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3 years ago

Write a linear function ff with f(4)=−3f(4)=−3 and f(0)=−2f(0)=−2

1 answer:

7 0

<span>The line goes through the points (4,-3) and (0,-2), so its gradient is (-2-(-3))/(0-4) = -1/4. The y-intercept is -2 because f(0)=-2. So the equation of the line is y = -1/4 x - 2.This can be represented by the linear function f(x) = (-1/4)x - 2. The composite function ff is given by ff(x) = f(-1/4 x - 2) = (-1/4)((-1/4)x - 2) - 2 = x/16 - 3/2.</span>

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You split $2000 between two savings accounts. Account a pays 2% annual interest and account b pays 5% annual interest. After one

sdas [7]

0.05x+0.02(2000-x)=77.5
0.05x+40-0.02x=77.5
(0.05-0.02)x=77.5-40
0.03x=37.5
X=37.5/0.03=1250 at 2%
2000-1250=750 at 5%

8 0

3 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

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For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0

3 years ago

On a day, a video was posted online, 5 people watched the video. The next day the number of viewers had doubled. This tend conti

aleksklad [387]

Answer:

Day 13

Step-by-step explanation:

The question can be represented with a geometric sequence

Where,

a = 5

Common ratio , r = 2

v(t) = a * r^(t - 1)

v(t) = 5 * 2^(t-1)

20,000 = 5 * 2^(t-1)

Divide both sides by 5

20,000 / 5 = 5 * 2^(t-1) / 5

4,000 = 2^(t-1)

2^12 = 2^(t-1)

12 = t - 1

12 + 1 = t

t = 13 days

What would be the day when more than 20,000 people will see the video

Check:

v(t) = 5 * 2^(t-1)

= 5 * 2^(13-1)

= 5 * 2^12

= 5 * 4,096

= 20,480

Therefore,

The day when more than 20,000 people will see the video is day 13

6 0

3 years ago

An object moves at a constant rate. The graph shows the object's height at different times. Which equation represents the situat

zlopas [31]

<span>d. y = 15x 900</span>

Although you do not give the rates I assume the answer is D as it is the only positive multiplication. You cannot have negative height.

3 0

4 years ago

Read 2 more answers

Solve the following using pemdas 3 x 9 + 7

MAXImum [283]

27+7 = 34

so, the answer is 34

8 0

2 years ago