The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,00
0 miles and a population standard deviation of 4,000 miles. Suppose you bought a set of four tires; what is the likelihood the mean tire life of these four tires is more than 66,000 miles?
1 answer:
Answer:
0.0013
Step-by-step explanation:
To do this, we need to use a normal distribution table with Z score values, like the one I'm attaching here.
Now, the expression to calculate the Z value is the following:
Z = x - μ / (σ/√n)
Where:
μ: mean
σ: standard deviation
x: value required
n: sample population
Now that we have the data, let's calculate the Z value:
Z = 66,000 - 60,000 / (4000/√4)
Z = 3
Now, let's look at the table to get the value that belongs to this Z score. According to the table, it's 0.0013
Therefore, the likelihood would be 0.0013
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Answer:
Bottle containing cranberry juice has a greater percent of water.
Step-by-step explanation:
A 60 ounce bottle of grape juice contains 33 ounces of water.
So the percentage of water in the bottle =
=
= 55%
Similarly 50 ounces bottle of cranberry juice contains 30 ounces of water.
So the percentage of water in the bottle =
=
= 60%
Therefore, bottle containing cranberry juice has a greater percent of water.
Answer:
minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in
Step-by-step explanation:
given data
volume = 150 in³
base material costing = 5 cents/in²
front cost = 10 cents/in²
remainder sides cost = 2 cents/in²
to find out
the dimensions that will minimize the cost
solution
we consider here length = x and breadth = y and height = z
and
area of base = xy
area of front = xz
and area of remaining side = xz + 2yz .....................1
so
cost of base will be = 5xy
cost of front = 10xz
cost of remaining side = 2 ( xz+ 2yz)
and
total cost will be
total cost TC = 5xy + 10xz + 2 ( xz+ 2yz)
total cost TC = 5xy + 10xz + 2xz+ 4yz
total cost TC = 5xy + 12xz + 4yz ....................2
and total volume will be = xyz
150 = xyz
z = .......................3
now put z value in equation 2
total cost TC = 5xy + 12xz + 4yz
total cost TC = 5xy + 12x + 4y
total cost TC = 5xy + +
...........4
now differentiate TC w.r.t x and y
TC (x) = 5y -
TC (x) = 5x -
now equating with 0 these
5y - = 0
x² =
and
5x - = 0
x =
solve we get
y = 10.2598
x = 3.4199
now put x and y value in equation 3
z =
z =
z = 4.2750
so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in