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:
C
Step-by-step explanation:
sqrt(a^7) = sqrt(a^6 * a)
Answer:
15,236 sq in
Step-by-step explanation:
First, you have to split the net into three rectangles and two triangles
step 1
triangular base area
1/2 bh = 1/2 (44 in)(64 in)
= 1/2 (2,816 sq in)
= 1,408 sq in.
2 x 1,408 sq in = 2,816 sq in.
step 2
end rectangle area
lw = (68 in)(69 in)
= 4,692 sq in
2 x 4,692 sq in = 9,384 sq in
Now find the area of the middle rectangle
lw - (69 in) (44 in) = 3,039 sq in.
then add all the areas together
2,816 + 9,384 + 3,036 = 15,236 sq in
Hope this helps :)
1st number = x
2nd number = 3x
3rd number = 2x-10
x +3x + 2x-10 =92
6x - 10 = 92
6x =102
x = 102/6
x = 17
1st number = 17
2nd number = 17*3 = 51
3rd number = 17 *2 = 34-10 = 24
17 + 51 + 24 = 92
3 numbers are 17, 51 and 24
