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Maurinko [17]
3 years ago
5

In this truss bridge, AAOB ABDC. AO=5 meters, BD=10 meters, AB=5 meters. What is the length of BC? A. 5 m B. 8 m C. 10 m D. 15 m

Mathematics
1 answer:
Olenka [21]3 years ago
5 0

Consider \Delta AOB\cong \Delta BDC.

Given:

\Delta AOB\cong \Delta BDC, AO=5\ m,BD=10\ m, AB=5\ m.

To find:

The length of BC.

Solution:

We have,

\Delta AOB\cong \Delta BDC

We know that the corresponding parts of congruent triangles are congruent (CPCTC).

AB=BC                   (CPCTC)

5\ m=BC

The length of BC is 5 m. Therefore, the correct option is A.

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7 0
2 years ago
Tommy has 6 more than 3 times the amount of money in his bank account than Judy. Judy has $7500 more than a number in her accoun
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Answer:

Amount of money Judy has in her account is \$(7500+n).

Amount of money Tommy has in his account is  \$(22506+3n)

Step-by-step explanation:

To find : Amount of money in Judy account and amount of money in Tommy account:

Solution:

Given:

Judy has $7500 more than a number in her account.

Let the number be 'n'.

So we can say that;

Amount of money in Judy account is equal to 7500 plus number.

framing in equation form we get;

Amount of money in Judy account = \$(7500+n)

Hence Amount of money Judy has in her account is \$(7500+n).

Now Given:

Tommy has 6 more than 3 times the amount of money in his bank account than Judy.

So we can say that;

Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.

framing in equation form we get;

Amount of money in Tommy account = 3(7500+n)+6 =22500+3n+6=\$(22506+3n)

Hence Amount of money Tommy has in his account is  \$(22506+3n).

8 0
3 years ago
A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean(\mu) = 500 and a standard deviation(\sigma) =  100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = \dfrac{Y-\mu}{\sigma}

Z-score = \dfrac{625-500}{100}

⇒ Z-score = \dfrac{125}{100}

⇒Z-score =1.25

Therefore , the required z-score = 1.25

6 0
3 years ago
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