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11 days ago

If f(1) = 160 and f(n + 1) = –2f(n), what is f(4)?

2 answers:

8 0

Greetings,

f(1)=160
f(2)=160*(-2)
f(3)=160*(-2)²
f(4)=160*(-2)^3=-1280

The function can be expressed as f(n)=160*(-2)^(n-1)

Svet_ta [4.3K]11 days ago

6 0

Result:

-1280

Detailed explanation:

I just completed the test.

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NEED HELP ON NUMBER 13 AND 14

AnnZ [3900]

Answer:

Question 13: For age groups y=1 and y=1.3, the response time is 8 microseconds.

Question 14: The club experienced losses between 11.28 and 4.88 years.

Step-by-step explanation:

Question 13:

The equation that gives the response rate R of 8 microseconds can be expressed as

$8=y^4 +2y^3 - 4y^2 -5y +14.$

Upon graphing this, we determine the solutions to be

$y=1;$ $y=1.302;$ $y=-2;$ $y=-2.302.$

We consider only positive values of y applicable in real-life scenarios.

Thus, the response is 8 microseconds solely for the age groups y=1 and y=1.3.

Question 14:

The football club incurs losses when $p(t)$

$t^3 -14t^2 +20t +120$

Graphing this inequality reveals the solutions to be

$t$ and $4.88$

As only positive values for t are relevant in practical situations, we accept the second solution.

Hence, the club faced losses during the years $4.88$

5 0

5 days ago

3.7.6 .WP A state runs a lottery in which six numbers are randomly selected from 40 without replacement. A player chooses six nu

PIT_PIT [3949]

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

$P(X=4)=(^6_4)P({1, 1, 1, 1, 0, 0})=\frac{6!}{4!(6-4)!}(\frac{6}{40} \frac{5}{39} \frac{4}{38} \frac{3}{37} \frac{34}{36} \frac{33}{35} )=2.19\times 10^{-3}$

3 0

8 days ago

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [3900]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

7 0

11 days ago

Read 2 more answers

While standing on a highway overpass, Jennifer wonders what proportion of the vehicles that pass on the highway below are trucks

PIT_PIT [3949]

Answer: Acquire a systematic sample by selecting every 20th vehicle that travels past (irrespective of the lane or direction).

Step-by-step explanation:

8 0

9 days ago

The endpoints of CD are C(3,−5) and D(7, 9).<br> The coordinates of the midpoint M of CD are?

Inessa [3926]

Answer:

M(5,2)

Step-by-step explanation:

We apply the midpoint formula:

$( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} )$

The arithmetic average of the x and y coordinates is needed.

The coordinates for points C and D are C(3,−5) and D(7, 9).

By substituting these values, we obtain:

$( \frac{7 + 3}{2}, \frac{9 + - 5}{2} )$

$( \frac{10}{2}, \frac{4}{2} )$

$(5, 2)$

Thus, the midpoint coordinates M of segment CD are (5,2)

3 0

11 days ago