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vodomira [7]
3 years ago
7

Use the distributive property to simplify the expressions

Mathematics
1 answer:
Bogdan [553]3 years ago
5 0
#1 is B. #2 is C. #3 is A
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I need help finding this answer to this inequality -10[9-2x]-x≤2x-5
Ad libitum [116K]

Answer:

x≤5

Step-by-step explanation:

-10(9-2x)-x≤2x-5

-90+20x-x≤2x-5

19x-2x≤90-5

17x≤85

x≤85/17

x≤5

8 0
3 years ago
The average number of Saturday night shootings is 5 a night. What’s the chance that there will be a need for exactly 3 shootings
pishuonlain [190]

Answer:

0.14

Step-by-step explanation:

Using the poisson probability relation :

P(x = x) = (λ^x * e^-λ) ÷ x!

From the question ; mean, λ = 5 ; x = 3

Hence,

P(x = 3) = (5^3 * e^-5) ÷ 3!

P(x = 3) = (125 * 0.0067379) / 6

P(x = 3) = 0.8422375 / 6

P(x = 3) = 0.140

5 0
2 years ago
Write the digit on the blank which occupies the place value​
nasty-shy [4]

Answer:

guganitckihkgucifufu

6 0
2 years ago
-5
Iteru [2.4K]

Answer:

a

Step-by-step explanation:

4 0
3 years ago
A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0
3 years ago
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