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Alex73 [517]
3 years ago
11

Angle A is the complement of angle B.

Mathematics
1 answer:
love history [14]3 years ago
5 0

Answer:

C) cos A=sin B

Step-by-step explanation:

A is complement of B: A+B=90⁰

Sin(A)=sin(90-B)=cos(B)

Cos(A)=cos(90-B)=sin(B)

So c)

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erma4kov [3.2K]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
What is 3y^2-y-2 in factored form?
GaryK [48]

BY USING MIDDLE TERM SPLITTING,

3y^2-(3y-2y)-2

(3y^2-3y)+(2y-2)

3y(y-1)+2(y-1)

(3y+2)(y-1)

HOPE THIS WILL HELP U

7 0
4 years ago
PLEASE HELP ME I DONT UNDERSTAND.
Bingel [31]

width = 7cm

Length= 41cm

width=x

length=3x+20

96=3x+20+3x+20+x+x

96=8x+40

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5 0
3 years ago
Read 2 more answers
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
Describe any domain restrictions that make sense for each of these scenarios: The price of a stock fluctuates several times thro
mrs_skeptik [129]

Answer:

  • a) Restriction: days of the year (a discrete time)

  • b) The time the ball is on the air: t = 0, until the balls touchs the ground (any real value from t = 0 to t when the ball hits the ground)

  • c) Any positive real value: x ≥ 0

Explanation:

The <em>domain </em>is the set of elements that can be inputs of the <em>function</em>.

<u><em>1. The price of a stock fluctuates several times throughout the year as a function of time.</em></u>

The input variable is the time through out the year.

If you assume that the fluctuaion is recorded once a day, and not every second, the logical assumption is that the time is a day of the year.

Thus, the restricion is the days of the year.

You could represent them in different forms, like DD/MM/YY, or even as a number from 1 to 365 if it covers just one year.

<u><em></em></u>

<u><em>2. A ball is hit into the air by a baseball player, rises until it reaches its highest point, and then falls into left field.</em></u>

The domain is the time the ball is on the air, which may be any real value from t = 0, until the balls touchs the ground.

It is not restricted to integer numbers.

<u><em>3. The volume of a spherical balloon increases as the balloon is blown up and decreases as air is let out of it.</em></u>

The inputs of the function are amounts of air, which may be measured as number of moles.

The number of moles can be any real value equal or greater than zero.

Thus, the domain is the positive real values x ≥ 0.

8 0
3 years ago
Read 2 more answers
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