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saveliy_v [14]
4 years ago
10

What times what equals 92

Mathematics
2 answers:
nevsk [136]4 years ago
5 0
46 times 2 is equals 92
Strike441 [17]4 years ago
5 0
2*46=92
1*92=92
.
Z
Z
Z
Z
Z

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Answer:

Part A) distance covered by the aligned red cells is: 180000 Km

Part B) 8000000 bacteria (8 million bacteria)

Step-by-step explanation:

Part A)

Let's first find the total number of red cells that the person has in the 5 liters, knowing that there are 4,500,000 of them in a cubic millimetre:

Since 1 liter contains 1,000,000 cubic millimetres, then, in 5 liters there must be 5 times that number of cubic millimetres. That is a total of 5,000,000 cubic millimetres.

Since in each cubic millimetre there are 4,500,000 red cells, then the number of red cells in the 5 liters must be the product of these two quantities:

4,500,000 x 5,000,000 = 22,500,000,000,000 red blood cells.

Now, to find the distance they will cover aligned one after the other, knowing that their average diameter is 0.008 mm is:

22,500,000,000,000 x 0.008 mm= 180000000000 mm

22.5\,\,10^{12} * 8\,\,10^{-3}=1.8\,\,10^{11} mm

In order to convert this distance into Km as requested, we use the fact that 1 mm = 0.000001 Km, or using exponents in base ten:

1 mm = 10^{-6}\,\,Km

Then, the distance covered in kilometers is:

1.8\,\,10^{11}\,*\,10^{-6} \,\,Km=1.8\,\,10^{5}\,\,Km=180000\,\,Km

Part B)

The vaccine contains a total of  10^8  bacteria per cm^3

In order to find how many bacteria we have in 80 mm^3 of the vaccine, we first need to convert 80 mm^3  into cm^3:  

80\,\, mm^3 = 0.08 \,\,cm^3

Then, if in one cm^3 there is 10^8  bacteria, in 0.08\,\,cm^3 there would be: 0.08\,*\,10^8 bacteria. That is: 8\,\,10^6 bacteria = 8000000 bacteria (eight million bacteria)

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How much money is shown?
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3 years ago
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Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time
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Answer:

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

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New machine

Sample mean                  x₁ =    25

Sample variance               s₁  = 27

Sample size                       n₁  = 45

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Sample mean                    x₂ =  23  

Sample variance               s₂  = 7,56

Sample size                       n₂  = 36

Test Hypothesis:

Null hypothesis                         H₀             x₂  -  x₁  = d = 0

Alternative hypothesis             Hₐ            x₂  -  x₁  <  0

CI = 90 %  ⇒  α =  10 %     α = 0,1      z(c) = - 1,28

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z(s)  =  ( x₂  -  x₁ ) / √s₁² / n₁  +  s₂² / n₂

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n₁  = 45    ⇒   s₁² / n₁    = 16,2

s₂  = 7,56   ⇒    s₂²  = 57,15

n₂  = 36     ⇒    s₂² / n₂  =  1,5876

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The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

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