HEP BEING TIMED! The length of pregnancies in giraffes is normally distributed with a mean length of 430 days and a standard deviation of 9 days. If a random sample of 25 giraffes is taken, what is the probability the sample mean length will be between 429 days and 432 days?

Give your answer to four decimal places

Step-by-step explanation:

Let x be the random variable representing the length of pregnancies in giraffes. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = standard deviation

n = number of Samples

From the information given,

µ = 430

σ = 9

n = 25

the probability that the sample mean length will be between 429 days and 432 days is expressed as

P(429 ≤ x ≤ 432)

For x = 429,

z = (429 - 430)/(9/√25) = - 0.56

Looking at the normal distribution table, the probability corresponding to the z score is 0.288

For x = 432,

z = (432 - 430)/(9/√25) = 1.11

Looking at the normal distribution table, the probability corresponding to the z score is 0.867

Therefore,

P(429 ≤ x ≤ 432) = 0.867 - 0.288 = 0.579

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

part d

step-by-step explanation:

theresa talked her parents into getting sd cards for her phone and her brother’s phone. inserting an sd card into a phone gives it more storage. they both get 8-gigabyte sd cards. how many bits of storage come with each sd card? type your answer in both scientific and standard notation.