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sladkih [1.3K]
3 years ago
5

How is estimation helpful when adding and subtracting decimals

Mathematics
2 answers:
Usimov [2.4K]3 years ago
6 0
It's helpful because let's say you have $3.23 and you need to give someone $1.00 then you would need to subtract then what way you will know what you need to do and how many money u have left
Gnesinka [82]3 years ago
4 0
Its a lot easier to do 1.200 - .200. Than to do 1.248- .239. How I see it, estimating always gives you something to fall back on or to help see if your exact answer is accurate
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What is that product of 2x + y and 5x -y + 3
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Each term in the second factor is multiplied by each term in the first factor, as shown

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A random sample of 144 recent donations at a certain blood bank reveals that 81 were type A blood. Does this suggest that the ac
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Answer:

Null hypothesis:p=0.4  

Alternative hypothesis:p \neq 0.4

z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98  

p_v =2*P(z>3.98)=0.0000689  

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

Step-by-step explanation:

Information given

n=144 represent the random sample taken

X=81 represent the number of people with type A blood

\hat p=\frac{81}{144}=0.5625 estimated proportion of  people with type A blood

p_o=0.4 is the value that we want to verify

\alpha=0.01 represent the significance level

z would represent the statistic

p_v{/tex} represent the p value Hypothesis to testWe want to test if the percentage of the population having type A blood is different from 40%.:  Null hypothesis:[tex]p=0.4  

Alternative hypothesis:p \neq 0.4  

the statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98  

Now we can calculate the p value with this probability taking in count the alternative hypothesis:

p_v =2*P(z>3.98)=0.0000689  

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

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