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Rufina [12.5K]
3 years ago
7

The ages

Mathematics
2 answers:
Zepler [3.9K]3 years ago
7 0

Answer:

Mean: 46 Median: 45 Mode: 45

Step-by-step explanation:

Mean: 41+45+39+56+48+45+42+34+47+62+35+58 = 552 then, 552/12 = 46

Median: Put numbers in order from least to greatest and cross one out on each side until you get to the middle number. 45 and 45 were the last two numbers to be crossed out, the middle of those would simply be 45!

Mode: 45 appears twice while the rest of the numbers only appeared once.

I hope this helped, good luck! :)

dolphi86 [110]3 years ago
6 0
Same the it mean is 45,median is ,47,and mode it is 34
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350 yards to 420 yards
djverab [1.8K]
70 yards hoping this helps
5 0
3 years ago
Please helpppp !!!!!!!!!!! Will mark Brianliest correct answer !!!!!!!!!!!!!!!!!!!
BlackZzzverrR [31]

Answer:

Step-by-step explanation:

This is a duplicate of one I have done for you

BF = 3*HF

BF = 3*6

BF = 18

6 0
3 years ago
Find the future value of 575 at 5.5% compounded quarterly for 5 years. Round to the nearest cent
Readme [11.4K]

Answer:

Future value = $755.61 ( to the nearest cent)

Step-by-step explanation:

The formula for calculating the future value of an invested amount compounded periodically for a number of years is given as:

FV = PV (1+\frac{r}{n} )^{n*t}

where:

FV = future value = ???

PV = present value = $575

r = interest rate in decimal = 5.5% = 0.055

n = number of compounding periods per year = quarterly = 4

t = time of investment = 5 years

∴ FV = 575 (1+\frac{0.055}{4} )^{4*5}

FV = 575 (1+0.01375)^{20}\\FV = 575 (1.01375 )^{20}\\FV = 575 * 1.3141\\FV = 755.607

∴ Future value = $755.61 ( to the nearest cent)

4 0
3 years ago
Durant and Westbrook are playing a three-point challenge game to improve long distance shooting. Each time each of them makes a
muminat

i think its C

:) :) :) :) :) :) :)

3 0
2 years ago
During the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this
alex41 [277]

Answer:

Step-by-step explanation:

Given that during  the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this period, P(t) remained close to the solution of the initial value problem.

\frac{dP}{dt} =0.03135P =0.0001489P^2, P(0) = 3.9

a) 1930 population is the population at time t = 40 years taking base year as 40

We can solve the differential equation using separation of variables

\frac{dP}{0.03135P – 0.0001489P^2 } =dt\\\frac{dP}{-P(0.03135 – 0.0001489P } =dt

Resolve into partial fractions

\frac{31.8979}{P} -\frac{0.00474}{0.0001489P-0.03135}

Integrate to get

ln P -0.00474/0.0001489  (ln (0.0001489P-0.03135) = t+C

ln P -31.833  (ln (0.0001489P-0.03135) =t+C

\frac{P}{( ( (0.0001489P-0.03135)^{31.833}  } =Ae^t

Limiting population would be infinity.

6 0
3 years ago
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