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irinina [24]
3 years ago
11

3. In the fig. AD = DC and AB = BC. Prove that ΔADB = ΔCDB​

Mathematics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

The two column proof is presented as follows;

The given parameters are;

\overline {AD} = \overline {DC} and \overline {AB} = \overline {BC}

Statement               {}                     Reason

\overline {AD} = \overline {DC} and \overline {AB} = \overline {BC}       {}       Given

\overline {BD}  ≅  \overline {BD}               {}                    Reflexive property

\overline {BD}  =  \overline {BD}               {}                     By the definition of congruency

ΔADB ≅ ΔCDB             {}                By Side-Side-Side (SSS) rule of congruency

ΔADB = ΔCDB              {}                 By the definition of congruency

If the three sides of one triangle are congruent to the corresponding three sides of another triangle, then both triangles are said to be congruent according to the SSS rule of congruency

Step-by-step explanation:

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Step-by-step explanation:

2 times 2=4    4 times 5=20   20 times 5/5 times 2=10 then add a zero

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Step-by-step explanation:

First you take away the original amount away from the remaining amount.

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Then that's your answer!

Her call lasted 22 minutes.

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Step-by-step explanation:


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3 years ago
The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10
Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

N\sim Pois(0.2)

P(N=r)=\frac{e^{-np}(np)^r}{r!}

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}

P(N=0)=\frac{e^{-0.2}}{1}

P(N=0)=e^{-0.2}

P(N=0)=0.8187

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute r\geq 2 in formula,

P(N\geq 2)=1-P(N

P(N\geq 2)=1-[P(N=0)+P(N=1)]

P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]

P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]

P(N\geq 2)=1-[0.8187+0.1637]

P(N\geq 2)=1-0.9825

P(N\geq 2)=0.0175

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0
3 years ago
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