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Alika [10]
3 years ago
10

Lim (1 + 1/x)^(3x) as x->infinity

Mathematics
1 answer:
kobusy [5.1K]3 years ago
4 0
Lim ln([(x+1)/x]^3x) as x ->.infinity =lim ln([(x+1)^(3x)]/[x^(3x)]) as s->infinity =lim ln((x+1)^(3x))-ln(x^(3x)) = infinity - infinity
your answer is e3 but you can use l'hopital if you liketake the log, get 3xln(1+1/x)which is in the form ∞×0 then use the usual trick of rewriting as ln(1+1/x)/1/3x
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Find the area of a circle with radius 8yd.
olya-2409 [2.1K]

Answer:

Area of a circle = 200.96 yd²

Step-by-step explanation:

Area of a circle = πr²

Radius, r = 8 yd

π = 3.14

Area of a circle = πr²

= 3.14 * (8 yd)²

= 3.14 * 64

= 200.96 yd²

Area of a circle = 200.96 yd²

5 0
3 years ago
Find the percent decrease. Round to the nearest percent.
xenn [34]

Answer:

It should be -3.125%

Step-by-step explanation:

(124 - 128)/128 × 100

-4/128 × 100 = -3.125%

6 0
3 years ago
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Vinil7 [7]

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a

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8 0
3 years ago
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G(t) = -9t - 4 <br> g( ) = 23
Katarina [22]

Answer:

G(t) = -9t - 4

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Step-by-step explanation:

G(t) which = -9t then u - 4 g( ) from 9t which also equals 23

I hope this helps!

7 0
3 years ago
The mean work week for engineers in start-up companies is claimed to be about 63 hours with a standard deviation of 5 hours. Kar
Phantasy [73]

Answer:

a. Mean

b. μ = μ₀

c. 11.5%

d. Yes

Step-by-step explanation:

The given parameters are;

The mean work week for engineers, μ₀ = 63 hours

The standard deviation, σ = 5 hours

The number of engineers in Kara's sample, n = 10 engineers

The responses given by the 10 engineers are;

70; 45; 55; 60; 65; 55; 55; 60; 50; 55

a. The given information which the newly hired is hoping to find out that it is not is the <em>mean</em>

<em />

b. The null hypothesis which is that the company claim is correct, is therefore;

Null hypothesis, H₀; μ = μ₀ = 63 hours

c. Kara's mean, \overline x is found as follows;

\overline x = (70 + 45 + 55 + 60 + 65 + 55 + 55 + 60 + 50 + 55)/10 = 57

\overline x = 57 hours

The Z-score is therefore;

Z=\dfrac{\overline x-\mu }{\sigma }

Z = (57 - 63)/5 = -1.2

From the Z-table, we have;

The p-value for P(\overline x ≤ 57) =  P(Z ≤ -1.2) = 0.11507

The probability as a percentage given to one decimal place, that the mean would be as low as Kara's mean, P(\overline x ≤ 57) = 11.5%

d. Given that the percent percentage chance (11.5%) that the mean could be as low as hers (\overline x = 57 hours) is higher than 10%, she should accept the claim.

3 0
3 years ago
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