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2 years ago

For the annual rate of change of -31% find the corresponding growth or decay factor

1 answer:

3 0

Firstly, the negative sign indictaes that this is a decay. Secondly, it is decaying by 31% (or 31 per 100) per annum, or 0.31, however, I dont think any of these answers are right. certainly C and D represent growth as they are 131% and 169% (1.31 and 1.69 respectively), B represents a decrease of 4% or 0.04, and A represents a decrease of 87% or 0.87, which as you can see does not correspond to to a decrease of -0.31or -31%, are you sure you have supplied all the information in the question(such as a starting point). If you have then I would refer to teacher/tutor for further examination.

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Tina has 60 grams of popcorn. She wants to give all of her popcorn to her 4 friends. She says there are 2 ways that she can give

timurjin [86]

Answer:

you can give 15 grams to each friend or you can give 30 grams to 2 friend and share with each other.

Step-by-step explanation:

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2 years ago

Guillermo is a professional deep water free diver. His altitude (in meters relative to sea level), xxx seconds after diving, is

fgiga [73]

Answer: 140

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the area of △FGH to the nearest tenth of a square meter? The image is of a triangle GHF with base GH length 2m, FG is 2.

Gnesinka [82]

First, we are going to use the law of cosines to find the length of the line segment FH:
$FH= \sqrt{2.5^{2}+2^{2}-(2)(2.5)Cos(121)}$
$FH= \sqrt{2.5^{2}+2^{2}-5Cos(121)}$
$FH=3.2$

Next, we are going to use the semi-perimeter formula: $s= \frac{GH+FG+FH}{2}$
$s= \frac{2+2.5+3.2}{2}$
$s= \frac{7.7}{2}$
$s=3.9$

Now that we have the semi-perimeter of our triangle, we can find its area using Heron's formula:
$A= \sqrt{s(s-GH)(s-FG)(s-FH)}$
$A= \sqrt{3.9(3.9-2)(3.9-2.5)(3.9-3.2)}$
$A= \sqrt{3.9(1.9)(1.4)(0.7)}$
$A=2.7m^{2}$

We can conclude that the area of the triangle <span>GHF is 2.7 </span> $m^{2}$ .

3 0

2 years ago

A town has a population of 5000 and grows 3.5% every year. To the nearest tenth of a year, how long will it be until the populat

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

This is an exponential function. In order to find the answer to the question, we need to first determine what the equation is that models this information. The standard form for an exponential function is

$y=a(b)^x$ where a is the initial value and b is the growth/decay rate. If the starting population is 5000, then

a = 5000

If the population is growing, that means that it retains 100% of the initial population and is added to by another 3.5%. So in a sense the population grows 100% + 3.5% = 103.5% or, in decimal form, 1.035. So

b = 1.035

Our function is

$y=5000(1.035)^x$ where y is the ending population and x is the number of years it takes to get to that ending population. We want to know how long, x, it will be til the population reaches 7300, y.

$7300=5000(1.035)^x$ and we need to solve for x. The only way to do that is by using logs. I'll use natural logs for this.

Begin by dividing both sides by 5000 to get

$1.46=1.035^x$ and take the natural log of both sides:

$ln(1.46)=ln(1.035)^x$

The power rule for natural logs is that we can now bring the exponent down in front of the ln to get:

$ln(1.46)=xln(1.035)$ To solve for x, we now divide both sides by ln(1.035):

$\frac{ln(1.46)}{ln(1.035)}=x$

Do that division on your calculator and get that

x = 11.0 years.

That means that 11 years after the population was 5000 it will be expected to reach 7300 (as long as the growth rate remains 3.5%)

5 0

2 years ago

An attempt to establish a video call via some social media app may fail with probability 0.1. If connection is established and i

xxMikexx [17]

Answer:

(1). y = x ~ Exp (1/3).

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

Step-by-step explanation:

Kindly check the attachment to aid in understanding the solution to the question.

So, from the question, we given the following parameters or information or data;

(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.

(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "

(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."

(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.

8 0

1 year ago