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:
4
Step-by-step explanation:
since the fraction is 1/4 you can multiply 4 by 4 and get sixteen and 4x1=4
79.90*0.8= 63.92
Answer
63.92
The slope of first line is m 1= ( 1 - 0 ) / ( - 2 - ( - 6 ) );
m = 1 / 4 ;
Because the second line is perpendicular to the first line, m2 * m1 = - 1 ;
m2 * ( 1 / 4 ) = -1;
m2 = - 4 ;
The equation of the second line is : y - 3 = ( - 4 )( x - 2 );
y - 3 = -4x + 8 ;
y = -4x + 11.