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3 years ago

Which equation represents exponential decay?

Mathematics

2 answers:

timama [110]3 years ago

8 0

Answer:

i would say A) But what you gotta do is look up the word of exponential decay and whatever it says will help you with The answer

kotykmax [81]3 years ago

8 0

Answer:

The Answer is: B.

Step-by-step explanation:

The form of the equation is:

f(x) = ab^x

When b is greater than 0 and less than 1, the function is an Exponential Decay Function.

B. is f(x) = (.08)^x

In this case, a = 1

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What times 18 gives you a factor of negative 12?

kozerog [31]

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2 years ago

Solomon can buy four sodas for $6 how much would 6 sodas cost

Bad White [126]

Answer:

6 soda's will cost $9.

Step-by-step explanation:

4 soda's costs $6.

Find the amount 2 soda's will cost. Divide 2 from both sides.

(4)/2 = 2

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3 years ago

Read 2 more answers

What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i

Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

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Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

$P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}$

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

$P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\$

$P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\$