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wlad13 [49]
2 years ago
11

Share 1200kg in the ratio 3:37

Mathematics
2 answers:
bearhunter [10]2 years ago
5 0

Answer:

3+37=40

3/40×1200=90kg

37÷40×1200=1110kg

Step-by-step explanation:

''.''

igor_vitrenko [27]2 years ago
4 0

Answer:

3x+37x=40x\\40x=1200kg\\x=\frac{1200}{40} \\x=30kg\\

3x=30kg*3=90kg\\37x=30kg*37=1110kg

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Solve the following equation: <br><br> x^3 -168 = [24(x-2)] / 2<br><br> PLEASE REPLY ASAP
Nataly_w [17]
Hi Larry

x^3 - 168 = [ 24(x-2)] / 2
x^3 - 168 = (24x - 48)/2
x^3 - 168 = 12x - 24
Subtract 12x - 24 to both sides
x^3 - 168 - (12x - 24) = 12xx - 24 - (12x - 24)
x^3 - 12x - 144 = 0
Now, factor the left sides
(x - 6)(x^2 + 6x + 24) = 0
Set factors equal to 0
x - 6 = 0 or x^2 + 6x + 24 = 0
x = 0 + 6 or x^2 + 6x + 24 - 0
x = 6
Answer : X = 6

Good luck !
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3 years ago
What is 190×76=? <br> Please help!
mart [117]
190*76 is equal to 14440.
6 0
3 years ago
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three points have coordinates A(-1,2),B(3,10)and C(p,8).Find the value (s) of p if AC is perpendicular to BC.
Reil [10]

Step-by-step explanation:

All steps are in pic above.

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4 0
2 years ago
A poll found that 53% of a sample of 1170 teens in a certain large country go online several times a day. (Treat this sample as
posledela

Answer:(a) margin error = 2.4%

(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.

(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%

Step-by-step explanation:

Using the proportion formulae

Margin error = z √p(1-p)/n

n= 1170, p = 53% = 0.53, 1-p = 0.47

and the z value at 90% C.I = 1.645

M error= 1.645 √0.53×0.47/1170

Margin error = 0.024 = 0.024 ×100

Margin error = 2.4%

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In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.

8 0
3 years ago
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so that x = 2.

6 0
2 years ago
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