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OverLord2011 [107]
2 years ago
7

(1 + )1⁄3 + (1 − )1⁄3 = 21⁄3

Mathematics
1 answer:
ryzh [129]2 years ago
8 0
Ok what are you confused on
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Are these lines parallel explain?
nata0808 [166]

Answer: Yes

Step-by-step explanation: hope it helps

6 0
2 years ago
Read 2 more answers
Simplify. (12c + 3d + 2) + (4c + 8d + 9)
Dovator [93]
16c+11d+11 
:-) good luck 
6 0
3 years ago
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Stewart said that 30-10p does not equal 5(2p-6) for any value of p. Do you agree with Stewart? Justify Stewart's answer, or expl
charle [14.2K]
Ya he is correct !

for any value of P, both can never be equal !
as, 5(2p-6) = 10p-30 .....so even if we put any value , negative or positive, (-30) can not be changed anyway !

thus 30-10p can never be equal to 5(2p-6) !!
5 0
3 years ago
El recorrido de una etapa de una vuelta ciclística tiene una longitud de 213 kilometros y un ciclista recorre 2/3 del trayecto e
Maurinko [17]

the question in English

The route of a stage of a cycle tour is 213 km long and a cyclist covers 2/3 of the route in five hours.

a) How many kilometers are left to finish the stage?

b) If he continues with the same average speed, how much time does he have left to finish the stage?

Step 1

we know that

1) Total kilometers of a stage is 213\ km

2) The cyclist covers \frac{2}{3} of the route

so

the kilometers remaining to finish the stage is equal to

(1-\frac{2}{3})*213= \frac{1}{3}*213\\ \\=71\ km

therefore

<u>the answer   Part a) is </u>

71\ km

Step 2

<u>Find the average speed of the cyclist</u>

we know that

the speed is equal to

speed=\frac{distance}{time}

we have

time=5\ hours\\ distance=\frac{2}{3}*213=142\ km

substitute

speed=\frac{142}{5}=28.4\frac{km}{h}

Step 3

<u>Find the time remaining to finish the stage</u>

we know that

speed=\frac{distance}{time}

Solve for the time

time=\frac{distance}{speed}

we have

distance=71\ km\\ speed=28.4\frac{km}{h}

substitute

time=\frac{71}{28.4}=2.5\ hours

therefore

<u>the answer Part b) is</u>

2.5\ hours


8 0
3 years ago
The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni
aleksandr82 [10.1K]

Answer:

a) population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44

b)sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3

   sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70

c) both the means of the two samples underestimate the population mean.

   the mean pulse rate of sample 1 underestimates the population mean.

   the mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

i) population size, N = 9

a) population\hspace{0.1cm} mean = \frac{64 + 77 + 89 + 69 + 89 + 65 + 88 + 69 + 87}{9} = 77.44

b)sample\hspace{0.1cm} mean 1 = \frac{65 + 69 + 77}{3} = 70.3

   sample\hspace{0.1cm} mean 2 = \frac{64 + 69 + 77}{3} = 70

c) both the means of the two samples underestimate the population mean.

   the mean pulse rate of sample 1 underestimates the population mean.

   the mean pulse rate of sample 2 underestimates the population mean.

3 0
3 years ago
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