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Sedbober [7]
3 years ago
11

Please help!!!! I'm struggling each says fraction,demical and word form

Mathematics
2 answers:
pav-90 [236]3 years ago
5 0
What each are their any numbers or anything?
Alina [70]3 years ago
4 0
Fraction form is number over number ex..1/2 decimal form is number . Number ex- 2.1
Word form is taking like a number and writing it in a word ex.. 10 would be ten
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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
The probability of winning a game is 15%. If you play 20 times, how many times should you expect to win?
alex41 [277]
You should expect to win at least 3 times.
Hope this helps!
3 0
3 years ago
Read 2 more answers
Describe the end behavior of the given function <br> f(x)=9x^3-x^2-8x+6<br>​
Mekhanik [1.2K]
As x approaches -inf f(x) -> -inf
and as x approaches inf, f(x) approaches +inf

Mark brainliest please
6 0
3 years ago
How do you work these out im confused
Vitek1552 [10]

\boxed {45)} Multiply 520 students by 20% (0.20).

520 * 0.20 = 104

Multiply 104 students by 12.5% (0.125).

104 * 0.125 = 13

13 students were on the wrestling team.

\boxed {46)} dimes (d) = n + .80, nickels (n) = n

Make the dimes and nickels add up to $1.40 in an equation.

(n + .80) + n = 1.40

Combine like terms.

2n + .80 = 1.40

Subtract .80 from both sides.

2n = 0.60

Divide both sides by 2.

n = 0.30

Divide this by 0.05 (how many nickels are in a dollar).

0.30/0.05 = 6

d = n + 0.80 ⇒ 0.30 + 0.80 = 1.10

Divide this by 0.10 (how many dimes are in a dollar).

1.10/0.10 = 11

There are 6 nickels and 11 dimes in the bank.

4 0
3 years ago
5x squared plus 4x plus 40
ollegr [7]
No solution . This problem makes no sense .
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