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Morgarella [4.7K]
3 years ago
6

Find the area of the rhombus. d1 = 12 m; d2 = 20 m

Mathematics
2 answers:
kati45 [8]3 years ago
5 0
Check the picture below.

so, assuming d1 and d2 are the diagonals of it, bearing in mind that the diagonals of a rhombus bisect each, cut in equal halves, then we can get the area of one of those 4 congruent triangles in the rhombus, notice each triangle has a base of 6 and a height of 10.

\bf \stackrel{\textit{area of the 4 triangles}}{4\left[ \cfrac{1}{2}(6)(10) \right]}

lesantik [10]3 years ago
5 0

Answer: 120\ m^2

Step-by-step explanation:

We know that the area of a rhombus is given by :-

\text{Area}=\dfrac{1}{2}\times d_1\times d_2, where d_1\ and \ d_2 are the diagonals of the rhombus.

Given:

d_1=12\ m\\\\d_2=20\ m

Then, the area of given  rhombus will be :-

\text{Area }=\dfrac{1}{2}\times12\times20\\\\\Rightarrow\text{area}=120\ m^2

Hence, the area of rhombus = 120\ m^2

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Answer:

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Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432

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a) We can express the probability that at most 12 ties are too tight as:

P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1

The probability that at most 12 ties are too tight is P=1.

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