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Citrus2011 [14]
3 years ago
7

What greatest common factor should be used to reduce the fraction 12 over 40

Mathematics
2 answers:
Gala2k [10]3 years ago
5 0

Answer:

10

Step-by-step explanation:u

Lera25 [3.4K]3 years ago
4 0

Answer:

4

Step-by-step explanation:

4 x 3 = 12

4 x 10 = 40

4 goes into 12 3 times

4 goes into 40 10 times

your GCF (Greatest Common Factor) = 4

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4k-6=-2k-16-2
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3 years ago
How to simplify this?<br>(3a-b)+2a
taurus [48]

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It took my question down twice here's a real math question I guess
aliina [53]

Answer:

n-3

Interval notation: (-\infty, -10)\cup(-3,\infty)

Step-by-step explanation:

<u>First inequality:</u>

<u />n+8

Therefore, this inequality restricts:

n \in \mathrm{R};\: n

<u>Second inequality:</u>

< 8+n-3

Therefore, this inequality restricts:

n \in \mathrm{R};\: n>-3

Therefore, with both of these restrictions together, we have:

\fbox{$n \in \mathrm{R}; n-3$}\\\mathrm{or\:}\fbox{$n-3$}\\\mathrm{or\:}\fbox{$(-\infty, -10)\cup(-3,\infty)$}.

4 0
3 years ago
Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t
hram777 [196]

Answer:

a.\mu=15

b.\mu=7.8586 \ and  \ \mu=22.1414

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

P(X=k)=C_k^n.p^k.(1-p)^{n-k}

Let n be the number of trials,n=100

and p be the probability of success,p=15\%

The mean of a binomial distribution is the probability x sample size.

\mu=np=100\times0.15=15

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as:\sigma=\sqrt npq\\q=1-p

Therefore, \sigma=\sqrt(100\times0.015\times0.85)=3.5707

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c.  x=45 is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0
4 years ago
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