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kati45 [8]
3 years ago
8

Interest of 3 million dollars 100% and 50 years please help

Mathematics
2 answers:
FromTheMoon [43]3 years ago
5 0

Answer:

$3,377,699,720,527,868,854,272  

Step-by-step explanation:

Hope this helps

dlinn [17]3 years ago
4 0

Answer:

$3,377,699,720,527,868,854,272  

Step-by-step explanation:

Hope this helps

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What is the range of the function below?
sergij07 [2.7K]

Answer:

C

Step-by-step explanation:

f(x)=72-4x

72-4x=0

4x=72

x=18

f(x)<=18

3 0
3 years ago
Help please ((: ughhh
Dvinal [7]

1/2............................................ I hope this helps



4 0
3 years ago
E<img src="https://tex.z-dn.net/?f=e%5E%7Bx%7D%20-%20e%5E%7B-2%7D%20%3D-2" id="TexFormula1" title="e^{x} - e^{-2} =-2" alt="e^{x
pav-90 [236]

$e\cdot e^x -e^{-2}=-2$

$\implies e^{x+1}=e^{-2}-2$

note that RHS is negative. (because e with negative exponent is less than 1)

and LHS is always positive.

so there cannot be any solution

3 0
3 years ago
A traffic light at a certain intersection is green 50% of the time, yellow 10% of the time, and red 40% of the time. A car appro
kati45 [8]

Answer:

6.4\times 10^{-5} = 0.000064 = 0.0064\%.

Step-by-step explanation:

Probability that the car encounters a green light on the first day: 50 \% = 0.5.

To meet the question's conditions, the car needs to encounter another green light on the second day. Given that the colors of the light on each day are "independent," the chance that there's a green light followed by another green light will be

(0.5) \times 0.5 = 0.25.

  • Condition is met on the first two days and green light on the third day: (0.5 \times 0.5) \times 0.5 = 0.125.
  • Condition is met on the first three days and green light on the fourth day:   (0.5 \times 0.5 \times 0.5) \times 0.5.

To meet the condition on the fifth day, there needs to be a yellow light. The probability that the condition is met on the first four days and on the fifth day will be (0.5 \times 0.5 \times 0.5 \times 0.5) \times 0.1 = 0.5^{4} \times 0.1.

To meet the condition on the sixth day, all prior days should meet the conditions. Besides, there needs to be a red light on the sixth day. (0.5^{4} \times 0.1) \times 0.4

  • Seventh day: (0.5^{4} \times 0.1 \times 0.4 ) \times 0.4
  • Eighth day: (0.5^{4} \times 0.1 \times 0.4^2 ) \times 0.4
  • Ninth day: (0.5^{4} \times 0.1 \times 0.4^3 ) \times 0.4
  • Tenth day: (0.5^{4} \times 0.1 \times 0.4^4 ) \times 0.4 = 0.5^{4} \times 0.1 \times 0.4^{5}

The question asks that the condition be met on all ten days. As a result, the probability of meeting the condition will be equal to the probability on the tenth day: 0.5^{4} \times 0.1 \times 0.4^{5} = 6.4\times 10^{-5} = 0.000064 = 0.0064\%.

6 0
3 years ago
Read 2 more answers
POSSIBLE
Levart [38]
I believe the answer is supplementary
7 0
3 years ago
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