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aleksklad [387]
3 years ago
15

PLEASE HELP

Mathematics
2 answers:
sveta [45]3 years ago
7 0
The mid point of srgment isT (3,2)
X= (6-2) + 2
6-2=4+2=6

Y=(4-6)+6
4-6=-2+6=4
The common denominator for 6 and 4 is 2. 2 into 6 is 3 and 2 into 4 is 2 which = T (3,2)
m_a_m_a [10]3 years ago
3 0
The midpoint T equals to 3 and 4
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A numerical data set has 9 values. How many of these values does the data
Paraphin [41]
I think it’s D) 9 because when you find the mean of a set of values you are taking into account and adding all of the numbers, and then dividing them by how many numbers there is, so the mean would be summarizing all of them, hope this helps
5 0
3 years ago
Quadrilateral MNQP has vertices as shown. If the figure is translated 4 units left and 3 units down, what are the coordinates of
PilotLPTM [1.2K]
<h3>Answer:    (-3, 0)</h3>

Explanation:

Point N is at (1,3)

We apply the rule (x,y) \to (x-4,y-3) which will translate the point 4 units to the left and 3 units down.

The old x coordinate x = 1 becomes x-4 = 1-4 = -3

The old y coordinate y = 3 becomes y-3 = 3-3 = 0

The point N(1,3) moves to N ' (-3, 0)

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2 years ago
A pattern follows the rule "Starting with two, every consecutive line has a number one more than the previous line."
mars1129 [50]
There will be eight marbles on the 9th line
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3 years ago
Find the balance in the account after the given period.
Likurg_2 [28]

Answer:

2240

Step-by-step explanation:

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6 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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