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slamgirl [31]
3 years ago
14

The height of a weather balloon as it is launched can be represented by an exponential model in the form y = a* b^x where y is t

he height in km after x minutes. After 2 minutes, the recorded height is 3 km, and after 3 minutes, it is 9 km. Which of the following represents another point on the graph of this exponential model?
Group of answer choices

(1,-3)

(1, 1)

(4, 15)

(4, 16)
Mathematics
1 answer:
Reil [10]3 years ago
4 0

Answer:

Step-by-step explanation:

4,15

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What does 0.47 million mean
Alex777 [14]

Answer:

b

Step-by-step explanation:

3 0
3 years ago
PLEASE PLEASE HELP!!!GIVING BRAINLEIST!! RESPOND IN 5 MIN FOR POINTS!!❤️
Mamont248 [21]

Step-by-step explanation:

The error carolyn has made is that she hasn't subtracted properly.

2x + 3y - (2x +6y) = 2x - 2x + 3y - 6y = -3y.

Carolyn probably didn't use parentheses...

To do the thing in substitution, we single out x or y in any equation. In this case, i'm gonna single out the x in the 2nd eqn. In other words, i'll make x the subject of the 2nd eqn.

2x + 6y = 10

2x         = 10 - 6y

x           = 5 - 3y

Now we simply plug in 5 - 3y wherever x apperas in the 1st equation.

2 (5-3y) + 3y = 8

10 - 6y + 3y   = 8

10 - 3y           = 8

-3y                 =  -2

y                    = 2/3

Then x = 5 - 3*2/3 = 3

x=3

y=2/3

Hope it helps:)              

8 0
3 years ago
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte
Mashcka [7]

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0
3 years ago
Find the probability that the person is frequently or occasionally involved in charity work.
Schach [20]
Given the table below which shows the result of a survey that asked 2,881 people whether they are involved in any type of charity work.

\begin{tabular}&#10;{|c|c|c|c|c|c|}&#10; &Frequently&Occassionally&Not at all&Total\\[1ex]&#10;Male&227&454&798&1,479\\&#10;Female &205&450&747&1,402\\&#10;Total&432&904&1,545&2,881&#10;\end{tabular}

Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being&#10; \ female \ or \ not \ being \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = &#10;\frac{2200}{2881}=\bold{0.764}



Part C:

If a person is chosen at random, the probability that the person is male or frequently involved in charity work is given by

P(being&#10; \ male \ or \ being \ frequently \ involved)\\ \\= &#10;\frac{1479}{2881} + \frac{432}{2881}-\frac{227}{2881} = &#10;\frac{1684}{2881}=\bold{0.585}



Part D:

If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being&#10; \ female \ or \ not \ being \ frequently \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = &#10;\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
4 0
3 years ago
Enter the value of x that makes the equation 5+3x-37=6x+23-8x
Tamiku [17]

Answer:

x=-7

Step-by-step explanation:

6 0
2 years ago
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