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4 years ago

1.8×10^9 in standard form

Mathematics

2 answers:

prisoha [69]4 years ago

7 0

1800000000
the nine means to the ninth power

marusya05 [52]4 years ago

4 0

The answer is as follows: 1,800,000,000

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The data set below has an outlier.

nika2105 [10]

Data set: 2, 10, 10, 11, 11, 12, 12, 12, 13, 14, 14

range: 14 - 2 = 12
median: 12
lower quartile = 10
interquartile range = 4

Data set w/o outlier 2: 10, 10, 11, 11, 12, 12, 12, 13, 14, 14

range: 14 - 10 = 4
median: 12
lower quartile: 10.5
interquartile range = 3.5

The value that change the most by removing the outlier is A.) THE RANGE.

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3 years ago

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At camp, ¼ of the children are divided into 3 equal groups for crafts. What fraction of the children at camp are in EACH group?

zloy xaker [14]

Answer:

3/4?

Step-by-step explanation:

maybe not but i think this is it

7 0

3 years ago

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Determine the first four terms of the sequence in which the nth term is

Licemer1 [7]

Answer:

The correct answer option is: $\frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,\frac{1}{6}$ .

Step-by-step explanation:

We know that the $nth$ term $a_n$ for an arithmetic sequence is given by:

$a_n=\frac{(n+1)!}{(n+2)!}$

where $n$ is the number of the position of the term.

We are supposed to find the first four terms of the sequence so we will substitute the values of $n$ from 1 to 4 in the given formula to get:

1st term:

$a_1=\frac{(1+1)!}{(1+2)!}=\frac{1}{3}$

2nd term:

$a_2=\frac{(2+1)!}{(2+2)!}=\frac{1}{4}$

3rd term:

$a_3=\frac{(3+1)!}{(3+2)!}=\frac{1}{5}$

4th term:

$a_4=\frac{(4+1)!}{(4+2)!}=\frac{1}{6}$

8 0

3 years ago

Which expression is equivalent to 3√64k12?

jasenka [17]

The first one because when you simplify it equals to 3√2x2x2x2x2x2xk12, so then you can take out two 2s and 4ks.

3 0

4 years ago

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The lifetime of a certain brand of battery is known to have a standard deviation of 9 hours. Suppose that a random sample of 150

Kobotan [32]

Answer:

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645\frac{9}{\sqrt{150}} = 1.2$

The lower end of the interval is the sample mean subtracted by M. So it is 40.5 - 1.2 = 39.3 hours

The upper end of the interval is the sample mean added to M. So it is 40.5 + 1.2 = 41.7 hours

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.

7 0

3 years ago