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Veronika [31]
3 years ago
9

A biologist wants to estimate the median with of over two hundred oak trees in a forest​

Mathematics
2 answers:
zmey [24]3 years ago
8 0

Answer:

47 cm

Step-by-step explanation:

its really simple just put them in from least to greatest then cross one off from each side until you get your middle number.

3241004551 [841]3 years ago
3 0

Answer:

the answer is 45

Step-by-step explanation:

i took the quiz

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Each tape diagram represents I whole.
Monica [59]

Hi! From what I know, your answer should be 5n = 1.75.

This is because all 5 of those n's ultimately equal the large box of 1.75 on the top. Think of it as fractions. if we replaced 1.75 with 1, those n's would each be 1/5.

4 0
3 years ago
Find a polynomial function with real coefficients that has zeros-3, 2i
Jlenok [28]
Check google it might have it
8 0
3 years ago
Find 5x − 3y − z if x = −2, y = 2, and z = −3 <br> Select one: a. −19 b. 19 c. 13 d. −13
NeTakaya

Answer:

-13

Step-by-step explanation:

5x − 3y − z

substitute x = −2, y = 2, and z = −3

5(-2) -3(2) - (-3)

multiply

-10 -6 +3

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7 0
4 years ago
The density of an object can be found by taking its mass and dividing by its volume write an equation to represent the relations
lukranit [14]
The density of an object is what i can’t see the question
4 0
3 years ago
The hours a week people spend answering and sending emails is normally distributed with a standard deviation of Ï = 150 minutes.
storchak [24]

Answer:

sample size n would be 149305 large

Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval

Step-by-step explanation:

Given that;

standard deviation α = 150 min

confidence interval = 99%

since; p( -2.576 < z < 2.576) = 0.99

so z-value for 99% CI is 2.576

E = 1 minutes

Therefore

n = [(z × α) / E ]²

so we substitute

n = [(2.576 × 150) / 1 ]²

n = [ 386.4 ]²

n = 149304.96 ≈ 149305

Therefore sample size would be 149305 large

Value of n is too high, that would be the practical problem with attempting to find this confidence interval

4 0
3 years ago
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