All categories

2 years ago

Determine if the following triangles are congruent by SSS or SAS.

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

6 0

Answer:

They are congruent by SAS (Side Angle SIde) test or congruence criterion

Since,

In triangle ABD and triangle BCD

AB = BC (Given)

BD = DB(Common Side)

Angle ABD = Angle CBD

Therefore triangle ABD and triangle BCD by SAS (Side Angle SIde) test or congruence criterion

You might be interested in

Calculate the number of atoms, please help in struggling

Veseljchak [2.6K]

Answer:

I think there are 13 atoms present.

Step-by-step explanation:

Please mark brainliest

8 0

2 years ago

The average of a list of 4 numbers is 90.0. A new list of 4 numbers has the same first 3 numbers as the original list, but the f

Thepotemich [5.8K]

Answer:

the average of this new list of numbers is 94

Step-by-step explanation:

Hello!

To answer this question we will assign a letter to each number for the first list and the second list of numbers, remembering that the last number of the first list is 80 and the last number of the second list is 96

for the first list

$\frac{a+b+c+80}{4} =90$

for the new list

$\frac{a+b+c+96}{4} =X$

To solve this problem consider the following

1.X is the average value of the second list

2. We will assign a Y value to the sum of the numbers a, b, c.

a + b + c = Y to create two new equations

for the first list

$\frac{y+80}{4} =90$

solving for Y

Y=(90)(4)-80=280

Y=280=a+b+c

for the second list

$\frac{y+96}{4} =X\\$

$\frac{280+96}{4} =X\\x=94$

the average of this new list of numbers is 94

4 0

2 years ago

Read 2 more answers

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$