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monitta
2 years ago
15

if agents usually drinks three 8oz cups of drip coffee each day, how many grams of caffeine will she consume in one week?

Mathematics
1 answer:
Ivenika [448]2 years ago
4 0

Answer:

1890 mg or 1.89 grams

Step-by-step explanation:

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20 points
Bezzdna [24]

Answer:

32%

Step-by-step explanation:

P(G/S) = P(G&S)/P(S)

= (0.32×0.5)/0.5

= 0.32

32%

5 0
3 years ago
PLEASE HELP!! and, thanks so much @sparkssparkles001, @justicehall2002, @tiffanieonderdock123, @graywoods for answering the past
natali 33 [55]

Answer:

C) 120 square feet

Step-by-step explanation:

Note that the bed is in the shape of a <em>triangle</em>. You are also finding the area of the bed (Use the formula Area = 1/2(base * height)).

The base = 20, height = 12. Plug in the corresponding numbers to the corresponding variable.

A = 1/2(20)(12)

Simplify.

A = 1/2(20)(12)

A = (20 * 12)/2

A = (10 * 12)

A = 120

C) 120 square feet is your answer.

~

*Good luck!

4 0
3 years ago
What is the solution to the inequality 1/5x-4&lt;-2/5x​
timurjin [86]

Answer:

I'm sure the answer to this is: x<20/3

6 0
2 years ago
Read 2 more answers
5)Express in standard form: (i)4809000 (ii)0.00849​
zysi [14]

Answer:

I hope it helps .in case you don't understand any part ,feel free to ask.

6 0
2 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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