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3 years ago

11

For every 6kg a child's weight, pediatricians prescribe 17mg of acetaminophen. If Lexie weighs 48 kg, how many milligrams of the

acetaminophen will the pediatrician prescribe?

1 answer:

Inessa05 [86]3 years ago

3 0

The answer will be 136mg
48/6=8
8*17=136mg

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The reciprocal of half a number increased by half the reciprocal of the number is 1/2. Find the number

klemol [59]

Answer:

5

Explanation:

Let the number equal x. Half the number is then

x

2

and the reciprocal of that is

2

x

The reciprocal of the number is

1

x

and half that is

1

2

x

then

2

x

+

1

2

x

=

1

2

4

x

+

x

2

x

2

=

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2

10

x

=

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x

2

2

x

2

−

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x

=

0

2

x

(

x

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Zero is not viable solution as its reciprocal is infinity. The answer is therefore

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5

7 0

3 years ago

If logb 5=0.41, find logb 12

Hitman42 [59]

<span>Determine which one of the following is the correct application of the change of base formula?

a)log3 14= log10 14/3
b)lo4 17= log10 4/ log10 17
c)log8 3= log10 3/log10 8
d)log9 32 = log10 32-log10 9

The change of base formula is as follows:
log to the base 10 of x
log to the base b of x = ----------------------------------
log to the base 10 of b

Compare this to answer choice C:

</span><span>c)log8 3= log10 3/log10 8 Here x = 3 and b = 8. This is the only correct choice.</span><span>

</span>

6 0

3 years ago

The area of a circle is equal to the area of a square measuring 5 centimeters on each side find the radius of the circle

Gre4nikov [31]

The area os the square is 25 cm sq so the radius of the circle should be about 2.82 cm

4 0

3 years ago

F. If you make $28 dollars and hour and work for 32 hours in a week. How much money do you Gross ?

shtirl [24]

Answer:

F. 896 G.627.2

Step-by-step explanation:

Take 896 and multiply it by 70% if u want u can use a calculator.

8 0

2 years ago

Read 2 more answers

2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,00

svlad2 [7]

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

55% of the students live within five miles of you, thus $p = 0.55$ .

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$

$p = 0.091125(0.55) = 0.0501$

0.0501 = 5.01% probability that you need to contact four people.

Item b:

The expected number of trials in the binomial distribution until q successes is given by:

$E = \frac{q}{p}$

In this problem, $p = 0.55$ , and 1 trial, thus $q = 1$ , hence:

$E = \frac{1}{0.55} = 1.82$

You expect to contact 1.82 students until you find one who lives within five miles of you.

Item c:

The standard deviation of the number of trials until q successes are found is given by:

$S = \frac{\sqrt{q(1 - p)}}{p}$

Hence, since $q = 1, p = 0.55$ :

$S = \frac{\sqrt{0.45}}{0.55} = 1.22$

The standard deviation is of 1.22 students.

Item d:

This probability is $P(X = 3)$ when $n = 5$ , hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.55)^{3}.(0.45)^{2} = 0.3369$

There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

Item e:

The expected value of the binomial distribution is:

$E(X) = np$

Hence, since $n = 5, p = 0.55$ :

$E(X) = 5(0.55) = 2,75$

It is expected that 2.75 students live within five miles of you.

A similar problem is given at brainly.com/question/25343741

7 0

3 years ago