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Alborosie
3 years ago
13

A consumer wanted to compare two types of battery storage. One type is refrigerated and the other is room temperature. It is a c

ommon belief that refrigerated batteries last longer. This belief is tested by placing 10 fully charged batteries into each type of storage and then measuring the percentage of charge remaining after thirty days. A coin tip was used to determine which type of storage each battery would be placed in first. Results indicated that there was no difference in the two types of storage Complete parts (a) through in below.
What type of experimental design is this?
A) Matched pairs design
B) Completely randomized design
C) Randomized block design
D) Case-control study
Mathematics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

<u>A) Matched pairs design</u>

<u>Step-by-step explanation:</u>

In<em> </em><em>experimental design</em>, the match pairs experimental design is one that often involves only two treatment conditions; which in this case are the 'refrigerated and the room temperature battery storage types'.

Thus, these two treatment conditions form a matched pair because we are told that 10 fully charged batteries are placed into each type of storage.

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Remember to simplify your fraction***
romanna [79]

Answer:

1. 2/6 or 1/3

2.  4/6 or 2/3

3. 3/6 or 1/2

Step-by-step explanation:

I hope this helps I did the ones I knew

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2 years ago
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An unknown number y is 20 more than an unknown number x. The number y is also x less than 2. The equations to find x and y are s
zloy xaker [14]

Answer:

b)Add the equations to eliminate x.

Step-by-step explanation:

y= x+20

y= - x+2

Result of the addition: (-x cancels x)

2y = 22

y =11

11=x +20            -9 = x

11=9+2

11 = -9 +20

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3 years ago
Which is the greater than or less than sign I cannot remember it is confusing sorry if you help thank you!
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What is the fraction for five over two
nignag [31]

Answer:

So normally you would say it as

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                  /       \

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3 0
3 years ago
Type the correct answer in each box. Use numerals instead of words.
lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a = 5/4, b =0 and c = -17

x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17    

8(-2) - y = -17    

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

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3 years ago
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