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avanturin [10]
2 years ago
13

A=V1-V0/t solve for V1

Mathematics
2 answers:
enyata [817]2 years ago
8 0
For this case we have the following equation:
 A =  \frac{v1-v0}{t}
 From here, we must clear the value of v1.
 For this, we follow the following steps:
 1) Pass variable t to multiply:
 A*t  = v1 - v0
 2) Add v0 on both sides of the equation:
 A*t+v0=v1-v0+v0
 3) Rewrite the expression:
 A*t+v0=v1
 Answer:
 
The resolved expression for v1 is:
 
v1 = A*t+v0
MA_775_DIABLO [31]2 years ago
6 0
 A = V1 - V0/t.....multiply both sides by t
At = V1 - V0....add V0 to both sides
At + V0 = V1
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A university is applying classification methods in order to identify alumni who may be interested in donating money. The univers
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Answer:

Accuracy = 0.81

Sensitivity = 0.93

Specificity = 0.81

Precision = 0.047

Step-by-step explanation:

Given the confusion matrix :

Actual_______ Donation___ No Donation

Donation______ 268 (TP) _______ 20 (FN)

No Donation ___5375 (FP) _____23439 (TN)

Accuracy is calculated as :

(TP + TN) / (TP+TN+FP+FN)

(268 + 23439) / (268 + 23439 + 5375 + 20)

ACCURACY = (23707 / 29102) = 0.81

Sensitivity (True positive rate) :

TP ÷ (TP + FN)

268 ÷ (268 + 20)

268 ÷ 288 = 0.93

Specificity (True Negative rate) :

TN ÷ (TN + FP)

23439 ÷ (23439 + 5375)

23439 ÷ 28814

= 0.81

Precision :

TP ÷ (TP + FP)

268 ÷ (268 + 5375)

268 ÷ 5643

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Biologists conducted a study to investigate the flying velocity of mosquitoes both before and after feeding. The following scatt
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It represents a mosquito that flew very fast after feeding relative to all other mosquitoes.

Step-by-step explanation:

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4. Suppose that Peculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply thr
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Answer:

Peculiar purples would be more abundant

Step-by-step explanation:

Given that eculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply through a mechanism in which each single  bacterial cell splits into four. Time taken for one split is 12 m for I one and 10 minutes for 2nd

The function representing would be

i) P=P_0 (4)^{t/12} for I bacteria where t is no of minutes from start.

ii) P=P_0 (4)^{t/10} for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

a) Here P0 =3, time t = 60 minutes.

i) I bacteria P = 3(4)^{5} =3072

ii) II bacteria P = 3(4)^{4} =768

b) Since II is multiplying more we find that I type will be more abundant.

The difference in two hours would be

3(4)^{10}- 3(4)^{8} =2949120

c) i) P=P_0 (4)^{t/12} for I bacteria where t is no of minutes from start.

ii) P=P_0 (4)^{t/10} for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

d) At time 36 minutes we have t = 36

Peculiar purples would be

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e) when splits into 2, we get

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Hope this helps :)
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hope this helped :)

Step-by-step explanation: there is none

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2 years ago
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