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frozen [14]
3 years ago
10

Write 4.029 in expanded form

Mathematics
1 answer:
ruslelena [56]3 years ago
7 0
The answer for that would be: 4+.02+.09
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Any help greatly appreciated
REY [17]

Answer:

Option B. 8.6%

Step-by-step explanation:

Simple index of two stocks: i=?

i=[(5,000*5.1+2,500*7.45)/(5,000*4.5+2,500*7.25)-1]*100%

i=[(25,500+18,625)/(22,500+18,125)-1]*100%

i=[(44,125)/(40,625)-1]*100%

i=[1.086153846-1]*100%

i=[0.086153846]*100%

i=8.6153846%

i=8.6%

8 0
3 years ago
Which of the following is an example of the compliment rule being applied to
UNO [17]

Answer:

Hello There!!

Step-by-step explanation:

I think the answer is B. The probability of not drawing a club or jack from a deck of cards.

hope this helps,have a great day!!

~Pinky~

6 0
3 years ago
Read 2 more answers
It take Andre 4 minutes to swim 5 laps.
Liula [17]

Answer:

a. 1.25 laps per min

b. .8 min per lap

c. 17.6 minutes

Step-by-step explanation:

5 0
2 years ago
What is the side length a in the triangle below?
grigory [225]

Answer:

a=5

Step-by-step explanation:

a^2+b^2=c^2

c^2-b^2=a^2

13^2-12^2=a^2

169-144=a^2

25=a^2

5=a

8 0
3 years ago
Suppose that a random sample of 10 adults has a mean score of 62 on a standardized personality test, with a standard deviation o
aniked [119]

Answer:

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

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.

So it 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{4}{\sqrt{10}} = 2.08

The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

5 0
2 years ago
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