Decrease £189 by 38%

The surface areas of two similar solids are 169 m2 and 81 m2. the volume of the larger solid is 124.92 m3. which proportion corr

ikadub [295]

By finding the scale factor, we will see that the volume of the smaller solid is 86.75 m³.

<h3>How to get the volume of the smaller solid?</h3>

If the solids are similar, then there is a scale factor between the two. Then the relation between the areas is equal to the scale factor squared, and the relation between the volumes is equal to the scale factor cubed.

This means that if the areas are 169 m² and 81 m², then we can write:

169 m² = (k²)*81 m²

Solving for k, we get:

k = √(169 m²/81 m²) = 1.44

Then if the volume of the large solid is 124.92m³ we can write:

124.92m³ = k³*V

Replacing k and solving for V we get:

124.92m³ = (1.44)³*V

(124.92m³/ (1.44)³) = V = 86.75 m³

If you want to learn more about scale factors:

brainly.com/question/3457976

#SPJ4

6 0

2 years ago

Read 2 more answers

If a is one-factor x, what is the other factor

NARA [144]

B-y is the another factor

5 0

2 years ago

Help me please, its really easy.

notka56 [123]

Hey there. To find the answer solve 8^3=512.

Count the rows and columns of each to get C as your answer.

3 0

3 years ago

Read 2 more answers

The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor

nadya68 [22]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: