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I am Lyosha [343]
1 year ago
12

Insert 2 geometric means between -32 and 4

Mathematics
1 answer:
denis23 [38]1 year ago
6 0

Answer:

16, -8

Step-by-step explanation:

Let the terms of the geometric sequence be a, ar, ar^2, ar^3.

We know that a=-32 and thus, -32r^3=4 \implies r=-1/2.

This means ar=(-32)(-1/2)=16 and ar^2=-32(1/2)^2=-8.

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you and your go shopping for gift wrap one day. You buy 3 rolls of wrapping paper and 5 bows for a total of $32. Your friend get
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A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21
sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{340119}{10} = 34011.9

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

S.D = \sqrt{\frac{2711418821}{9}} = 17357.09

Confidence interval:

\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621

34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)

7 0
3 years ago
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