Greatest common factor of 45 51 63

1 answer:

8 0

$45|3\\15|3\\.\ 5|5\\.\ 1|\\45=\fbox3\times3\times5\\\\51|3\\17|17\\.\ 1|\\51=\fbox3\times17\\\\63|3\\21|3\\.\ 7|7\\.\ 1|\\63=\fbox3\times3\times7\\\\GCF(45;\ 51;\ 63)=\fbox3$

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A list of numbers is shown; 1 2 3 4 5 6 7 8 9 10

ahrayia [7]

Answer:

5.5

Step-by-step explanation:

We have a list of 10 numbers (n = 10) ordered from lower to higher.

1 2 3 4 5 6 7 8 9 10

The median is the value separating the higher half from the lower half of a data sample. Since n is even, there is no middle value and the median is defined as the mean between the two middle values, in this case, 5 and 6. Then,

median = 5 + 6 / 2 = 5.5

5 0

3 years ago

5. A company sells small, colored binder clips in packages of 20 and offers a money-back guarantee if two or more of the clips a

SOVA2 [1]

Answer:

a) Binomial.

b) n=20, p=0.01, k≥2

The probability hat a package sold will be refunded is P=0.0169.

Step-by-step explanation:

a) We know that

the defective probability is constant and independent.
the sample size is bigger than one subject.

The most appropiate distribution to represent this random variable is the binomial.

b) The parameters are:

Sample size (amount of clips in the package): n=20
Probability of defective clips: p=0.01.
number of defective clips that trigger the money-back guarantee: k≥2

The probability of the package being refunded can be calculated as:

$P(x\geq2)=1-(P(x=0)+P(x=1))\\\\\\P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}\\\\\\P(x=0) = \dbinom{20}{0} p^{0}q^{20}=1*1*0.8179=0.8179\\\\\\P(x=1) = \dbinom{20}{1} p^{1}q^{19}=20*0.01*0.8262=0.1652\\\\\\P(x\geq2)=1-(0.8179+0.1652)=1-0.9831=0.0169$