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Snezhnost [94]
3 years ago
10

Geometric mean of 8 and 5

Mathematics
1 answer:
Papessa [141]3 years ago
3 0

Answer:

40

Step-by-step explanation:

8 16 24 32 40

5 10 15 20 25 30 35 40

They both have the number 40

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Factor completely.
Marta_Voda [28]
<span>x^2 - 3x - 28 = </span><span>(x - 7)(x + 4)

answer
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</span><span>(x - 7)(x + 4)
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8 0
3 years ago
Five Cs Bank issues credit cards to its customers. In the past 3.4% of the out-of-state nonfraudulent credit card transactions w
kolezko [41]

Answer:

0.0838 (8.62%)

Step-by-step explanation:

defining the event G= an out-of-state transaction took place in a gasoline station , then the probability is

P(G) = probability that the transaction is fraudulent * probability that took place in a gasoline station given that is fraudulent + probability that the transaction is not fraudulent * probability that took place in a gasoline station given that is not fraudulent =  0.033 * 0.092 + 0.977 * 0.034 = 0.0362

then we use the theorem of Bayes for conditional probability. Defining also the event F= the transaction is fraudulent  , then

P(F/G)=P(F∩G)/P(G) =  0.033 * 0.092 /0.0362 =  0.0838 (8.62%)

where

P(F∩G)= probability that the transaction is fraudulent and took place in a gasoline station

P(F/G)= probability that the transaction is fraudulent given that it took place in a gasoline station

7 0
2 years ago
Read 2 more answers
Which is the function represented by the table?
gregori [183]
I’m pretty sure it’s B
7 0
3 years ago
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Easy Question giving 80 points
Bas_tet [7]

Answer:

1=30

2=90

3=21

4=39

5=39

6=24

7=24

8=66

9=66

10=21

11=60

12=120

13=60

14=120

15=48

16=105

17=27

18=129

19=51

20=129

21=51

Step-by-step explanation:

please please please pleas mark me brainliest that was really hard

5 0
3 years ago
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For each of the following vector fields
olga nikolaevna [1]

(A)

\dfrac{\partial f}{\partial x}=-16x+2y

\implies f(x,y)=-8x^2+2xy+g(y)

\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y

\implies\dfrac{\mathrm dg}{\mathrm dy}=10y

\implies g(y)=5y^2+C

\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}

(B)

\dfrac{\partial f}{\partial x}=-8y

\implies f(x,y)=-8xy+g(y)

\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x

\implies \dfrac{\mathrm dg}{\mathrm dy}=x

But we assume g(y) is a function of y alone, so there is not potential function here.

(C)

\dfrac{\partial f}{\partial x}=-8\sin y

\implies f(x,y)=-8x\sin y+g(x,y)

\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y

\implies\dfrac{\mathrm dg}{\mathrm dy}=4y

\implies g(y)=2y^2+C

\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}

For (A) and (C), we have f(0,0)=0, which makes C=0 for both.

4 0
3 years ago
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