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2 years ago

Which equation has the greatest product? A 5x 6 B 7x 3 C 9x 3 D 3x 8

Mathematics

1 answer:

borishaifa [10]2 years ago

3 0

Answer:

A: 5x6

Step-by-step explanation:

5x6=30

7x3=21

9x3=27

3x8=24

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Joshua goes to a local farmers market and see sinuses apples $9.80 per 10 pound bag which is a unit rate of the price per pound

horrorfan [7]

Answer:

$0.98

Step-by-step explanation:

Given data

Cost of sinuses apples = $9.8 for 10 pound

We want to find the cost per pound

Hence

if 10 pounds cost $9.8

1 pound will cost x

cross multiply

x= 9.8/10

x= $0.98

Hence a pound will cost $0.98

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2 years ago

What is the quotient?

ryzh [129]

Answer:

just post them all in one question instead of a ton of different questions. it's annoying.

Step-by-step explanation:

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2 years ago

To get a free meal at his favorite restaurant, Tom needs to spend $50 or more at the restaurant. He has already spent $30.25. Ho

MrRissso [65]

Answer:

50-30.25=19.75

Step-by-step explanation:

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2 years ago

6.Pirate Pete wants to give each of his pirate friends an equal number of gold

Leya [2.2K]

Answer:

Step-by-step explanation:

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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

sasho [114]

Answer:

Prof B and Prof D

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

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

$P_A=\frac{13}{100}$

$P_A=0.13$

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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2 years ago