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denpristay [2]
3 years ago
15

a child is to receive a dose of 0.5 teaspoon of cough medicine every 6 hours. if the bottle contains 24 doses, how many days wil

l the medicine last?
Mathematics
1 answer:
Basile [38]3 years ago
8 0
Well, to find this out, lets find out how much is taken each day:

6 hours x 4 = 24 hours (Or one day)

So because we multiplied the hours by 4, we also multiply the dosage by 4:

.5 ts x 4 = 2 ts

That means every day, 2 teaspoons is taken by the child.

Then we figure out how many teaspoons of medicine there is in 24 doses:

If .5 ts = 1 dose, then 12 ts= 24 doses because half a teaspoon is 1 dose.

So if there are 12 ts and and 2 ts are taken each day, then that means the final answer is:

The medicine will last 6 days.

Hope this helps!


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Which fraction is the smallest? A. 5⁄6 B. 2⁄3 C. 3⁄5 D. 11⁄15
BabaBlast [244]

Answer:

C

Step-by-step explanation:

5/6=0.83

2/3=0.66

3/5=0.60

11/15=0.73

by changing them to a decimal you get a better look at the amounts they are work and it puts them in a common value.From this you can see that 3/5 is the lowest fraction form the groups so your answer would be <u>C</u>

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3 years ago
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What integer values satisfy both inequalities? − 2 &lt; x &lt; 3 − 2 ≤ x &lt; 2
NARA [144]

Given:

The inequalities are:

-2

-2\leq x

To find:

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We have,

-2

-2\leq x

For -2, the possible integer values are

x=-1,0,1,2          ...(i)

For -2\leq x, the possible integer values are

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The common values of x in (i) and (ii) are

x=-1,0,1

Therefore, the integer values -1, 0 and 1 satisfy both inequalities.

6 0
3 years ago
What is the slope of the lines that are parallel to the line defined by the equation y = -x – 4?
Svet_ta [14]

Answer:

slope = -1

Step-by-step explanation:

parallel lines have the same slope,

y = -x – 4

=> y = -1(x) -4

          ↑ slope

y = mx + b

     ↑ slope

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3 years ago
Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join
zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx  = 1

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y)  and replacing a and b with their respective values, we have

\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx  = 1

Since c is a constant, we can bring it out of the integral sign; to give us

c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx  = 1

Open the bracket

c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx  = 1

Integrate with respect to y

c\int\limits^3_0 {\frac{xy^{2}}{2}  +\frac{xy^{3}}{3} } \, dx (0,3}) = 1

Substitute 0 and 3 for y

c\int\limits^3_0 {(\frac{x* 3^{2}}{2}  +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2}  +\frac{x * 0^{3}}{3})} \, dx = 1

c\int\limits^3_0 {(\frac{x* 9}{2}  +\frac{x * 27}{3} ) - (0  +0) \, dx = 1

c\int\limits^3_0 {(\frac{9x}{2}  +\frac{27x}{3} )  \, dx = 1

Add fraction

c\int\limits^3_0 {(\frac{27x + 54x}{6})  \, dx = 1

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Rewrite;

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The \frac{1}{6} is a constant, so it can be removed from the integral sign to give

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Integrate with respect to x

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Substitute 0 and 3 for x

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\frac{c}{6} *  \frac{81 * 9 - 0}{2}    = 1

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Multiply both sides by \frac{12}{729}

c    =  \frac{12}{729}

c    =  0.0165 (Approximately)

8 0
3 years ago
Can someone plz help I do not know what I am doing
PolarNik [594]

the answer is D-n is greater then or equal to nine.

because the red dot starts on nine and goes up its greater than nine and because its on nine its also going to equal it. I had a problem getting the signs messed up when I learned it but I remember the sign always eats the greater number. I hope this answer could help and I'm sorry if the answer is incorrect.

3 0
3 years ago
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