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Zielflug [23.3K]
3 years ago
9

Naruto quiz first to get right will win

Mathematics
2 answers:
bonufazy [111]3 years ago
6 0

Answer:

1 lamen

2 onigiri

3 hashirama

4 uchiha

5 konoha

6 probably rasengan

7 kushina uzumaki and minato namikaze

8 uchiha itachi

9 7

10 boruto and himawari

11 10/10 he was born at the day that nine tails attacked konoha

12 13 in classic and 16 in shippuden

Step-by-step explanation:

Over [174]3 years ago
4 0

Answer:

Step-by-step explanation:

ramen

tomatoes

Hashirima

uchiha

konoha

Rasenshuriken

Minato and Kushina

itachi

naruto ep 6

boruto himiwari

oct 10  

15 16 shippueden

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If the least value of n is 3, which inequality best shows all the possible values of n?
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Answer:

1) B. n ≥ 3

2)A. 5x – 9

Step-by-step explanation:

1)

If the least value of n is 3 then all the values of n must be greater or equal to 3

Thus the correct option is B :n ≥ 3

2)

Given:

Milo's age is 9 years less than 5 times Pearl's age

Pearl's age is x years

5 times Pearl's age= 5x

then milo's age= 5x-9

so the correct option is A. 5x – 9 !

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What is 45.3% expressed as a decimal?
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3 years ago
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7 0
3 years ago
A couple plans to have children until they get a​ girl, but they agree they will not have more than three​ children, even if all
barxatty [35]

Answer:

a)Let X be number of children

X=1,2,3

P(X=1)= 0.47

P(X=2) = 0.2491

P(X=3) = 0.942923

b) 3.7970

c) 1.6232

Step-by-step explanation:

The complete question is:

A couple plans to have children until they get a​ girl, but they agree they will not have more than three​ children, even if all are boys. Assume that the probability of having a girl is 47.00​%. ​

a) Create a probability model for the number of children​ they'll have.

X=1,2,3

​P(X=1)=??

P(X=2)= ??

P(X=30=???

​(Round to four decimal places as​ needed

​b) Find the expected number of children.

E(X)= ???

​c) Find the expected number of boys​ they'll have.

Expected number of boys= ???

Solution:

Probability of a girl= 0.47

Probability of a boy= 0.53

a) P(X=1)= 0.47

P(X=2) = 0.47× 0.53= 0.2491

P(X=3)= 0.47× 0.53× 0.53 + 0.53× 0.53× 0.53

             = 0.942923

b) E(number of children)= 1× P(X=1) + 2 ×P(X=2) + 3 × (PX=3)

                                       = 3.796969

c) Y: number of boys

P(Y=1)= 0.53×0.47= 0.2491

P(Y=2) = 0.53×0.53×0.47=0.46375

P(Y=3)= 0.53× 0.53× 0.53= 0.148875

E(Y)= P(Y=1)×1 + P(Y=2)×2 + P(Y=3)×3

       = 0.148875×3 +0.46375×2+0.2491 ×1

      = 1.6232

5 0
3 years ago
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