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bearhunter [10]
3 years ago
7

Lots of points being given out! Plz answer a REAL answer otherwise I will report! Also giving out brainly-thing to the person wh

o does the most! I need help... Plz... (I rarely need help... some of these I could do, but I wanted to give a lot of points...)

Mathematics
2 answers:
pickupchik [31]3 years ago
7 0
1) C
2) C
3) B
4) C
5) C
6)C
7)A?
8)B

hope this helps
Vsevolod [243]3 years ago
7 0
1.
1st option is correct because it goes inifnitly left and right
the 2nd one is a ray, that starts at a certain point and goes to infinity
3rd is just a line segment



2.
that's a ray, starting at Q and passing through and passing P
answer is 3rd option


3.
line segment, so no arrows
2nd option


4.
comlement is what it adds to to get 90
supplement is what it adds to to get 180
90-53=copmlement=37
180-53=supplement=127
3rd option


5.
acute means less than 90 degrees
that is 3rd angle

6. right angle so right
also 2 equal sides so isocelsese
3rd opeion

7.
alll angles are less than 90, so acute
also, it isn't isoolsees since the angles ar all different
1st option

8.
110>90 so obtues
and isocoleses since 2 equal sides
2nd option
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Answer:

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Step-by-step explanation:

Question a:

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 - 0.98}{2} = 0.01

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 - 0.01 = 0.99, so Z = 2.327.

Now, find the margin of error M as such

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

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M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

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

0.0001 = 2.327\frac{0.0003}{\sqrt{n}}

0.0001\sqrt{n} = 2.327*0.0003

\sqrt{n} = \frac{2.327*0.0003}{0.0001}

(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2

n = 48.73

Rounding up

49 measurements are needed.

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How to solve this problem?
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