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mart [117]
3 years ago
13

Consider the experimental situation described below. Ability to grow in shade may help pines in the dry forests of Arizona resis

t drought. How well do these pines grow in shade? Investigators planted pine seedlings in a greenhouse in either full light or light reduced to 5% of normal by shade cloth. At the end of the study, they dried the young trees and weighed them. Identify the experimental unit(s) or subject(s)?
a) shade cloth

b) pine tree seedlings

c) drought resistance

d) greenhouses

e) rainy seasons
Mathematics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

The experimental subjects are pine tree seedlings .

Step-by-step explanation:

The physical entity to which the treatment is assigned at random is known as experimental unit .  

The same treatment cannot be received by two experimental units, they have to different treatments. The selection of experimental units plays an important role to determine the p-values in the statistical analysis. The experimental units are divided into different homogeneous groups called blocks.

From the question, the experimental subjects identified are pine tree seedlings.

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3 years ago
I need help on this, please hurry and thank you!
inna [77]

Answer:

A.\ a_n=n^2+1

Step-by-step explanation:

Check:

a_n=n^2+1\\\\a_1=1^2+1=1+1=2\qquad CORRECT\\\\a_2=2^2+1=4+1=5\qquad CORRECT\\\\a_3=3^2+1=9+1=10\qquad CORRECT\\\\a_4=4^2+1=16+1=17\qquad CORRECT\\\\a_5=5^2+1=25+1=26\qquad CORRECT

Used PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

First Power, next Addition

4 0
3 years ago
Read 2 more answers
6n + 9 = -3 - 21 - 6
denis23 [38]
6n+9=-3-21-6
subtract the numbers:-3-21-6=-30
6n+9=-30
subtract 9 from both sides
6n+9-9=-30-9
simplify
6n=-39
divide both sides by 6
\frac{6n}{6}= \frac{-39}{6}
n= \frac{-13}{6}
decimal n=-6.5
Hope this helps :)
3 0
3 years ago
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
6 0
3 years ago
Anybody? I'm pretty sure calculators don't do it right so..please no calculators. ;-;
Dima020 [189]

Answer:

.109 !!

Hope this helps!!

8 0
2 years ago
Read 2 more answers
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