All categories

2 years ago

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:

You might be interested in

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<-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