<u>Given</u>: The number of woodpeckers in a forest have
a mean = 20 , mode= 20 , and standard deviation = 5
<u>Computation:</u>
The range of data between 2 standard deviations from the mean is given by :-
Mean ± 2 (Standard deviation)
i.e. ( Mean- 2 (Standard deviation), Mean + 2(Standard deviation)
i.e. (20-2(5), 20+2(5))
i.e. (20-10, 20+10)
i.e. (10, 30)
Required range = (10, 30) which contains approx. 95% of the data ( By 68-95-99.7 rule)
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Area of a parallelogram = b × h
Let the heights of first and second parallelogram be h and h' respectively.
Given that,
h = ½h'
Area of first parallelogram= bh ...(i)
Area of second parallelogram = bh'
but, h = ½h'
Replacing h by ½h' in equation (i)
Therefore, area of first parallelogram = ½bh'
Ratio - ½bh' : bh'
= 1:2
Answer:
4.5z^(3.5) -6.5z^(-1.5)
Step-by-step explanation:
I find it easier to write this as the sum of two terms and use the power rule.
f = z^(4.5) +13z^(-0.5)
f' = 4.5z^(3.5) -6.5z^(-1.5)
![f'=\dfrac{9z^5-13}{2z\sqrt{z}}](https://tex.z-dn.net/?f=f%27%3D%5Cdfrac%7B9z%5E5-13%7D%7B2z%5Csqrt%7Bz%7D%7D)
For this case we have that the initial and final peach diameters are:
![D_ {i} = 98 \ mm\\D_ {f} = 188,869 \ mm](https://tex.z-dn.net/?f=D_%20%7Bi%7D%20%3D%2098%20%5C%20mm%5C%5CD_%20%7Bf%7D%20%3D%20188%2C869%20%5C%20mm)
According to the difference percentage we have:
![\frac {188,869-98} {98} = 0.927](https://tex.z-dn.net/?f=%5Cfrac%20%7B188%2C869-98%7D%20%7B98%7D%20%3D%200.927)
That is to say, the peach grew by almost 93%. That is, almost twice its diameter.
Answer:
the peach grew by almost 93%. That is, almost twice its diameter.
Answer:
answer below
Step-by-step explanation:
square and rectangle