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3 years ago

15

There are 3 red and 9 blue counters in a box. A counter is chosen and then replaced. If you did this 144 times, how many times w

ould you expect to get a blue?

1 answer:

anygoal [31]3 years ago

3 0

If you do it 144 times that is (3+9)*12
This means that 12 times 3 red and 9 blue counters will be chosen that means that 9 blue counters will be chosen 12 times. 9*12 is 108.
This means blue counters will be chosen 108 times.
This can be checked as 108/3 would give you the number of red counters which is 36 and 108+36 is 144

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2 years ago

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Answer:

50%

Step-by-step explanation:

68-95-99.7 rule

68% of all values lie within the 1 standard deviation from mean $(\mu-\sigma,\mu+\sigma)$

95% of all values lie within the 1 standard deviation from mean $(\mu-1\sigma,\mu+1\sigma)$

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$\mu = 55\\\sigma = 4$

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Refer the attached figure

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3 years ago

There were 900 students enrolled in a high school in 2009 and 1,500 students enrolled in the same high school in 2012. The stude

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Answer:

(A)

$(t_0,P_0)=(0,900)$

$(t_1,P_1)=(3,1500)$

(B)

$m=200$

(C)

$P=200t+900$

(D)

The high school's enrollment in 2017 is 2500

Step-by-step explanation:

Let's assume time starts since 2009

so, In 2009 , t=0

P=900

In 2012,

t=2012-2009=3

P=1500

(A)

So, we have points as

$(t_0,P_0)=(0,900)$

$(t_1,P_1)=(3,1500)$

(B)

we can use slope formula

$m=\frac{P_2-P_2}{t_2-t_1}$

we can plug values

$m=\frac{1500-900}{3-0}$

$m=200$

we know that

slope is rate of change of P with respect to time

so, slope means increase in population is 200 per year

(C)

we can use point slope form of line

$y-y_1=m(x-x_1)$

$m=200$

$(t_0,P_0)=(0,900)$

$P-900=200(t-0)$

$P=200t+900$

(D)

In 2017 ,

t=2017-2009=8

we can plug t=8

and we get

$P=200(8)+900$

$P=2500$

The high school's enrollment in 2017 is 2500

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