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icang [17]
3 years ago
8

1.278.05 - 43.78 help me​

Mathematics
1 answer:
bixtya [17]3 years ago
5 0

Answer:

its -42.50195

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An oil company is considering 2 sites on which to drill, described as follows:
Olegator [25]
A)site A has a larger probablility of finding oil
B) the difference between amount of profitmade is 30 million.

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3 years ago
2.Solve x squared - 4x by Factoring
777dan777 [17]

Answer:

Step-by-step explanation:

hello:

x²-4x=0  means : x(x-4)=0

x=0 or x-4=0

conclusion : x=0 and x= 4   ...answer : B

8 0
3 years ago
Given f (x) = -4x + 4, solve for x when f (x) = 0
Kazeer [188]

Answer:

x = 1

Step-by-step explanation:

Given:

  • f(x) = -4x + 4
  • f(x) = 0

We are asked to solve for x when the function is equal to zero.

<u>We should have</u>: 0 = -4x + 4

<u>Solve</u>

1. Subtract 4 from both sides

0 - 4 = -4x + 4 - 4

-4 = -4x

2. Divide both sides by -4

-4 ÷ -4 = -4x ÷ 4

1 = x

6 0
2 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
2 years ago
What's the problem with this chemical equation: H2 + 02 --&gt; H20?
jarptica [38.1K]

Answer:

there is an additional oxygen in the product

3 0
3 years ago
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