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murzikaleks [220]
2 years ago
13

Calculati suma n+p stiind ca mn+mp=147si m =7

Mathematics
1 answer:
Anuta_ua [19.1K]2 years ago
8 0

Answer:  

n+p=21


Step-by-step explanation:

7n+7p=147si

7(10.5) +7(10.5)=147si

73.5+73.52=147si


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On a coordinate plane, 2 polygons are shown. Polygon A B C D has points (1, negative 6), (5, negative 6), (6, negative 2), and (
irga5000 [103]

Answer:

The correct option is;

Use a scale factor of 2

Step-by-step explanation:

The parameters given are;

A = (1, -6)

B = (5, -6)

C = (6, -2)

D = (0, -2)

A'' = (1.5, 4)

B'' = (3.5, 4)

C'' = (4, 2)

D'' = ( 1, 2)

We note that the length of side AB in polygon ABCD = √((5 -1)² + (-6 - (-6))²) = 4

The length of side A''B'' in polygon A''B''C''D'' = √((3.5 -1.5)² + (4 - 4)²) = 2

Which gives;

AB/A''B'' = 4/2 = 2

Similarly;

The length of side BC in polygon ABCD = √((6 -5)² + (-2 - (-6))²) = √17

The length of side B''C'' in polygon A''B''C''D'' = √((4 -3.5)² + (2 - 4)²) = (√17)/2

Also we have;

The length of side CD in polygon ABCD = √((6 -0)² + (-2 - (-2))²) = 6

The length of side C''D'' in polygon A''B''C''D'' = √((4 -1)² + (2 - 2)²) = 3

For the side DA  and D''A'', we have;

The length of side DA in polygon ABCD = √((1 -0)² + (-6 - (-2))²) = √17

The length of side D''A'' in polygon A''B''C''D'' = √((1.5 -1)² + (4 - 2)²) = (√17)/2

Therefore the Polygon A B C D can be obtained from polygon A''B''C''D'' by multiplying each side of polygon A''B''C''D'' by 2

The correct option is therefore;

Use a scale factor of 2.

3 0
2 years ago
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Which steps can be used to solve StartFraction 6 Over 7 EndFraction x + one-half = StartFraction 7 Over 8 EndFraction for x? Che
34kurt

Answer:

Subtract One-half from both sides of the equation.

Divide both sides by 6/7

Multiply both sides by 7/6

Step-by-step explanation:

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2 years ago
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The law of cosines is a2+b2-2abcosC=c2 find the value of 2abcosC
quester [9]
Isolating 2abCos(c) on one side of the equation and using the given values of a, b and c we can find the answer to this question as shown below:

a^{2} + b^{2} -2ab*cos(C)=c^{2}  \\  \\ 
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Two production lines are used to pack sugar into 5 kg bags. Line 1 produces twice as many bags as does line 2. One percent of ba
enyata [817]

Answer:

P(Bag is Defective) = 0.0167

Step-by-step explanation:

Line 1 produces twice as many bags as line 2. Let x be the number of bags produced by line 2.

No. of bags produced by line 2 = x

No. of bags produced by line 1 = 2x

Probability that the bag has been produced by line 1 can be written as:

P(Line 1) = No. of bags produced by line 1/Total no. of bags

             = 2x/(x+2x)

             = 2x/3x

P(Line 1) = 2/3. Similarly,

P(Line 2) = x/3x

P(Line 2) = 1/3

1% bags produced by line 1 are defective so the probability of line 1 producing a defective bag is:

P(Defective|Line 1) = 0.01

3% of bags from line 2 are defective, so:

P(Defective|Line 2) = 0.03

b. The probability that the chosen bag is defective can be calculated through the conditional probability formula:

P(A|B) = P(A∩B)/P(B)

<u>P(A∩B) = P(A|B)*P(B)</u>

The chosen defective bag can be either from line 1 or from line 2. So, the probability that the chosen bag is defective is:

P(Bag is Defective) = P(Defective and from Line 1) + P(Defective and from Line 2)

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                                = P(Defective|Line 1)*P(Line 1) + P(Defective|Line 2)*P(Line 2)

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P(Bag is Defective) = 0.0167

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mariarad [96]
You divide 288 by 6 and get the answer of 48
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