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3 years ago

HELP QUICKLY!!!! WILL MARK BRAINLIEST!!!

2 answers:

4 0

The volume of a cone is equal to (pi*r^2*h)/3

(pi*25*2)/3= 52.35987...

Density = mass/volume

d= 6 grams/ 52.35987 cm^3= 0.11459...

Final answer: D

stepan [7]3 years ago

4 0

Answer:

$0.11g/cm^3$

Step-by-step explanation:

We are given that

Mass of cone=6g

Radius of cone=5 cm

Height of cone=2 cm

We have to find the density of cone.

Formula:Volume of cone= $\frac{1}{3}\pi r^2 h$

Volume of cone= $\frac{1}{3}\times 3.14\times 25\times 2=52.33 cm^3$

We know that

Density = $\frac{mass}{volume}$

Density of cone= $\frac{6}{52.33}=0.11g/cm^3$

Answer: $0.11g/cm^3$

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Solve. 164+56-[3+4+1]=

alukav5142 [94]

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3 years ago

The lengths of two sides of a triangle are 4 cm 4 cm and 20 cm . 20 cm. Identify the correct pair of measures between which the

Gennadij [26K]

Answer:

I think what you meant to say is "If a triangle has 2 sides that are 4 cm and 20 cm, what are the possible lengths of the third side?"

Basically, when given 2 sides of a triangle, the third side must be GREATER than the DIFFERENCE of the other 2 sides and LESS than the SUM of the other 2 sides.

difference = 16 cm

sum = 24 cm

So, the third side must be greater than 16 cm and less than 24 cm

Step-by-step explanation:

4 0

3 years ago

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s

user100 [1]

Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{1.7}{\sqrt{36}} = 0.73$

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

4 0

2 years ago

Point C is on the segment AB which has endpoints A(-1, 0) and B(3, 8). Point C is three times as far from point A as it is from

Viefleur [7K]

Answer:

C = (2, 6)

Step-by-step explanation:

The coordinates of point C can be found as the weighted average of the endpoint coordinates. The weights are the reverse of the relative segment lengths.

For AC : CB = 3 : 1, we have ...

C = (A +3B)/(1+3) = ((-1, 0) +3(3, 8))/4 = (-1+9, 0+24)/4

C = (2, 6)

8 0

3 years ago

Two workers finished a job in 12 days. How long would it take each worker to do the job by himself if one of the workers needs 1

Lorico [155]

Two workers finished a job in 7.5 days.

How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?

let t = time required by one worker to complete the job alone

then

(t+8) = time required by the other worker (shirker)

let the completed job = 1

A typical shared work equation

7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1

multiply by t(t+8), cancel the denominators, and you have

7.5(t+8) + 7.5t = t(t+8)

7.5t + 60 + 7.5t = t^2 + 8t

15t + 60 = t^2 + 8t

form a quadratic equation on the right

0 = t^2 + 8t - 15t - 60

t^2 - 7t - 60 = 0

Factor easily to

(t-12) (t+5) = 0

the positive solution is all we want here

t = 12 days, the first guy working alone

then

the shirker would struggle thru the job in 20 days.

8 0

3 years ago

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