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Furkat [3]
2 years ago
7

Find the domain of the graphed function.

Mathematics
1 answer:
saw5 [17]2 years ago
7 0

Answer:

Step-by-step explanation:

The domain is an x thing. We are being asked to determine what x values to function takes on. We see that the blue dot on the left is at the x-value of 2. It doesn't go any farther to the left of 2. Since the dot is closed, the inequality symbol will have the "equal to" line underneath.

The function then continues along the x axis and stops at 5, closed hole. This means that the function does not go past the x value of 5 but it is included in our domain.

2 ≤ x ≤ 5. That is choice B.

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The length of a rectangle is 5 times its width. If the length is decreased by 3 meters, and the width is increased by 10 meters,
adelina 88 [10]

Answer:

   The original rectangle had a width of 30 m and a length of 150 m.

Step-by-step explanation:

6 0
3 years ago
Bill had a $7 coupon for the purchase of any item. He bought a dvd recorder that was on sale forlu
IgorC [24]

Answer:

5 maybe

Step-by-step explanation:

6 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
Which expression is the factored form of −1.5w+7.5 ?
prohojiy [21]

Answer:

c.

Step-by-step explanation:

the most that can be factored out is -1.5,

-1.5 * w = -1.5w

-1.5 * -5 = 7.5

-1.5w + 7.5, so it's

-1.5(w-5)

8 0
3 years ago
HELLLLP BEST ANSWER GETS BRAINALIST!!!!!
emmasim [6.3K]

36/15 = 8/a

multiply a .. on both sides

a×36/15= 8

multiply 36/15 reciprical on both sides

a= 8 × 15/36

a= 120/36

a= 3.33

5 0
3 years ago
Read 2 more answers
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