Answer:
d
Step-by-step explanation:
Answer:
The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.
Step-by-step explanation:
We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.
Let
= <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>
The z-score probability distribution for the sample mean is given by;
Z =
~ N(0,1)
where,
= population mean length of rods = 259.2 cm
= standard deviaton = 2.1 cm
n = sample of steel rods = 17
Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(
> 259 cm)
P(
> 259 cm) = P(
>
) = P(Z > -0.39) = P(Z < 0.39)
= <u>0.65173</u>
The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.
Answer: y = 5/2x + 6
Step-by-step explanation:
y - y₁ = m(x - x₁)
y - (-9) = 5/2(x- (-6))
y + 9 = 5/2(x+ 6)
y +9 = 5/2x + 15
y = 5/2x + 15 - 9
y = 5/2x +6