The list below shows the number of minutes Addison spent reading on each of six days. 90, 60, 89, 94, 60, 93

Mathematics

1 answer:

CaHeK987 [17]3 years ago

4 0

Mean and Median best describe the typical number of minutes she spent reading each day.

You might be interested in

Use the distributive property to write an equivalent expression <br><br> 5(6y - 5b + 1)

pav-90 [236]

Answer: 30y - 25b +5

Step-by-step explanation:

(look at attachment image fir explanation)

6 0

3 years ago

Help with#14 please!!!

Alisiya [41]

<h3> $▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$ </h3><h3 /><h3><u>Question</u> : 14</h3>

let's solve for f :

$\dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b}$

$\dfrac{1}{f} = \dfrac{b + a}{ab}$

Taking reciprocal of both sides :

$f = \dfrac{ab}{a + b}$

7 0

3 years ago

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$