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3 years ago

What simplified ratio correctly compares 10,000 centimeters to 1000 meters?

Mathematics

2 answers:

zloy xaker [14]3 years ago

7 0

10:1

Explanation: Divide 1000 meters by 1000 to decrease it to 1 so 10000 divided by 1000 is 10.

mina [271]3 years ago

5 0

ANSWER

$1: 10$

EXPLANATION

We want to simplify, the ratio 10,000 centimeters to 1000 meters.

Thus,

$10,000cm:1000m$

We need to convert to the same unit and simplify.

Recall that,

$100cm = 1m$

So the above ratio can be rewritten as,

$100m:1000m$

Or

$10,000cm:100,000cm$

We simplify now to get,

$1:10$

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Ask Your Teacher The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. (a) Use differentials t

SCORPION-xisa [38]

Answer:

a) 28,662 cm² max error

0,0111 relative error

b) 102,692 cm³ max error

0,004 relative error

Step-by-step explanation:

Length of cicumference is: 90 cm

L = 2*π*r

Applying differentiation on both sides f the equation

dL = 2*π* dr ⇒ dr = 0,5 / 2*π

dr = 1/4π

The equation for the volume of the sphere is

V(s) = 4/3*π*r³ and for the surface area is

S(s) = 4*π*r²

Differentiating

a) dS(s) = 4*2*π*r* dr ⇒ where 2*π*r = L = 90

Then

dS(s) = 4*90 (1/4*π)

dS(s) = 28.662 cm² ( Maximum error since dr = (1/4π) is maximum error

For relative error

DS´(s) = (90/π) / 4*π*r²

DS´(s) = 90 / 4*π*(L/2*π)² ⇒ DS(s) = 2 /180

DS´(s) = 0,0111 cm²

b) V(s) = 4/3*π*r³

Differentiating we get:

DV(s) = 4*π*r² dr

Maximum error

DV(s) = 4*π*r² ( 1/ 4*π*) ⇒ DV(s) = (90)² / 8*π²

DV(s) = 102,692 cm³ max error

Relative error

DV´(v) = (90)² / 8*π²/ 4/3*π*r³

DV´(v) = 1/240

DV´(v) = 0,004

3 0

3 years ago

Read 2 more answers

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .