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3 years ago

5

At a butcher shop, Hilda bought 6 1/4 lb of Beef and some pork. She left with 13 1/20 lb of meat. Express the number of pounds

of pork she bought using a decimal so Hilda bought ? lb of pork

2 answers:

lina2011 [118]3 years ago

8 0

Hilda bought 15.11 lb of pork.

Step-by-step explanation:

Cost of Beef that Hilda bought = 4 17/20 = 4.85 lb

Cost of the total meat she bought = 19 24/25 = 19.96 lb.

To find the number of pounds of pork she bought, subtract.

Number of pounds of pork she bought = 19.96 - 4.85 = 15.11 lb.

Murrr4er [49]3 years ago

6 0

Answer:

Hilda bought 6 1/5 pound of pork

As a decimal 6.2 pounds of pork

Step-by-step explanation:

Make the denominator of 6 1/4 equivalent to ?/20

1/4 = ?/20

Multiply numerator and denominator by 5

5/20

So the amount of beef she bought would be 6 5/20

Change both fractions to an improper fraction

6 5/20 = 125/20 13 1/20 = 261/20

Subtract 125/20 from 261/20

261/20 - 125/20 = 136/20

<h2>Remember to keep the denominator the same.</h2>

Turn 136/20 to a proper fraction by dividing 20 from 136

136/20 = 6 8/20

Simplify

6 8/20 = 6 1/5

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2 years ago

The chair of the operations management department at Quality University wants to construct a p-chart for determining whether the

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Prof B and Prof D

Step-by-step explanation:

Step-by-step explanation:

From the question we are told that

Sample size 100

Instructor Number of Failures

Prof. A 13

Prof. B 0

Prof. C 11

Prof. D 16

Confidence level= 0.95

From Z table

Z=1.96

Generally proportion for failure is mathematically given as

$Proportion\ of\ Failure\ P=\frac{Number\ of\ Failure\ N}{Sample\ size\ S}$

$P=\frac{N}{S}$

For Prof A

$P_A=\frac{N_A}{S}$

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For Prof B

$P_B=\frac{N_B}{S}$

$P_B=\frac{0}{100}$

$P_B=0$

For Prof C

$P_C=\frac{N_C}{S}$

$P_C=\frac{11}{100}$

$P_C=0.11$

For Prof D

$P_D=\frac{N_D}{S}$

$P_D=\frac{16}{100}$

$P_D=0.16$

Generally the average proportional failure is mathematically given as

$P_a_v_g=\frac{0.13+0+0.11+0.16}{4}$

$P_a_v_g=0.10$

Therefore having this we calculate for the control limits

Generally the upper control limit UCl is mathematically given as

Upper control limits:

$UCL =P_a_v_g +Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }$

$UCL =0.1 +1.96\sqrt{\frac{0.1(1-0.1) }{100}}$

$UCL =0.1 +1.96*0.03$

$UCL =0.1 +0.0588$

$UCL =0.1588$

Generally the upper control limit UCl is mathematically given as

Lower limit control limits:

$UCL =P_a_v_g _Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }$

$UCL =0.1 -1.96\sqrt{\frac{0.1(1-0.1) }{100}}$

$UCL =0.1 -1.96*0.03$

$UCL =0.1 -0.0588$

$UCL =0.0412$

Therefore the Range is 0.0412 to 0.1588

Given the range 0.0412 to 0.1588 Prof B and Prof D are outside the Range making them the correct options

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3 years ago