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zhannawk [14.2K]
4 years ago
13

Which is a positively skewed distribution

Mathematics
2 answers:
Soloha48 [4]4 years ago
6 0

Answer:

c.

Step-by-step explanation:

vovangra [49]4 years ago
5 0

Answer:

c

Step-by-step explanation:

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In a pair of similar polygons, corresponding angles are congruent.
SashulF [63]
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3 0
4 years ago
Read 2 more answers
Jonah’s dog walking service went so well that he decided to do it again the following summer. This summer, however, Jonah will o
SOVA2 [1]

Answer:

The correct options are;

1) Continue walking 5 dogs per week, but increase his rate to $5 per dog

4) Double the amount of dogs walked per week but  keep the same rate of $3 per dog

Step-by-step explanation:

The parameters given are;

Jonah is hoping to earn $200 from 8 weeks of dog walking

Therefore, Jonah has to make $200/8 per week or $25 per week

1) Continue walking 5 dogs per week, but increase his rate to $5 per dog

With the above strategy, Jonah will make $5 × 5 = $25 per week which will amount to $25 × 8 = $200 in 8 weeks total

2) Walking 5 dogs per week at $4 per dog = $20 per week and 8 × $20 = $160 in 8 weeks

3) Walking 8 dogs per week at $3 per dog = $24 per week and 8×$24 = $192 in 8 weeks

4) Double the amount of dogs walked per week to 5×2 or 10 dogs per week but keep the same rate of $3 per dog would give him 10 × $3 = $30 per week and 8 × $30 = $240 in 8 weeks

5) Double the amount of dogs walked per week to 5×2 or 10 dogs per week and cut his rate to $2 per dog would give him 10 × $2 = $20 per week and 8 × $20 = $160 in 8 weeks

Therefore, the strategies that would allow Jonah to reach his $200 goal in 8 weeks are;

1) Continue walking 5 dogs per week, but increase his rate to $5 per dog

4) Double the amount of dogs walked per week but  keep the same rate of $3 per dog.

7 0
4 years ago
Pint-Sized Portraits has 6 different backgrounds and 3 poses in which to photograph children. How many different pictures are th
pentagon [3]

It’s 18 since it’s just 6(3)

7 0
4 years ago
Read 2 more answers
The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

6 0
3 years ago
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