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emmasim [6.3K]
3 years ago
15

I need measures for both angle 2 and 3 please help

Mathematics
2 answers:
olchik [2.2K]3 years ago
7 0

Answer:

44 and 22

Step-by-step explanation:

DedPeter [7]3 years ago
5 0

Answer: number 1 is 44 and number 2 is 22 hope it helps

Step-by-step explanation:

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3. There are four teams in the raquetball league on the island of Allendine. In a season, they play each other once. Three point
andrew-mc [135]

Answer and Step-by-step explanation:

Team: Points. Results fetching points

Arcadia. 3+1. 1 win & 1 draw

Brextone. 3+1. 1 win & 1 draw

Cremore 1+1. 2 draws

Dunross. 3+3. 2 wins

a) Based on the previous data, we can conclude that the number of matches drawn was 4

b) Dunross defeated cremore since cremore did not win any league game and dunross reported no ties

c) Yes, you can get a prediction of the matches with their respective results as follows

Brextone drew with Cremore

Dunross beat Arcadia

Arcadia beat Brextone

Arcadia drew with Cremore

Dunross beat Cremore

7 0
3 years ago
Read 2 more answers
What goes into 125 and 560 evenly
nikitadnepr [17]
The only numbers that go into 125 and 560 evenly are 1 and 5
8 0
3 years ago
(X-7)(3x+2) need answer!!
grin007 [14]

Answer: the answer is

=3x^{2}−19x−14  

Step-by-step explanation:  (x−7)(3x+2)

=(x+−7)(3x+2)

=(x)(3x)+(x)(2)+(−7)(3x)+(−7)(2)

=3x2+2x−21x−14

=3x2−19x−14

7 0
3 years ago
Can you help me with part b of problem 25
neonofarm [45]
C is the correct answer for the problem hope it helps :)
6 0
3 years ago
Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, th
LuckyWell [14K]

Answer:

Step-by-step explanation:

Given that Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that each of their offspring will develop the disease is approximately 0.25.

Let X be the no of children having this disease out of 4 occasions

Then X has two outcomes and each trial is independent of the other trial

Hence X is binomial with n =4, and p = 0.25

a) P(X=4) = 4C4 (0.25)^4 = 0.003906

b) P(X=1) = 4C1 (0.25)(0.75)^3\\= 0.4219

c) P(4th child/3 children did not) = 0.25 since independent.

4 0
3 years ago
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