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3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Answer:

Step-by-step explanation:

hi there, i am hooman :)

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For brainliest please help!

Advocard [28]

Answer:

negative multiples of 3 less than -3

Step-by-step explanation:

if m <9, then

m ={8,7,6,5,4,3,2,1,0,-1,-2,....,..........}

And |6m-55|

So is said that,

when, m= 8

6(8)-55=-6

when, m=7

6(7)-54=-12

when, m = 6

6(6)-54=-18

Since the trend is negative multiples of 3 less than -3

this is my answer

8 0

3 years ago

What formula would you use to calculate the mass of fermium-253 remaining after d days? f(d)=400(0.833)d f(d)=400(0.794)d f(d)=4

seraphim [82]

Answer:

f(d)=400(0.794)^d

Step-by-step explanation:

4 0

3 years ago

A line passes through (10,-1) and has a slope of -1/2 whats the equation of the line

8090 [49]

Answer:

Y=-1/2x+4

Step-by-step explanation:

-The slope is -1/2 so you just plug it in where the slope would go with an x in front

-all you need now is your y-intercept which can be found by plugging in the provided info to your basic form y=mx+b

-You can take the provided point at (10,-1) and work out what the intercepts is

-1=-1/2(10)-b

-1=-5-b

4=b

8 0

3 years ago

14. Suppose that 1 out of every 10,000 doctors in a certain region is infected with the SARS virus; in the same region, 20 out o

hram777 [196]

Answer:

The person in the at-risk population is much more likely to actually have the disease

Step-by-step explanation:

The probability of a randomly selected doctor having the disease is 1 in 1,000 (P(I)=0.0001).

The probability that a doctor is infected with SARS, given that they tested positive is:

$P(I|+)=\frac{P(I)*0.99}{P(I)*0.99+(1-P(I))*0.01}\\P(I|+)=\frac{0.0001*0.99}{0.0001*0.99+(1-0.0001)*0.01}\\P(I|+)=9.9*10^{-3}$

The probability of a randomly selected person from the at-risk population having the disease is 20 in 100 (P(I)=0.20).

The probability that a person in the at-risk population is infected with SARS, given that they tested positive is:

$P(I|+)=\frac{P(I)*0.99}{P(I)*0.99+(1-P(I))*0.01}\\P(I|+)=\frac{0.2*0.99}{0.2*0.99+(1-0.2)*0.01}\\P(I|+)=0.962$

Therefore, the person in the at-risk population is much more likely to actually have the disease

3 0

3 years ago

PLEASEEEEEEEEE HELPPPPPPPP!!!!!!!

ANTONII [103]

Second and 4th are your answers!

8 0

3 years ago

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