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Kobotan [32]
2 years ago
12

The navy contract was for 2,314 devices and took 23 workers two weeks (40 hours per week) to complete. The army contract was for

4,707 devices that were produced by 16 workers in three weeks. What's the productivity of the army contract for the input of labor hours
Mathematics
1 answer:
AlekseyPX2 years ago
3 0

Answer:

0.408 hour / device

Step-by-step explanation:

As we know The army contract was for 4,707 devices that were produced by 16 workers which means 1 worker can produce: \frac{4707}{16} ≈294 devices in 3 weeks.

<=> 294 devices= 3*40 hours

<=> 294 devices = 120 hours

<=> 1 device = 120/294 = 20/49 ≈ 0.408 hour

So it takes 0.408 hour for a worker to finish producing 1 devicein the army contract

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For this case we have the following function:

f (x) = - x ^ 2 + 13x-36

To find the zeros of the function we make y = 0and solve for "x", then:

0 = -x ^ 2 + 13x-36

We multiply by -1 on both sides of the equation:

0 = x ^ 2-13x + 36

We factor the equation, for this we look for two numbers that, when multiplied, result in 36 and when added, result in -13. These numbers are -9 and -4.

(-9) * (- 4) = 36\\-9-4 = -13

Thus, the factored equation is:

(x-9) (x-4) = 0

Therefore, the roots are:

x_ {1} = 9\\x_ {2} = 4

Answer:

x_ {1} = 9\\x_ {2} = 4

6 0
3 years ago
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kkurt [141]

Answer:

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

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qth term of given arithmetic progression is b

a_{q}=A+(q-1)D=b and

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We have to prove that

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

Now to prove LHS=RHS

Now take LHS

\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)

=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}

=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
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const2013 [10]
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500=400(1+r)^1
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A=p(1+r/100)^n
A=2500+100=2600, p=2000, n=1 year
plugging the values in the equation we obtain:
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