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Ganezh
5 days ago
9

El 1 de noviembre de 2005, la señora Rodgers invirtió $ 10,000 en un certificado de depósito a 10 años que pagaba interés a la t

asa anual de 4% compuesto continuamente. Cuando el certificado maduró el 1 de noviembre de 2015, ella reinvirtió el monto total acumulado en bonos corporativos, los cuales pagan una tasa de interés del 5% compuesto anualmente. ¿Cuál será el monto acumulado de la señora Rodgers el 1 de noviembre de 2020?
Mathematics
1 answer:
babunello [8.4K]5 days ago
3 0
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A small-appliance manufacturer finds that the profit P (in dollars) generated by producing x microwave ovens per week is given b
PIT_PIT [9117]

Answer:

The amount of ovens that must be produced in a week to earn a $1610 profit is 70.

Step-by-step explanation:

Given:

A small-appliance manufacturer determines the profit P (in dollars) from producing x microwave ovens weekly by the formula:

P=\frac{1}{10}x(300-x)

with 0 ≤ x ≤ 200.

The target profit is $1610

So, set P = 1610, then solve for x:

1610=\frac{1}{10}x(300-x)

Multiply both sides by 10:

1610\cdot \:10=\frac{1}{10}x\left(300-x\right)\cdot \:10

16100=x\left(300-x\right)

16100=300x-x^2

-x^2+300x-16100=0

Next, factor the quadratic:

(x-70)(x-230)=0

Solving for x gives:

x=70,x=230

Since x=230 is outside the domain 0 ≤ x ≤ 200, we discard it.

Hence, the valid solution is x=70.

Therefore, to achieve a $1610 profit, the manufacturer must produce 70 ovens weekly.


7 0
1 month ago
an auto transport truck holds 12 cars. A car dealer plans to bring in 1006 new cars in June and July. if an auto transport truck
tester [8842]

Divide 1006 by 12

1006 / 12 = 83.833

This results in 83 complete trucks

83 * 12 = 996

Hence, 10 cars will be in the final truck

5 0
17 days ago
Read 2 more answers
Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t
AnnZ [9099]

Answer:   a) \bold{\dfrac{3}{16}}     b) \bold{\dfrac{1}{36}}

Step-by-step explanation:

a) To achieve an even total, there are 3 possible combinations:

1) Even, Even, Even, Even     \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

2) Even, Even, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

3) Odd, Odd, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

Order is irrelevant

Summing these yields your final result: \dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}

b) If one die shows a 2 and another a 3, while the remaining two can show any digits, there’s only one way to get a 2, one way for a 3, and six potential numbers for each of the other two dice.

\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}

3 0
19 days ago
The system of equations can be solved using linear combination to eliminate one of the variables. 2x − y = −4→10x − 5y = −20 3x
PIT_PIT [9117]
The correct choice is option D. The given equations are:...[1]...[2] Multiply equation [1] by 5 on both sides; we have...[3]. By using the elimination method, we can add equations [2] and [3] to eliminate y and determine x, resulting in... Dividing both sides by 13 yields x = 3. Substituting x back into equation [1] results in 2(3) - y = -4, which simplifies to 6 - y = -4. After subtracting 6 from both sides, we find -y = -10. Dividing through by -1 gives y = 10. Hence, the solution is (3, 10). Consequently, a valid equation that can replace 3x + 5y = 59 in the original set while still yielding the same result is 13x = 39.
6 0
9 days ago
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By selling a school blazer for Rs 1215, the shopkeeper earns a profit of 8%. Find
Svet_ta [9500]

Detailed explanation:

Thus,

100% plus 8% equals 108%

108% equals 1215

1% corresponds to 11.25

Therefore, 100% amounts to 1125 rs

4 0
23 days ago
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