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Fittoniya [83]
2 years ago
7

Expand this question ...

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

Step-by-step explanation:

hope this helps

Leviafan [203]2 years ago
3 0

Answer:

(4p^2-28pq)

Step-by-step explanation:

You have to times 4p by p meaning you get 4p squared as p multiplied by p gets you a squared. for the second half, 4x7= 28, and p multiplied by q is pq so it would be 28pq

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What is 11183333.3 in scientific notation
umka2103 [35]
11183333.3/10^7 = 1.11x10^-7.
8 0
3 years ago
The graph represents the data cost for monthly internet service for a cell phone. Which function, C(x), respresent the monthly c
Tanya [424]

Answer:

Correct option: first one

Step-by-step explanation:

The graph has 3 different parts:

Cost = 15 when the Gigabytes used is lesser than or equal 2,

Cost from 20 to 40 when the Gigabytes used is greater than 2 and lesser than or equal 6.

An increase of 4 gigabytes caused an increase of 20 in the cost, so the slope is 20/4 = 5, and the y-intercept is:

20 = 5*2 + b -> b = 10

Cost = 50 when the Gigabytes used is greater than 6.

So the correct option is the first one

3 0
3 years ago
ANSWER QUICKLY! I GIVE BRAINLIEST!!
julsineya [31]

Answer:

(10,27)

Step-by-step explanation:

Hope this helps!!! :)

8 0
3 years ago
Factorize<br> ху<br> + 3x +3y<br> + 18
noname [10]

Answer:

(x+6)(y+3)

Step-by-step explanation:

  1. xy + 3x + 6y + 18
  2. x(y + 3) + 6(y+3)
  3. <u>(</u><u>x</u><u>+</u><u>6</u><u>)</u><u>(</u><u>y</u><u>+</u><u>3</u><u>)</u>
3 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
3 years ago
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