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Nesterboy [21]
3 years ago
14

Suppose an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that

will have the best fuel economy. The four grades are regular, economy, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. The miles per gallon were recorded for each grade. At the 0.05 level, what is the critical value of F used to test the hypothesis that the miles per gallon for each fuel are the same? Kilometers per Liter Regular Economy Premium Super Premium 39.31 36.69 38.99 40.04 39.87 40.00 40.02 39.89 39.87 41.01 39.99 39.93
Mathematics
1 answer:
lapo4ka [179]3 years ago
4 0

Answer:

this is very confusing to understand that you want.

Step-by-step explanation:

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Round 3.751 to the nearest tenth.
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Answer: It's 3.8


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Mark wants to know how many families in neighborhood plan to attend the neighborhood party
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is this an actual question? or you jus sayin stuff

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The diameter of a ping pong ball is 4.0cm. find the length in kilometers of a row of 6.02×1023 ping pong balls.
Vikentia [17]
The length is determined by multiplying the number of balls with the diameter. The answer that would come out of this is in units of centimeters.

    L = (6.02 x 10^23 balls)(4 cm/ball)
    L = 2.408 x 10^24 cm

Then, we use the proper conversion factor to convert centimeter to kilometer.

     L = (2.408 x 10^24 cm)(1 m/100 cm)(1 km/1000 m)
       L = (2.408 x 10^24)(10^5)
        

The answer to this item is 2.408 x 10^19 km. 
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3 years ago
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Andy and Jordan's are playing a game that involves tossing two dice numbered one to six, and recording the sum of the two how ma
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there are 11 diffrent sums and the greatest sum is 12

Step-by-step explanation:

5 0
3 years ago
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
3 years ago
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