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tia_tia [17]
3 years ago
11

7 in 9 in 5 in what is the volume

Mathematics
1 answer:
Alinara [238K]3 years ago
7 0

Answer:

157.5

Step-by-step explanation:

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A certain drug treatment cures 90% of cases of hookworm in children.18 Suppose that 20 children suffer- ing from hookworm are to
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Answer:

Step-by-step explanation:

Here X the no of children suffering from hookworm treated and recovered is binomial because

a) Each child is independent of the other

b) There are only two outcomes

c) The probability for any child to recover is constant as 90% = 0.9

X is Bin (20,0.9)

a) Prob that all 20 will be cured

=P(X=20)=0.9^{20} \\=0.1216

b) Prob all but 1 will be cured=P(x=1) = 20C1 (0.9) (.1)^{19} \\=1.8(10^8)

c) Prob exactly 18 will be cured

=P(X=18)\\=20C18 (0.9)^{18} (0.1)^2\\= 0.2852

d) Prob exactly 90% will be cured.

=P(X=18)\\=20C18 (0.9)^{18} (0.1)^2\\= 0.2852

7 0
4 years ago
Joshua went kayaking. It cost him $ 15 to rent the kayak. It also cost $ 6 for each hour spent on the lake kayaking. Joshua spen
pishuonlain [190]

Step-by-step explanation:

The total bill will be the sum of the rent and the cost for the total time spent.

$15 + $6 * 2.5

$15 + $15

$30

8 0
3 years ago
In a survey of women in a certain country the mean height was 62.9 inches with a standard deviation of 2.81 inches answer the fo
Natasha2012 [34]

The question is incomplete. The complete question is :

In a survey of women in a certain country ( ages 20-29), the mean height was 62.9 inches with a standard deviation of 2.81 inches.  Answer the following questions about the specified normal distribution.  (a) What height represents the 99th percentile?  (b) What height represents the first quartile?  (Round to two decimal places as needed)

Solution :

Let the random variable X represents the height of women in a country.

Given :

X is normal with mean, μ = 62.9 inches and the standard deviation, σ = 2.81 inches

Let,

$Z=\frac{X - 62.9}{2.81}$ , then Z is a standard normal

a). Let the 99th percentile is = a

The point a is such that,

$P(X

$P \left( Z < \frac{a-62.9}{2.81} \right) = 0.99$

From standard table, we get : P( Z < 2.3263) =0.99

∴ $\frac{(a-62.9)}{281} = 2.3263$

  $a= (2.3263 \times 2.81 ) +62.9$

     = 6.536903 + 62.9

     = 69.436903

     = 69.5 (rounding off)

Therefore, the height represents the 99th percentile = 69.5 inches.

b). Let b = height represents the first quartile.

It is given by :

P( X < b) =0.25

$P \left( Z < \frac{(b-62.9)}{2.81} \right) = 0.25$

From the standard normal table,

P( Z < -0.6745) =0.99

∴ $\frac{(b-62.9)}{2.81}= 0.6745$

$b=(0.6745 \times 2.81) +62.9$

  = 1.895345 + 62.9

   = 64.795345

   = 64.8 (rounding off)

Therefore, the height represents the 1st quartile is 64.8 inches.

8 0
3 years ago
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