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Serggg [28]
2 years ago
15

A certain medicine is given in an amount proportional to a patient's body weight. Suppose a patient weighing 104 pounds requires

130 milligrams of medicine. What is the weight of a patient who requires 278.75 milligrams of medicine?(you'll get 15 points)
Mathematics
1 answer:
sergeinik [125]2 years ago
7 0

Answer:

223

Step-by-step explanation:

i think : )

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Someone please help me!
Lina20 [59]

Answer:

The fraction 58/60 is equal to 29/30 when reduced to lowest terms. To find equivalent fractions, just multiply the numerator and denominator of that reduced fraction (29/30) by any interger number, ie, multiply by 2, 3, 10, 30... 58/60 is equivalent to 29/30 because 29 x 2 = 58 and 30 x 2 = 60

Step-by-step explanation:

Dont know if this helps but i hope it does!

5 0
2 years ago
Read 2 more answers
What is the slope of the line that passes through the points (-2,12) and (0,3)
ycow [4]

Answer:

-9/2

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-12)/(0-(-2))

m=-9/(0+2)

m=-9/2

4 0
2 years ago
What is the simplest form of the expression (14.2a + 9.8b) – (13.1b – 0.2a) – (3.7a + 4.8b)?
makvit [3.9K]

The first; 10.7a-8.1b

5 0
3 years ago
Read 2 more answers
1. Jenny bought a shirt for $30. The sales tax was 6%. What was the<br> total cost for the shirt?
ahrayia [7]

Answer:

$31.80

Step-by-step explanation:

8 0
3 years ago
In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal
KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

nCx=\frac{n!}{x!(n-x)!}

So, the number of ways in which we can select four students from a group of 19 students is:

19C4=\frac{19!}{4!(19-4)!}=3,876

On the other hand, the number of ways in which we can select four students with no math majors is:

(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680

(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360

(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16

Then, the probability that the group has no math majors is:

P=\frac{1820}{3876} =0.4696

The probability that the group has at least one math major is:

P=\frac{1680+360+16}{3876} =0.5304

The probability that the group has exactly two math majors is:

P=\frac{360}{3876} =0.0929

6 0
3 years ago
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