If x = 2, and y = 4, then x^y =

In a research laboratory, a scientist measures the mass of a 1-carat diamond and finds it to be 200 milligrams, measured to the

vlada-n [284]

Answer:

$D=3.5\ gr/cm^3$

Step-by-step explanation:

<u>Density</u>

The density of an object of mass m and volume V is given by

$\displaystyle D=\frac{m}{V}$

It can be expressed in common units like $gr/lt, Kg/m^3, gr/cm^3$ or any other combination of proper mass [M] by volume [V] units.

The data provided in the question is

$m=200\ mg = 0.2\ gr$

$V=0.057\ cm^3$

Thus, the measured density is

$\displaystyle D=\frac{0.2\ gr}{0.057 \ cm^3}$

$\boxed{D=3.5\ gr/cm^3}$

We have expressed the result with 1 decimal place because the mass was measured to the nearest hundred milligrams (or one-tenth grams). Any further decimal is senseless because that precision comes from calculations, not from measurements.

6 0

3 years ago

There is a 70% chance of rain tomorrow and a 20% chance that the volleyball game will be postponed. What is the probability that

lions [1.4K]

0.20/0.70 = 0.2857

28.6% chance

round the answer as needed.

7 0

3 years ago

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(30 points)PLEASE HELP! LOVE Y'ALL SO MUCH THX! If x=3b and y= 2/6b+4 , what is y in terms of x?

Anastaziya [24]

Answer:

y=(1/9)x+4

Step-by-step explanation:

x=3b

b=(1/3)x

y=2/6(1/3)x+4

y=(2/18)x+4

y=(1/9)x+4

8 0

3 years ago

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Can any one solve this.?

SpyIntel [72]

Answer:

x = 1

Step-by-step explanation:

Expand brackets

2x + 16 - 4x = 10x + 4

Rearrange so that x's are on one side and whole numbers are on the other

2x - 4x - 10x = 4 - 16

Simplify

-12x = -12

Make x on its own

x = 1

3 0

3 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

C)

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

3 years ago