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Scrat [10]
2 years ago
8

Ppppppppppllllllllls I need helppppppp​

Mathematics
1 answer:
Phantasy [73]2 years ago
5 0

Answer:

ΔABC≅ΔDEC by AAS

Step-by-step explanation:

You can use the AAS method of congruency.

Since you already have <BAC and <EDC congruent to eachother, and sides BC and EC congruent to each other, you only need that one remaining angle in between. <ACB can be proven congruent to <DCE by the Vertical Angles Theorem, and that gives you the AAS you need to prove that these two triangles are congruent

Hope this helped.

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Use any order, any grouping to write an equivalent expression: 3(2x) + 4y(5) + (4 x 2 x z)
enyata [817]

Answer:

We conclude that the equivalent expression is:

3\left(2x\right)+4y\left(5\right)+\left(4\times \:2z\right)=6x+20y+8z

Step-by-step explanation:

Given the expression

3\left(2x\right)+4y\left(5\right)+\left(4\times \:2z\right)

Remove parentheses: (a) = a

=3\times \:2x+4\times \:5y+4\times \:2z

Multiply the numbers: 3 ×  2 = 6

=6x+4\times \:5y+4\times \:2z

Multiply the numbers: 4 × 5 = 20

=6x+20y+4\times \:2z

Multiply the numbers: 4 × 2 = 8

=6x+20y+8z

Thus, we conclude that the equivalent expression is:

3\left(2x\right)+4y\left(5\right)+\left(4\times \:2z\right)=6x+20y+8z

8 0
3 years ago
1-SCPS
Anton [14]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
g A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with
rodikova [14]

Answer:

a) 85.31%

b) 56 minutes

c) 17 days

d) 0.1721

Step-by-step explanation:

In order to make the calculations easier, let's <em>standardize the curve</em> by doing the change

Z=\frac{X-\mu}{\sigma}

where \mu is the average trip-time and \sigma the standard deviation and

P(X≤ t) is the probability that the trip takes less than t  minutes.

a)

Here we are looking for the value of P(X>20).

For X=20 we have Z = (20-24)/3.8 = -1.05

So, we want the area under the standard normal curve for Z > -1.05

that we can compute either using a table or a computer and we find this area equals 0.8531

So, he arrives late to work 85.31% of the times.

(See picture 1 attached)

b)

In this case we are looking for a value t of time such that

P(X≥ t) = 20% = 0.2

So, we are seeking a value Z such that the area under the normal curve to the left of Z equals 0.2

By using a table or a computer we find Z = 0.842

(See picture 2 attached)

By inserting this value in the equation on standardization  

0.842=\frac{X-24}{38}\Rightarrow X=38*0.842+24=55.996

So 20% of the longest trips takes 55.996 ≅ 56 minutes

c)

Since the average of days he arrives late is 85.31%, it is expected that in 20 work days he arrives late 85.31% of 20, which equals 17 days.

d)

Since the probability that he does not arrive late is 1-0.8531 = 0.1469 and the probability of arriving early is independent of the previous trip,

the probability of arriving early n days in a row is

0.1496*0.1496*...*0.1496 n times and

the probability of being early most than 10 trips in 20 days is

(0.1469)+(0.1469)^2+(0.1469)^3+...+(0.1469)^10=0.1721

3 0
3 years ago
50 is 10 times as much as???
sweet-ann [11.9K]
50 is 10 times as much as 500
5 0
3 years ago
Read 2 more answers
What is the solution of the system? Use any method.
Bad White [126]

Answer:

c

Step-by-step explanation:

-3,2

thats the answer your welcome

6 0
3 years ago
Read 2 more answers
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