All categories

3 years ago

The weight of a rock sample is measured to be 2.5 ponds. What is the percent of error in the measurement?

Mathematics

1 answer:

e-lub [12.9K]3 years ago

5 0

That makes no sense bc you can't find the percent of error without another number

You might be interested in

20 points for one question. You know you want to.

spayn [35]

Answer:

The answer is A.

Step-by-step explanation:

We only have to test one coordinate to find the answer. The original coordinate for N is (-1, -1). Using the formula, N' is (0, -2).

8 0

3 years ago

Read 2 more answers

I need some help 30 points for anyone

puteri [66]

Answer:

If she wants to face the food court, she has to turn through an angle of __225°__ anti-clockwise.

3 0

3 years ago

Kiran has 10 1/2 feet of lumber and wants to make a bookshelfKiran needs two side panels that are each 5 2/3 feet. How much more

eimsori [14]

Total height of lumber, H = 10 1/2 feet = 21/2 feet .

Height of side panel, h = 5 2/3 feet = 17/3 feet .

Now,

Extra lumber required, L = 2 × Height of side panel - Total height of lumber

$L=[2\times (\dfrac{17}{3})]-\dfrac{21}{2}\\\\L = \dfrac{5}{6}\ feet$

Therefore, extra lumber required is $\dfrac{5}{6}$ feet.

Hence, this is the required solution.

4 0

3 years ago

−2 1/2 −5 3/4 The distance between the two numbers is

emmainna [20.7K]

the answer to this question is -7.25 or 7 1/4

5 0

3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: Rate of change measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

3 years ago