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3 years ago

8.Which statement best describes the conditions for

Mathematics

2 answers:

Kamila [148]3 years ago

8 0

Answer:

<h2>C. $0$ </h2><h2>Step-by-step explanation:</h2>

The given function is

$f(x)=a(b)^{x}$

Exponential functions which have the independent variable as an exponent.

Moreover, these functions can express decay or growth. If the exponential function is decreasing, that will express decay.

Specifically, the condition to have a decreasing exponential function is that $0$ , that is, the base of the power must be between zero and one.

This means that the right answer is C, because expresses this condition for exponential decay.

frutty [35]3 years ago

7 0

Answer:

You might be interested in

In 2018, the population will grow over 700,000.The population of a city in 2010 was 450,000 and was growing at a rate of 5% per

Firlakuza [10]

Answer:

The year is 2020.

Step-by-step explanation:

Let the number of years passed since 2010 to reach population more than 7000000 be 'x'.

Given:

Initial population is, $P_0=450,000$

Growth rate is, $r=5\%=0.05$

Final population is, $P=700,000$

A population growth is an exponential growth and is modeled by the following function:

$P=P_0(1+r)^x$

Taking log on both sides, we get:

$\log(P)=\log(P_0(1+r)^x)\\\log P=\log P_0+x\log (1+r)\\x\log (1+r)=\log P-\log P_0\\x\log(1+r)=\log(\frac{P}{P_0})\\x=\frac{\log(\frac{P}{P_0})}{\log(1+r)}$

Plug in all the given values and solve for 'x'.

$x=\frac{\log(\frac{700,000}{450,000})}{\log(1+0.05)}\\x=\frac{0.192}{0.021}=9.13\approx 10$

So, for $x > 9.13$ , the population is over 700,000. Therefore, from the tenth year after 2010, the population will be over 700,000.

Therefore, the tenth year after 2010 is 2020.

8 0

2 years ago

You have a piggy bank made of only dimes and quarters. There are 42 coins and their value is $6.75. How many dimes and quarters

babunello [35]

There are 23 quarters and 19 dimes

7 0

3 years ago

There are 40 runners in a race. How many ways can the runners finish first, second, and third?

k0ka [10]

Answer:

117600

Step-by-step explanation:

50 x 49 x 48 = 117,600 ways

You have 50 possible for 1st, 49 for 2nd, 48 for 3rd

Permutations 50P3 = 117,600

5 0

2 years ago

2) 83,97,85,84,96, 80,80,87,91<br> Mean median and mode

marusya05 [52]

$\text {Hi! Let's Solve this Problem!}$

$\text {Before you find the Mean, Medina and Mode you must put the numbers in order.}$

$\text {Before: 83, 97, 85, 84, 96, 80, 80, 87, 91}\\\text {After: 80, 80, 83, 84, 85, 87, 91, 96, 97}$

$\underline {\text {Mean}}$

$\text {To Find the Mean you must Add all the Numbers then Divide.}$

$\text {Add:}$ $\text 80+80+83+84+85+87+91+96+97=\fbox {783}}$

$\text {Divide:}$ $\text {783/9=}$

$\text {The Mean Is:}$

$\fbox {87}$

$\underline {\text Median}}$

$\text {To find the Median you find the number that's in the middle of the set.}$

$\text {Put your Left Pointer Finger on 83. Put your Right Pointer Finger on 91.}$

$\text {Move your Left Finger to the Right. Move your Right Finger to the Left.}$

$\text {Once your Left Finger makes it to 84 and your Right Finger makes it to 87}$ $\text {it shows that 85 is left in the middle.}$

$\text {The Median Is:}$

$\fbox {85}$

$\underline {\text {Mode}}$

$\text {Finding the Mode means finding the number}$ $\text { that shows the most in the number set given.}$

$\text {Since 80 is the number that is showing more than any other number}$ $\text {this tells us that 80 is our Mode.}$

$\text {The Mode Is:}$

$\fbox {80}$

$\text {Best of Luck!}$

4 0

3 years ago

Please help !! Picture provided !!!

adelina 88 [10]

Method One
Your calculator might be able to do this. Mine does it like this.
6
nCr
2
=
15

Method 2
You could simply set up 6C2

This gives you
$\frac{n!}{(n- r)! r!} = \frac{6!}{(6 - 2)!2!} = \frac{6!}{4! 2!} = \frac{6*5}{2*1} = 15$

Method Three
You only have to do this a couple of times to see how the cancellation works.

$\frac{6!}{4!2!} = \frac{6*5*\st{4*3*2*1}}{4*3*2*1 * 2*1}$
After all the cancellation takes place you have
6*5/2 = 15

7 0

2 years ago