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3 years ago

Make a box-and-whisker plot to represent the data set. video game prices: $29$29$50$39$45$20$40

Mathematics

1 answer:

charle [14.2K]3 years ago

7 0

Here is the box and whisker plot!
The median is 39.
The minimum is 20.
The first Quartile is 29.
The third Quartile is 45.
The maximum is 50.
Hope this helps!
Link - https://www.meta-chart.com/box-and-whisker

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Which choice is equivalent to the fraction below? 1/4

RoseWind [281]

The second one because 1/4 times 3/3 equals 3/12

4 0

3 years ago

Help please ASAP!!

Fittoniya [83]

Answer:

$\displaystyle 702$

Step-by-step explanation:

we are given a exponential function

$\displaystyle f(x) = {3}^{x + 1} - 22$

where x represents the number and f(x) represents the amount

we are also given that when x is 3 then f(x) is 59 likewise when x is 6 then f(x) is 2165

to figure out the average rate of change between 3 and 6 we can consider the average rate of change formula given by

$\displaystyle m = \frac{ f(x)_ {2} - f(x)_{1} }{ x_{2} - x_{1} }$

substitute what we have:

$\displaystyle m = \frac{ 2165 - 59 }{6 - 3 }$

simplify substitution:

$\displaystyle m = \frac{2106 }{3 }$

simplify division:

$\displaystyle m = 702$

hence, the average rate of change between 3 and 6 is 702

6 0

3 years ago

Read 2 more answers

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

5 0

3 years ago

5x-3(x-2)-x show your work

aliya0001 [1]

Answer: x+6

Step-by-step explanation:

5x-3(x-2)-x .......................... Question

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8 0

1 year ago

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The South Company cash account had a balance of $1,202.50 on January 31. A deposit, made on the 31st, in the amount of $100.05 w

labwork [276]

Answer:

$1326

Step-by-step explanation:

Follow the format of the reconciliation statement

The adjusted cash balances should match

Cash balance as per the bank statement, January 31 1421.95

Add: Deposit in transit 100.05

Deduct: Outstanding checks (196)

Adjusted cash balance 1326

Balance as per depositor's record, January 31 1202.5

Subtract: NSF check (18)

Subtract: Service charges (8.5)

Add: receivable collected 150

Adjusted cash balance 1326

The reconciled amount is $1326.

8 0

3 years ago