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vovikov84 [41]
3 years ago
12

Please help! It would be very much appreciated!

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
8 0
The answer is A) -26

hope i helped :))
LiRa [457]3 years ago
5 0
Hi, the answer is A.
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Chapter 1: Quadrati
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Answer:

mid point of AB (0.5 , 0)

Step-by-step explanation:

2x + 5y = 1      2x = 1 - 5y

y = 2xy+ 5 = (1 - 5y)*y + 5 = y - 5y² + 5

y-y = y - 5y² + 5 - y = - 5y² + 5

0 = - 5y² + 5

5y² = 5

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y = 1 or y = -1

if y = 1       x = 1/2 (1 - 5) = -2   .... point A (-2 , 1)

if y = -1      x = 1/2 (1 + 5) = 3   ..... point B (3 , -1)

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8 0
3 years ago
How many parts should be sampled in order to estimate the population mean to within 0.1 millimeter (mm) with 95% confidence? Pre
tekilochka [14]

Answer:

At least 1245 parts should be sampled.

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{1-0.95}{2} = 0.025

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 1-0.025 = 0.975, so z = 1.96

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.

How many parts should be sampled in order to estimate the population mean to within 0.1 millimeter (mm) with 95% confidence?

This is at least n parts, in which n is found when M = 0.1, \sigma = 1.8

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

0.1 = 1.96*\frac{1.8}{\sqrt{n}}

0.1\sqrt{n} = 1.96*1.8

\sqrt{n} = \frac{1.96*1.8}{0.1}

(\sqrt{n})^{2} = (\frac{1.96*1.8}{0.1})^{2}

n = 1244.67

Rouding up

At least 1245 parts should be sampled.

6 0
3 years ago
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Answer:

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What are you asking? 
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