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Amanda [17]
3 years ago
15

0.9(x+1.4)−2.3+0.1x=1.60.9(x+1.4)−2.3+0.1x=1.6

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
7 0
I assume that the given equation above is 0.9(x+1.4)−2.3+0.1x=1.6 and not 0.9(x+1.4)−2.3+0.1x=1.60.9(x+1.4)−2.3+0.1x=1.6, I think there is a typo error on this. Based on equation I assumed the answer is 2.64.Thank you for posting your question here, I hope my answer helps.
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Suppose that you take 1000 simple random samples from a population and that, for each sample, you obtain a 95% confidence interv
Bogdan [553]

Using the interpretation of a confidence interval, it is found that approximately 950 of those confidence intervals will contain the value of the unknown parameter.

A x% confidence interval means that we are x% confident that the population mean is in the interval.

  • Out of a large number of intervals, approximately x% will contain the value of the unknown parameter.

In this problem:

  • 95% confidence interval.
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0.95 x 1000 = 950

Hence, approximately 950 of those confidence intervals will contain the value of the unknown parameter.

A similar problem is given at brainly.com/question/24303674

3 0
2 years ago
100 Points and Brainly.
user100 [1]

function :  y = (-x) - 6

<u>Find x-intercept</u> :

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  • x = -6

<u>Find y-intercept</u> :

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  • y = -6

mark these two points on both the axis and draw a straight linear graph.

passes coordinates : (0, -6), (-6, 0)

6 0
2 years ago
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Un industrial obtiene un préstamo de $ 15.000, al 23% anual por el lapso de 5 años. Calcular el monto a la finalización de este
olasank [31]

Answer:

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Step-by-step explanation:

According to the problem, calculation of the given data are as follows,

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Time (t) = 5 years

Let this loan is compounding annually, then the amount after 5 years can be calculated as follows,

Final amount = P  (1 + ( r/t))^{t}

by putting the value in formula, we get

= $15,000 (  (1 + ( 0.23/5))^{5}

= $15,000 × 1.2521

= $18,781.5

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2 years ago
I don't understand the bottom part
frutty [35]
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OverLord2011 [107]
The baby drank 150 mL of milk.
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2 years ago
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