HURRY !!!<br> I really need the answer! And fast!

the density of a certain material is such that weighs 6900 grams for every 4.5 quarts of volume express this density in pounds p

dlinn [17]

Answer:

$d=101.4\ \text{pounds}/\text{feet}^3$

Step-by-step explanation:

Given that,

Mass of material, m = 6900 grams

Volume of the material, V = 4.5 quarts

We need to find the density in pounds per cubic foot

We know that,

1 pound = 453.59 grams

⇒ 6900 grams = 15.21 pounds

Also, 1 cubic foot = 29.92 quarts

4.5 quart = 0.15 feet³

Density = mass/volume

Or

$d=\dfrac{15.21\ \text{pounds}}{0.15\ \text{feet}^3}\\\\d=101.4\ \text{pounds}/\text{feet}^3$

Hence, the density of the material is $101.4\ \text{pounds}/\text{feet}^3$

3 0

3 years ago

Find the average distance each data value in the set is from the mean. Round your answer to the nearest tenth, if necessary. Pri

sineoko [7]

Answer:

The average distance of each observation from the mean is 0.

Step-by-step explanation:

We are given the following in the question:

$7, $20, $9, $35, $12, $15, $7, $10, $20, $25

Formula:

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{160}{10} =16$

Difference from the mean value of each term:

-9, 4, -7, 19, -4, -1, -9, -6, 4, 9

Average distances =

$\text{Average distance} =\displaystyle\frac{0}{10} =0$

Thus, the average distance of each observation from the mean is 0.

3 0

3 years ago

Simplify 5 with the power of 3 and 2.5 with the power of 2. Help fast please!!!

Mars2501 [29]

5^3=125; 2.5^2=6.25. So the answer to your first question is 125, and the second one is 6.25

6 0

3 years ago

Pls help me I am confused

Maslowich

Answer:

$\huge\boxed{A.\ \dfrac{565}{182}}$

Step-by-step explanation:

$\dfrac{8}{7}-(5x-3)=\dfrac{1}{5}x-12\\\\\dfrac{8}{7}-5x+3=\dfrac{1}{5}x-12\qquad|\text{multiply both sides by 35}\\\\35\cdot\dfrac{8}{7}-35\cdot5x+35\cdot3=35\cdot\dfrac{1}{5}x-35\cdot12\qquad|\text{simplify}\\\\5\cdot8-175x+105=7\cdot x-420\\\\40-175x+105=7x-420\\\\-175x+(40+105)=7x-420\\\\-175x+145=7x-420\qquad|\text{subtract 145 from both sides}\\\\-175x=7x-565\qquad|\text{subtract }7x\ \text{from both sides}\\\\-182x=-565\qquad|\text{divide both sides by (-182)}\\\\x=\dfrac{565}{182}$

8 0

2 years ago

Read 2 more answers

For what values of t on the interval [0, 21 ]is the instantaneous velocity positive (the projectile moves upward)?

OLEGan [10]

Answer:

0 <=t<=21

Step-by-step explanation:

Projectile is Moving upwards on an interval of (0 to 21), if we plot Velocity vs Time and denote positive y-axis above 0 and negative y-axis below 0(for velocity), then from 0 to 21 t projectile is moving upwards and has positive velocity, when the projectile reaches the top of it's motion and returns back down to ground it's velocity is negative and is plotted below the y =0 (note that is for t > 21).

hence for the interval 0 <=t <=21 the instantaneous velocity is positive (Note, instantaneous velocity is also the derivative of the velocity or the slope ).

8 0

2 years ago

HURRY !!! I really need the answer! And fast!