Identify the type of curve given by each equation.<br><br><br> r =

Every day, there are 3 times more likes on an internet video of a cat which is modeled by the function c(n) = (3)n − 1, where n

Kisachek [45]

Answer: Hi! the answer is D, now let's prove it:

Ok, let's analyze our problem; we know two facts:

1) the first day, the video has 143 likes

2) each day that pases, there are 3 times more likes.

this means, that the day 1, the video has 143 like, the day 2, has 3*143 = 429 likes, the day 3, it has 3*429 = 1287

now, this is f(n) = the number 143 multiplied by 3 (n - 1) times, where n is the amount of days.

the function that describes this is f(n) = $143*3^{n -1}$

when n = 1, f(1) = $143*3^{0}$ = 143

when n = 2, f(2) = $143*3^{1}$ = 143*3

and so on, so the correct answer is D.

Also you can check the other functions if you like:

A) c(1) = (3)(143)^{1-1} = 3, so A doesn't work for the first day.

B) c(1) = 143^{1-1} = 1, B neither works for the first day.

C) c(1) = (3)143 − 1 = 428. C neither works for the first day

6 0

2 years ago

The outside temperature at midnight today in Polina's hometown was 18 degrees Fahrenheit. The temperature increases by 1.2 degre

densk [106]

Answer:

only students who have recess at 12 p.m. or later may go out

Step-by-step explanation:

Let the time after midnight be represented by t, and the temperature be y. Since the temperature increases by 1.2 degrees Fahrenheit each hour, the time taken to reach 32 degrees Fahrenheit can be gotten from:

y = 1.2t + 18

put y = 32

32 = 1.2t + 18

1.2t = 14

t = 14/1.2 = 11.67 hours

Since the recess starts at half hour or hour, hence the recess would start at t = 12 hours, i.e. at 12 p.m.

Hence only students who have recess at 12 p.m. or later may go out

4 0

2 years ago

Adam is using the equation (x)(x + 2) = 255 to find two consecutive odd integers with a product of 255. When Adam solves the pro

kondaur [170]

$\bf (x)(x+2)=255\implies x^2+2x=255\implies x^2+2x-255=y
\\\\
\textit{setting y=0}\implies x^2+2x-255=0
\\\\
\textit{now, factoring that}
\\\\
(x+17)(x-15)=0\implies 
\begin{cases}
x=-17\\
x=15
\end{cases}$

notice the picture of the graph added here
low and behold, x = -17, y is 0, and x= 15, y is 0
the graph is touching the x-axis, an x-intercept
or so-called, a "solution"

3 0

2 years ago

Read 2 more answers

A rectangular board is 1.6 meters long and 1.2 meters wide. What is the area of the board in square millimeters? Do not round yo

Arisa [49]

<h2>Area of rectangular board is 1920000 square millimeters</h2>

Step-by-step explanation:

A rectangular board is 1.6 meters long and 1.2 meters wide.

Here we need to find the area of the board in square millimeters.

Length = 1.6 m= 1.6 x 1000 = 1600 mm

Width = 1.2 m= 1.2 x 1000 = 1200 mm

Area = Length x Width

Area = 1600 x 1200

Area = 1920000 square millimeters

Area of rectangular board is 1920000 square millimeters

6 0

2 years ago

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago

Identify the type of curve given by each equation. r = –12sin 2θ