Suppose a batch of metal shafts produced in a manufacturing company have a variance of 6.256.25 and a mean diameter of 206206 inches. If 9090 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.30.3 inches

Step-by-step explanation:

mean diameter = 206

variance = 6.25

standard deviation = √ variance

σ = √ 6.25 = 2.5

sample size n = 90

z = deviation from mean / (σ / √n)

= 0.3 / (2.5/ √90)

= .3 / .2635

= 1.1385

P value from the table =  (.87 - .5 ) x 2

= .37 x 2 .

or 74 % .

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol