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tekilochka [14]
3 years ago
5

Multiple choice: State any restrictions on the variables.

Mathematics
2 answers:
Semenov [28]3 years ago
7 0
I believe your answer is C. y ≠ -1. This is because if y was -1, then the denominator of the fraction would be -1 + 1 = 0, and you can't divide any value by 0.

I hope this helps!
a_sh-v [17]3 years ago
5 0

Answer:

I believe your answer is C. y ≠ -1. This is because if y was -1, then the denominator of the fraction would be -1 + 1 = 0, and you can't divide any value by 0.

I hope this helps!

Step-by-step explanation:

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Divide 3 1/3 by 2/3 bbbbbbbb
iris [78.8K]

Answer:

jgjhi

Step-by-step explanation:

vbnhjkui

5 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
Map pours 3 2/3 cup of orange juice into measuring cup from a large container then he poured 1 1/4 cups back into the container
Alecsey [184]

Answer:

2 5/12 cups

Step-by-step explanation:

Map pours 3 2/3 cup of orange juice into measuring cup from a large container then he poured 1 1/4 cups back into the container

The amount of juice remaining in the measuring cup is calculated as:

3 2/3 - 1 1/4

= 3 - 1 + ( 2/3 - 1/4)

LCD = Lowest Common Denominator is 12

= 2 + (4 × 2 - 3 × 1/12)

= 2 + (8 - 3/12)

= 2 + 5/12

= 2 5/12 cups

Hence, the amount of juice remaining in the measuring cup = 2 5/12 cups of juice

7 0
3 years ago
Please help.
serious [3.7K]

The number of throws that the player make X follows binomial distribution.

The probability that the player makes x throws out of eleven throws is

P(X=x)=C(x,11)0.5^x(0.5)^{11-x}=C(x,11)0.5^{11}

The probability that the player makes at most nine out of eleven free throws

is

\sum_{x=0}^9C(x,11)0.5^{11}=1-\sum_{x=10}^{11}C(x,11)0.5^{11}\\ =1-C(10,11)0.5^{11}-C(11,11)0.5^{11}=1-\frac{12}{2^{11}} \\ =1-\frac{3}{512} =\frac{509}{512}

Correct choice is (B).

8 0
3 years ago
What is 500,000,115 written in standard form
neonofarm [45]

<u>Answer:</u>

<u>The answer to five hundred million, one hundred fifteen in standard form is actually 500,000,115 because five hundred million in standard form is 500,000,000 and one hundred fifteen in standard form is 115.</u>

3 0
3 years ago
Read 2 more answers
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