Suppose that an anxiety measure has a mean of 114 and variance of 400.00. The researcher only wants to work with 56% of the lower participants. What is the largest average score the researcher should consider with a sample size of 35?

The largest average score the researcher should consider is 117

Step-by-step explanation:

Mean = xbar = 114

Standard deviation = σ = √(variance) = √(400) = 20

Let The standardized score that corresponds with lower 56% be x'.

Let the z value that corresponds to the probability of 0.56 be z'

We need to obtain the z value that corresponds with a Probability of P(z ≤ z') = 0.56

using the normal probability distribution table,

z' = 0.151

P(z ≤ z') = P(z ≤ 0.151) = 0.56

z' = (x' - xbar)/σ

0.151 = (x' - 114)/20

x' = 117

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

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step-by-step explanation:

at least i think its