The type of error that occurs if you fail to reject h0 when, in fact, it is not true is a type 2 error.
<h3>What is a type 2 error?</h3>
Type 2 error is when the null hypothesis is not rejected even though it is not false. Type 1 error is when the null hypothesis is rejected when it is true. A type 2 error leads to a false negative which is also known as an error of omission.
To learn more about type 2 error, please check: brainly.com/question/20914617
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Answer:
Length of the road parallel to the length of the park= 700m
Width of the road= 10m
Area of the road= l\times b=700m\times10m=7000m^2l×b=700m×10m=7000m
2
Length of the road parallel to the breadth of the park =300m
Width of the road= 10m
Area of this road= l\times b=300m\times10=3000m^2l×b=300m×10=3000m
2
Area of both the roads
=7000m^2+3000m^2-Area\ of\ the\ common\ portion\=7000m
2
+3000m
2
−Area of the common portion
=10000m^2-10m\times10m=10000m
2
−10m×10m
=10000m^2-100m^2=10000m
2
−100m
2
=9900m^2=0.99=9900m
2
=0.99
Area of the park = l\times bl×b
=700m\times300m=210000m^2=700m×300m=210000m
2
Area of the park excluding the roads
=210000m^2-9900m^2=210000m
2
−9900m
2
=200100m^2=20.01=200100m
2
=20.01
Answer:
(xM,yM)=(2,−8)
Step-by-step explanation:
hope i helped :) !
Given,
Diameter of the can = 3"
Height of the can = 7"
Looking at how the cans are arranged in the box, that is 4 x 5 (4 rows of cans [width] with 5 cans in each row [length])
The length of the box (L) = 5 cans multiplied by each can's diameter = 5 × 3" = 15"
The width of the box (W) = 4 cans multiplied by each can's diameter) = 4 × 3" = 12"
The height of the box (H) = 2 layers of cans = 2 cans multiplied by each can's height = 2 × 7" = 14"
Therefore, the volume of the box = Length (L) × Width (W) × Height (H) = 15" × 12" × 14" = 2520 inches³
Volume of the box = 2520 inches³
=====================================================
There is also an alternative method to calculate the volume of the box:
Consider each can. Although the can is cylindrical, each can would occupy the space required by a cuboid.
So, for each Cuboid space, the diameter of the can will be the length and width of the cuboid and the height of the can will be the height of the cuboid.
Therefore, for each can,
Length (L) = 3"
Width (W) = 3"
Height (H) = 7"
Volume occupied by one can (that is a cuboid) = L × W × H = 3" × 3" ×7" = 63 inches³
There are 40 such cans in total inside the box; therefore,
Volume of the box = 40 × 63 inches³ = 2520 inches³
