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Harman [31]
3 years ago
6

Elsa, Deon, and Carlos served a total of 107 orders Monday at the school cafeteria. Deon served 5 more orders than Elsa. Carlos

served 4 times as many orders as Elsa. How many orders did they each serve?
Mathematics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

Elsa - 17 orders

Deon - 22 orders

Carlos - 68 orders

Step-by-step explanation:

You need to set variables and create system of equations first.

Elsa=  e

Deon= d

Carlos = c

e+d+c=107

d=e+5

c=4e

Now you plug in the values into the equation x+y+z=116

e+(e+5)+(4e)=107

6e+5=107

6e=102

e=17

You plug in the value of x into the other equations.

e = 17

d= 17+5=22

c=4*(17)=68

Rearrange so it corresponds with days...

Elsa - 17 orders

Deon - 22 orders

Carlos - 68 orders

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3 years ago
11b-12&lt;-100
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Answer:

ok wait which symbol? this (<)?

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LHS=RHS

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3 years ago
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10 men can do the work in 8 days. how many men should be reduced to finish the work in 10 days​
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If Reducing 2 men will bbe enough too finish the work in 10 days

For rate of work use the formualar

rate of work = quantity of work / duration

10 men can do the work at a rate of  =  1 / 8

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x * ( 1 / 8 ) = 10 * ( 1 / 10 )

x = 1 * 8 = 8

therefore 8 men can do the work in 10 days

number of men reduced = 10 men - 8 men = 2 men

Hope it helped

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2 years ago
Which answer choice represents an equivalent number sentence that uses the commutative property to rewrite " 2 x 8 x 3"?
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8 0
3 years ago
. A shipyard makes a container ship that can withstand the total amount of weight W, which is normally distributed with mean of
Debora [2.8K]

Answer:

the maximum number of containers that the ship can load is 170

Step-by-step explanation:

Given the data in the question;

W ~ N( 600, 60 )

S ~ N( 4n, 0.4√n )

so

p( W > S ) = 0.90

⇒ P( W - S > 0) = 0.9 ------ let this be equation 1

now, since W and S are independent

Mean( W - S ) = Mean( W ) - Mean( S 0 = 600 - 4n

and

SD( W - S ) = √( var(W) + var(S) ) = √( 60² + 0.4²n)

hence;

W - S ~ N( 600 - 4n, √( 60² + 0.4²n) )

now, from equation one, P( W - S > 0) = 0.9

P( \frac{(W-S)-Mean(W-S)}{SD(W-S)} > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

from z- table

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = P( z >-1.282)  

\frac{4n - 600}{\sqrt{60^2 + 0.4^2n} } = -1.282 ------------------ let this be equation 2

now, we square both sides of equation 2

\frac{(4n - 600)^2}{60^2 + 0.4^2n} } = (-1.282)^2  

\frac{(4n - 600)(4n-600)}{3600 + 0.16n} } = 1.643524

we cross multiply

16n² + 360000 - 4800n = 1.643524( 3600 + 0.16n )

16n² + 360000 - 4800n = 5916.6864 + 0.26296384n

16n² + 360000 - 5916.6864 - 4800n - 0.26296384n = 0

16n² + 354083.3136 - 4800.26296384n = 0    

16n² - 4800.26296384n + 354083.3136 = 0  

solving the quadratic equation, we know that;

x = -b±√( b² - 4ac ) / 2a

so we substitute

x = [-(-4800.26296384) ±√( (-4800.26296384)² - (4 × 16 × 354083.3136)] / [2×16]

x = [ 4800.26296384 ±√( 23042524.522 - 22661332.0704 ] / 32

x = [ 4800.26296384 ±√(381192.4516) ] / 32

x = [ 4800.26296384 ± 617.4078 ] / 32

Hence;

x = [ 4800.26296384 - 617.4078 ] / 32 or  [ 4800.26296384 + 617.4078 ] / 32

x = 131   or  170

Therefore, the maximum number of containers that the ship can load is 170

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