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telo118 [61]
3 years ago
9

Can someone plz help me

Mathematics
1 answer:
meriva3 years ago
3 0
Faith has a ratio green:red of 24:6  which equals 4:1   which is >1

Faith is the answer
You might be interested in
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
Can someone help me please
laiz [17]

Area of trapezium= 1/2 × (a+b) × h

= 1/2 × (12 + 8) × 8

= 1/2 × 20 × 8

= 10 × 8

= 80 cm^3

Hope it helps you

3 0
3 years ago
Whats the answer to this problem ?
Hatshy [7]
Well, al you have to do is find the least common multiple of members and non-members.
members:9.50
non-members:14.75
is $126
5 0
3 years ago
in the school cafeteria students chose their lunch from 3 sandwiches, 3 soups, 4 salads, and 2 drinks. how many different lunche
Vesna [10]

Answer:

Answer:

3

×

3

×

4

×

2

=

72

Explanation:

Let's look at the 3 sandwiches and 3 soups first and then expand the calculation. There are 9 ways I can have one of the sandwiches and 1 of the soups:

⎛

⎜

⎜

⎜

⎜

⎝

0

Soup 1

Soup 2

Soup 3

Sandwich 1

1

2

3

Sandwich 2

4

5

6

Sandwich 3

7

8

9

⎞

⎟

⎟

⎟

⎟

⎠

And so we can see that we multiply the number of sandwiches and the number of soups to get the total number of ways to get one of each.

The same works for more categories of choices, and so we multiply the 3 sandwiches, the 3 soups, 4 salads, and 2 drinks to get:

3

×

3

×

4

×

2

=

72


7 0
3 years ago
3<br> 0 +<br> x+1<br> 2<br> ≤ −<br> 3x+1<br> 4
nlexa [21]

Answer:

<em>solve for x </em>

x   ≤  −  7

6 0
3 years ago
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