The perimeter = sum of all sides
= 120 + 80 + 50
= 250
So 250 - 3
247
Left space for gate
Now cost of fencing = Rs 20/per meter
= 247 × 20
= Rs 4,940
Now the area of the triangular park can be found using heron's formula
S = (a+b+c)/2
S = (120+80+50)/2
S = 250/2
S = 125
Now
Herons formula = √s(s-a)(s-b)(s-c)
√125(125-120)(125-80)(120-50)
√125(5)(45)(70)
√5×5×5×5×5×3×3×5×14
After Making pairs
5×5×5×3√14
375√14
Therefore 375√14m is the area of the triangular park
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Answer:
Step-by-step explanation:
OS ≅ OU, so we will set those 2 expressions equal to each other and solve for y:
6y = 42 so
y = 7
Same goes for OT and OV:
x + 5 = 23 so
x = 18 and
SU = 84
Answer:
The standard deviation of weight for this species of cockroaches is 4.62.
Step-by-step explanation:
Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.
To find : What is the approximate standard deviation of weight for this species of cockroaches?
Solution :
We have given,
Mean 
The sample mean x=55
A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.
i.e. P(X>55)=14%=0.14
The total probability needs to sum up to 1,



The z-score value of 0.86 using z-score table is z=1.08.
Applying z-score formula,

Where,
is standard deviation
Substitute the values,





The standard deviation of weight for this species of cockroaches is 4.62.
Answer:
Step-by-step explanation:
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