9.3 = 1.16 standard deviations from the mean.  This is not statistically unusual.

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.