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Mekhanik [1.2K]
2 years ago
15

HELP ME PLEASE 25 points

Mathematics
2 answers:
Aleks04 [339]2 years ago
8 0
Hello,

1) 2<2&2/3=2.6666...<3 Answer C (2 and 3)

2)-4<-3.75<3  Answer B ( A and B)

3)abs(B)=3/8 Answer B

4)an absolue value is always positive or null answer FALSE

5)|5-9|=|-4|=4 answer C

6)-|5|+|-5|=-5+5=0 answer C

7)I have understand AC as distance between A and C =2

AC-EF=2-1=1 Answer C

8) CF=2-(-2)=4 Answer D

9) NE=5-1=4 answer C

10)
x-7=6
==>x=1
Answer A
ira [324]2 years ago
7 0
Pg1
1. f,g
2. a,b
3. 3/8
pg2
1 false
2 -4
3 -10
4 -5
pg 3
1 0
2 6
3 13
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4 0
3 years ago
. The sandwich shop offers 8 different sandwiches. Jamey likes them all equally. He picks one randomly each day for lunch. Durin
Margarita [4]

Answer:

Step-by-step explanation:

From the given information:

A sandwich shop offers eight types of sandwiches, and Jamey likes all of them equally.

The probability that Jamey picks any one of them is 1/8

Suppose

X represents the number of times he chooses salami

Y represents the number of times he chooses falafel

Z represents the number of times he chooses veggie

Then  X+Y+Z ≤ 5 and;

5-X-Y-Z represents the no. of time he chooses the remaining

8 - 3 = 5 sandwiches

However, the objective is to determine the P[X=x,Y=y,Z=z] such that 0≤x,y,z≤5

So, since he chooses x no. of salami sandwiches with probability (1/8)x

and y number of falafel with probability (1/8)y

and for z (1/8)z

Therefore, the remaining sandwiches are chosen with probability \dfrac{5}{8} (5-x-y-z)

So. these x days, y days and z days can be arranged within five days in

= \dfrac{5!}{x!y!z!(5-x-y-z)!}

Thus;

P[X=x,Y=y,Z=z]=  \dfrac{5!}{x!y!z!(5-x-y-z)}  \times \dfrac{1}{8}x*\dfrac{1}{8}y* \dfrac{1}{8}z* \dfrac{5}{8}(5-x-y-z)

since 0 ≤ x, y, z ≤ 5 and x + y + z ≤ 5.

The distribution is said to be Multinomial distribution.

5 0
3 years ago
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