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AlladinOne [14]
3 years ago
6

7 is 25% of what number

Mathematics
2 answers:
Molodets [167]3 years ago
7 0
21 is the number that is it that's how I got the answer cause
Ipatiy [6.2K]3 years ago
3 0
7 is 25% of 28. Thats the answer

You might be interested in
Can someone help pls
vova2212 [387]

Answer:

-8

Step-by-step explanation:

slope is change in y divided by change in x

slope = (12-36)/(11-8)

slope = -24/3

slope = -8

5 0
3 years ago
a solution consisting of 52 mg of dopamine in 26 mL of solution at a rate of 10 mL/hr what is the flow rate in mg of dopamine pe
aivan3 [116]

Answer:

20 mg / hour

Step-by-step explanation:

The first step is to find the mg/mL of dopamine.

If there are 52 mg / 26 mL, then there are 2 mg / 1 mL (just reduce the fraction).

If we are losing 10 mL / 1 hour, and there are 2 mg / mL, then the flow rate of dopamine mg / hour is 20.

You can also do this using dimensional analysis.

\frac{52 mg}{26 mL} (\frac{10 mL}{1 hour})

Just cross out the units that cancel in the numerator and denominator (mL in this case), and you're left with mg / hour. Then multiply the numerators and divide by the denominators. You get the same answer.

4 0
4 years ago
2/3 is what percent of 1/4
asambeis [7]

Answer:

267%

Step-by-step explanation:

you need to divide the second number by the first. This will give you the percentage.

2/3 ÷ 1/4 = 8/3 = 2.67 = 267%

6 0
4 years ago
Joel ordered a set of beads. He received 40 beads, and 60% of them were blue and 40% of them are green. How many blue beads did
Elina [12.6K]

Answer:

the answer is 24 there you go hope it help

7 0
3 years ago
Nickel-63 is a radioactive substance with a half-life of about 100 years. An artifact had
Marina CMI [18]

Answer:

see below

Step-by-step explanation:

The equation for half life is

n = no  e ^ (-kt)

Where no is the initial amount of a substance , k is the constant of decay and t is the time

no = 9.8

1/2 of that amount is 4.9 so n = 4.9 and t = 100 years

4.9 = 9.8 e^ (-k 100)

Divide each side by 9.8

1/2 = e ^ -100k

Take the natural log of each side

ln(1/2) = ln(e^(-100k))

ln(1/2) = -100k

Divide each side by -100

-ln(.5)/100 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/100 t)

Approximating ln(.5)/100 =-.006931472

n = 9.8 e^(-.006931472 t)  when t is in years

Now changing to days

100 years = 100*365 days/year

36500 days

Substituting this in for t

4.9 = 9.8 e^ (-k 36500)

Take the natural log of each side

ln(1/2) = ln(e^(-36500k))

ln(1/2) = -36500k

Divide each side by -100

-ln(.5)/36500 = k

Our equation in years is

n = 9.8  e ^ (ln.5)/36500 d)

Approximating ln(.5)/365=-.00001899

n = 9.8 e^(-.00001899 d)  when d is in days

3 0
4 years ago
Read 2 more answers
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