Using the normal distribution relation, the probability that sample will exceed the weight limit is 0.004
<u>Using the relation</u> ::
The mean, μ = np = (162 × 19) = 3078
The standard deviation, σ = 28 × √19 = 122.049
<u>The Zscore</u> :
Zscore = (3401 - 3078) ÷ (122.049)
Zscore = 2.65
Hence,
P(Z > 2.65) = 1 - P(Z < 2.65)
Using a normal distribution table :
P(Z > 2.65) = 1 - 0.9959
P(Z > 2.65) = 0.004
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H = x, L = 1.5 x;
L + W + H = 6
x + 1.5 x + W = 6
W = 6 - 2.5 x
Dimensions of the camera in terms of x:
x, 1.5 x, 6 - 2.5 x.
V = L x W x H
V = x * 1.5 x * ( 6 - 2.5 x ) = 9 x² - 3.75 x³
V ` = 18 x - 11.25 x² ( V max is when V` = 0 )
18 x - 11.25 x² = 0
x ( 18 - 11.25 x ) = 0
11.25 x = 18
x = 18 : 11.25
x = 1.6
The dimensions are:
L = 2.4
W = 2.0
H = 1.6
Answer:
2 : 3
Step-by-step explanation:
Given:
The two cylinders are similar. So, their corresponding dimensions must be in proportion.
Therefore, the similarity ratio of the smaller to the larger similar cylinders is 2 : 3.
Circumference of circle is 62.8 inches.
Step-by-step explanation:
Using Formula
Circumference of circle = 2 π r
Substitute the values
Circumference of circle = 2 × 3.14 × 10 inches.
multiply the numbers
Circumference of circle = 62.8 inches
Hence, the circumference of circle is 62.8 inches.
