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lukranit [14]
3 years ago
8

Which of the following lines has a slope of -1/2? x + 2y = 0 x - 2y = 0 -x + 2y = 0

Mathematics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

A

Step-by-step explanation:

The correct answer is A, x+2y=0

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What is the range of the relation[(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]
vesna_86 [32]

The range is the set of values that the relation takes for the domain for which it is defined.

Is the set of values of the dependent variable.

In this case, we have the relation [(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]​.

The dependant variable takes the values: 0, -3, 4, 0 and -1. Some values are repeated.

We put the repeated values only once and sort to write the range.

Then, the range is R = {-3, -1, 0, 4}.

Answer: Range = {-3, -1, 0, 4}

4 0
1 year ago
Ginny has 16 chocolate bars. She gives away ¾ of her chocolate bars to friends. How many chocolate bars did Ginny give away?
Phantasy [73]

Answer:

She gave away 12, so she still has 4.

Step-by-step explanation:

16 times .75 equals 12.

6 0
3 years ago
Read 2 more answers
For the following right triangles:
CaHeK987 [17]

Let us first define Hypotenuse Leg (HL) congruence theorem:

<em>If the hypotenuse and one leg of a right angle are congruent to the hypotenuse and one leg of the another triangle, then the triangles are congruent.</em>

Given ACB and DFE are right triangles.

To prove ΔACB ≅ ΔDFE:

In ΔACB and ΔDFE,

AC ≅ DF (one side)

∠ACB ≅ ∠DFE (right angles)

AB ≅ DE (hypotenuse)

∴ ΔACB ≅ ΔDFE by HL theorem.

6 0
3 years ago
Whats the distance between {-4 , 12} and {18 , 12}
belka [17]
You would need to use the distance formula.
Distance = Sqrt((x2-x1)^2+(y2-y1)^2)
Plug in your numbers
Distance = Sqrt((12+4)^2+(12-18)^2)
Distance = 17.09
4 0
4 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
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