High-strength concrete is supposed to have a compressive strength greater than 6,000 pounds per square inch (psi). A certain type of concrete has a mean compressive strength of 7,000 psi, but due to variability in the mixing process it has a standard deviation of 420 psi, assuming a normal distribution.

What is the probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion?

Probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is 0.00866.

Step-by-step explanation:

We are given that High-strength concrete is supposed to have a compressive strength greater than 6,000 pounds per square inch (psi).

A certain type of concrete has a mean compressive strength of 7,000 psi, but due to variability in the mixing process it has a standard deviation of 420 psi, assuming a normal distribution.

Let X = certain type of concrete compressive strength

SO, X ~ Normal()

The z score probability distribution for normal distribution is given by;

                            Z  =    ~ N(0,1)

where, = mean compressive strength = 7,000 psi

            = standard deviation = 420 psi

SO, Probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is given by = P(X < 6,000 psi)

    P(X < 6,000 psi) = P( < ) = P(Z < -2.38) = 1 - P(Z 2.38)

                                                                    = 1 - 0.99134 = 0.00866

The above probability is calculated by looking at the value of x = 2.38 in the z table which has an area of 0.99134.

Hence, the probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is 0.00866.

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

ƒ = 0.03b

hope this ( :

step-by-step explanation: