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damaskus [11]
2 years ago
5

Someone please help :) Both pls :)) also ill try Mark u brainliest <33 please answer both :3

Mathematics
2 answers:
Butoxors [25]2 years ago
6 0

Answer:

5.56÷4=14

6. (2×27)-3×5-8=

1) 2×27=54

2)3×5=15

3)54-15=39

4)39-8=31

Step-by-step explanation:

hope it helped

Svetradugi [14.3K]2 years ago
5 0

Step-by-step explanation:

5. 56÷4 = 14

6. (2 × 27) - 3 × 5 - 8

54 - 15 - 8

39 - 8 = 31 ans.

Hope this answer helps you!

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What you must do in this case is to find the roots of the polynomial.
 We have then:
 x ^ 2 + 20x + 100 = 36
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 x ^ 2 + 20x + 64 = 0
 The values of the roots are:
 x1 = -4
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 Remember that the values of the roots are what make the polinome zero
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3 0
3 years ago
Alberta is going to take a 2-hour-26-minute flight from Chicago, Illinois, to
BigorU [14]

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Step-by-step explanation:

8 0
3 years ago
A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o
Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(XE(X_{k}) = n_{k} * p_{k}

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = \frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= \frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= \frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\ = 0.000001

c.

P(X₁≤2) = \frac{22!}{0!} * 0.08^{0} * (0.92)^{22} + \frac{22!}{1!} * 0.08^{1} * (0.92)^{21} + \frac{22!}{2!} * 0.08^{2} * (0.92)^{20} = 0.7442

I hope you have a SUPER day!

8 0
3 years ago
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