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Iteru [2.4K]
3 years ago
15

7) Given that Sn=3-2^(-4n)

Mathematics
1 answer:
joja [24]3 years ago
4 0

Answer:

Step-by-step explanation:

s_{n}=3-2^{-4n}\\s_{n+1}=3-2^{-4(n+1)}=3-2^{-4n} *2^{-4}\\s_{n+1}-s_{n}=3-2^{-4n}*2^-4-3+2^{-4n}

=2^{-4n}-2^{-4n}*2^{-4}\\=2^{-4n}(1-2^{-4})\\=2^{-4n}(\frac{2^4-1}{2^4})\\=2^{-4n}\frac{15}{16}\\U_{n+1}=15*2^{-4(n+1)}

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2 which of the following is a perfect cube<br> a) 400 -<br> D. 3375–<br> C 8000 -
allochka39001 [22]

Answer:

8000 is a perfect cube but 3375 can be also but i m a little sure that 8000 is a perfect cube

8 0
4 years ago
What is the solution to this system of linear equations? y − 4x = 7 2y + 4x = 2 (3, 1) (1, 3) (3, −1) (−1, 3)
ohaa [14]
Y - 4x = 7...y = 4x + 7
now sub 4x + 7 in for y in the other equation
2y + 4x = 2
2(4x + 7) + 4x = 2
8x + 14 + 4x = 2
12x = 2 - 14
12x = - 12
x = -1

y - 4x = 7
y - 4(-1) = 7
y + 4 = 7
y = 7 - 4
y = 3

solution is : (-1,3)
7 0
4 years ago
Read 2 more answers
The volume of a cylinder is 169 in.3 and the height of the cylinder is 1 in. What is the radius of the cylinder?
LenKa [72]

Answer:

4.71

Step-by-step explanation:

6 0
4 years ago
Can someone give me the answers I don't get this at all
Vikentia [17]
5 to 8 right or there more
4 0
3 years ago
Read 2 more answers
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.504 - 0.494}{0.002}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.496 - 0.494}{0.002}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

4 0
4 years ago
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