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Ivenika [448]
3 years ago
13

john likes to hike during the summer he can usually hike about 3 miles every hour but when it's really hot he swims in mountain

lake for an hour and a half every 4 miles how long does it take him to hike nine miles ?
Mathematics
1 answer:
MrRa [10]3 years ago
4 0
3 hours 3x3=9  9/3=3 hope i helped.
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This year the CDC reported that 30% of adults received their flu shot. Of those adults who received their flu shot,
Vlad [161]

Using conditional probability, it is found that there is a 0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

Conditional Probability

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 problem:

  • Event A: Person has the flu.
  • Event B: Person got the flu shot.

The percentages associated with getting the flu are:

  • 20% of 30%(got the shot).
  • 65% of 70%(did not get the shot).

Hence:

P(A) = 0.2(0.3) + 0.65(0.7) = 0.515

The probability of both having the flu and getting the shot is:

P(A \cap B) = 0.2(0.3) = 0.06

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.515} = 0.1165

0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

To learn more about conditional probability, you can take a look at brainly.com/question/14398287

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2 years ago
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Molly hikes 1/6 miles everyday to hike a total of 11/6 she would have to hike for how many days
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Molly would have to hike for 11 days for it to equal 11/6

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Find the quotient using polynomial long division:
Snowcat [4.5K]

(1) 50y^3=5y\cdot 10y^2, and

10y^2\cdot(5y-4)=50y^3-40y^2

Subtract this from 50y^3+10y^2-35y-7 to get a remainder of

(50y^3+10y^2-35y-7)-(50y^3-40y^2)=50y^2-35y-7

What we've done here is show that

\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+\dfrac{50y^2-35y-7}{5y-4}

We can keep going as long as the degree of the remainder's numerator is at least the same as the degree of its denominator.

Next, 50y^2=5y\cdot 10y, and

10y\cdot(5y-4)=50y^2-40y

Subtract this from the previous remainder to get a new remainder of

(50y^2-35y-7)-(50y^2-40y)=5y-7

which means

\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+10y+\dfrac{5y-7}{5y-4}

Finally, 5y=5y\cdot1, and

1\cdot(5y-4)=5y-4

Subtract this from the previous remainder to get

(5y-7)-(5y-4)=-3

So we end up with

\dfrac{50y^3+10y^2-35y-7}{5y-4}=\boxed{10y^2+10y+1}-\dfrac3{5y-4}

(the quotient is the expression in the box)

(2) Using the same process as before:

8m^4=2m\cdot4m^3

4m^3\cdot(2m+1)=8m^4+4m^3

(8m^4-4m^2+m+4)-(8m^4+4m^3)=-4m^3-4m^2+m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-\dfrac{4m^3+4m^2+m+4}{2m+1}

-4m^3=2m\cdot(-2m^2)

-2m^2\cdot(2m+1)=-4m^3-2m^2

(-4m^3-4m^2+m+4)-(-4m^3-2m^2)=-2m^2+m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-\dfrac{2m^2-m-4}{2m+1}

-2m^2=2m\cdot(-m)

-m\cdot(2m+1)=-2m^2-m

(-2m^2+m+4)-(-2m^2-m)=2m+4

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-m+\dfrac{2m+4}{2m+1}

2m=2m\cdot1

1\cdot(2m+1)=2m+1

(2m+4)-(2m+1)=3

\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=\boxed{4m^3-2m^2-m+1}+\dfrac3{2m+1}

8 0
3 years ago
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