The approximate probability that the weight of a randomly-selected car passing over the bridge is more than 4,000 pounds is 69%
Option C is the correct answer.
<h3>What is Probability ?</h3>
Probability is defined as the study of likeliness of an event to happen.
It has a range of 0 to 1.
It is given in the question that
The weights of cars passing over a bridge have a mean of 3,550 pounds and standard deviation of 870 pounds.
mean = 3550
standard deviation, = 870
Observed value, X = 4000
Z = (X-mean)/standard deviation = (4000-3550)/870 = 0.517
Probability of weight above 4000 lb
= P(X>4000) = P(z>Z) = P(z> 0.517) = 0.6985
The approximate probability that the weight of a randomly-selected car passing over the bridge is more than 4,000 pounds is 69%
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Total area = 104+24 = 128 sq.in , Option C is the right answer.
The missing figure is attached with the answer.
<h3>What is Area ?</h3>
The space occupied by a two dimensional figure is called Area.
The figure given in the question can be divided into two figure
A rectangle of 8 * 13 inch
and a triangle of 8 * 6 inch
The area of a rectangle is given by
Length * Breadth
8 * 13
= 104 sq.inch
The area of the triangle is given by
(1/2) * base * height
= (1/2) * 8 * 6
= 24 sq.inch
Total area = 104+24 = 128 sq.in
Therefore Option C is the right answer.
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Surface of a cubical box=6(side²)
1)We have to calculate the surface of this cubical box.
Rate=cost of painting / surface ⇒surface=cost of painting/rate
Data:
Rate=$15/m²
cost of painting=$1440
Surface=$1440/($15/m²)=96 m²
2)We find out the length of the side:
Surface of a cubical box=6(side²)
Data:
Surface of a cubical box=96 m2
Therefore:
96m²=6 (side²)
side²=96 m²/6
side²=16 m²
side=√(16 m²)=4 m
3) We find the volume of a cubical box.
volume=(side³)
volume=(4 m)³
volume=64 m³
Answer: the volume of this cubical box would be 64 m³.
Answer:
4.0506100 x 10^-1
Step-by-step explanation:
Because all you need to do is move the decimal point to the right once, you write 10^-1 to show that to get back to the original number, you just multiply 4.0506100 by -0.1
