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loris [4]
3 years ago
12

at a grocery store, an uncooked beef roast is on sale for $5.99/lb. At the same grocery store, prepared roast beef is available

at the deli for $2.99/100g. How much more expensive is the deli roast compared to the uncooked roast?
Mathematics
2 answers:
Anna11 [10]3 years ago
6 0

Answer:

.01669 $/g or $7.57 $/lb

Step-by-step explanation:

$2.99/100g = .0299 $/g

1 lb = 453.59237 g.

$5.99/453.59237 g= .0132$/g

-Dominant- [34]3 years ago
3 0

Answer:

$7.58/lb

Step-by-step explanation:

raw roast: $5.99 per lb

deli roast: $2.99 for 100g

We know the price per lb of the raw roast.

Let's find the price per lb of the deli roast.

1 lb = 454 grams

454 grams / 100 grams = 4.54

1 lb is 4.54 times 100 g

If we multiply the price of 100g of deli roast by 4.54, we get teh price per lb.

$2.99/lb * 4.54 = $13.57/lb

raw roast: $5.99/lb

deli roast: $13.57/lb

difference in price of 1 lb: $13.57 - $5.99 = $7.58

Answer: $7.58/lb

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Listed below are the numbers of heroic firefighters who lost their lives in the United States each year while fighting forest fi
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Answer:

Step-by-step explanation:

Hello!

Data

Year/Firefighters

2000: 20

2001: 18

2002: 23

2003: 30

2004: 20

2005: 12

2006: 24

2007: 9

2008: 25

2009: 15

2010: 8

2011: 11

2012: 15

2013: 34

<u>Common measures of variation</u>

<u>Variance</u>

S²= \frac{1}{n-1}[sumX^2-\frac{(sumX)^2}{n} ]

∑X= 20 + 18 + 23 + 30 + 20 + 12 + 24 + 9 + 25 + 15 + 8 + 11 + 15 + 34

∑X= 264

∑X²= 20² + 18² + 23² + 30² + 20² + 12² + 24² + 9² + 25² + 15² + 8² + 11² + 15² + 34²

∑X²= 5770

n= 14

S²= \frac{1}{13}[5770-\frac{(264)^2}{14} ]

S²= 60.90

<u>Standard deviation</u>

The standard deviation is the square root of the variance

S= √S²= √60.90= 7.80

<u>Coefficient of variation</u>

Is a relative standard deviation, it is defined as the division of the standard deviation (S) by the mean (X[bar)]. It has not units and is usually expressed in percentage. It shows the variability in relation to the mean.

C.V.= \frac{S}{X[bar]}*100 = \frac{7.80}{18.86}*100= 41.36%

The mean or average is a measurement of position and gives you an idea of the central value of the distribution of the data.

X[bar]= ∑X/n= 264/14= 18.86

<u>Range</u>

In the interval between the max and min values. It allows you to have an idea of the dispersion of the values.

R= Xmax - Xmin= 34 - 8= 26

Xmax= 34

Xmin= 8

<u>Inter Quartile Range</u>

Is the difference between Quantile 1 (C₁) and Quantile 3 (C₃). It shows the 50% mid values of the sample.

IQR= C₃ - C₁= 23.5 - 11.5 = 12

Quantile 1 (C₁) is the value that leaves 25% of the sample below and 75% of the sample above.

Pos1: n/4= 14/4= 3.5

The first quantile is the number between position 3 and position 4

8, 9, 11, 12, 15, 15, 18, 20, 20, 23, 24, 25, 30, 34

C₁= (11+12)/2= 11.5

Quantile 3 (C₃) is the value that leaves 75% of the sample below and 25% of the sample above.

Pos3: n*3/4= 14*3/4= 10.5

The third Quantile is between position 10 and 11

8, 9, 11, 12, 15, 15, 18, 20, 20, 23, 24, 25, 30, 34

C₃= (23+24)/2= 23.5

The measurements of variation don't allow you to know the origin of the data, that it is from consecutive years.

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