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Lapatulllka [165]
3 years ago
11

Ayuda es urgente ya lalalalalalalalalalal​

Mathematics
1 answer:
Crank3 years ago
8 0

Answer:

no se ve la imagen perdón no te puedo ayudar

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Marrisa sold a total of 18,200 worth of clothing last week at her store.If her commision is 4% of sales,how much commisssion did
dsp73

Answer:

She would earn 728 $ in commission

Step-by-step explanation:

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Use the following cell phone airport data speeds (Mbps) from a particular network. Find the percentile corresponding to the data
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I think it's 14.1 I hope my answer is correct

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3 years ago
The length of a rectangular garden is three feet less than twice it’s width. If the perimeter of the garden is 42 feet, what is
zzz [600]
<h3><u>The length is equal to 13.</u></h3><h3><u>The width is equal to 8.</u></h3>

l = 2w - 3

2l + 2w = 42

Because we have a value for l, we can plug it into the second equation to solve for w.

2(2w - 3) + 2w = 42

Distributive property.

4w - 6 + 2w = 42

Combine like terms.

6w - 6 = 42

Add 6 to both sides.

6w = 48

Divide both sides by 6.

w = 8

Now that we have a value for w, we can solve for the exact value of l.

l = 2(8) - 3

l = 16 - 3

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6 0
3 years ago
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weqwewe [10]
Answer = -10x-8y+7
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8 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
4 years ago
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