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ki77a [65]
3 years ago
5

What is f(r) = 60 - 2/3 r when r =12?

Mathematics
1 answer:
7nadin3 [17]3 years ago
5 0
F(r) = 60 - 2/3r        r = 12 
Substitute r for 12

f(12) = 60 - 2/3(12)                       12 * 2/3 = 8 

f(12) = 60 - 8 

f(12) = <u /><em><u>52</u></em><em><u /></em>  

52 is your answer

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7. Evaluate:<br> -8.24 + 3.7 =<br> Show work
MA_775_DIABLO [31]

Answer:

-5.46

Step-by-step explanation:

-8.24 + 3.7

3.7 can also be written as 3.70

-8.24 + 3.70

3.70

-8.24

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-5.46

The answer is -5.46

4 0
3 years ago
Ayako and Joshua have a total of 59 sweets between them.
andreyandreev [35.5K]
A+j=59 so j=59-a

and j=a-3

j=j so

59-a=a-3  subtract a from both sides

59-2a=-3  subtract 59 from both sides

-2a=-62  divide both sides by -2

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4 0
3 years ago
What is the solution of system of equations y=2x - 3.5 and x - 2y=-14
sattari [20]
Use substitution...
x-2(2x-3.5)=-14
x-4x+7=-14
-3x=-21
x=7

Plug in x into one of the equations to get y...
y=2(7)-3.5
y=10.5

(7,10.5)

Make sure to follow me!
8 0
3 years ago
Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a
Nana76 [90]

Answer:

0.00002 = 0.002% probability of actually having the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Having the disease

Probability of having a positive test:

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0.99 of 0.000001 positive. So

P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094

Probability of a positive test and having the disease:

0.99 of 0.000001. So

P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}

What is the probability of actually having the disease

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002

0.00002 = 0.002% probability of actually having the disease

6 0
3 years ago
The quotient of a and 5
Butoxors [25]

Answer:

\frac{a}{5}

Step-by-step explanation:

That would be a divide by 5: \frac{a}{5}

7 0
3 years ago
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