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4 years ago

Please help with 1 and 3! Please tell me the answer! I'm crying in confusion! :(

Mathematics

1 answer:

Aliun [14]4 years ago

6 0

Don't cry please :(

1:
To find the Average (mean). Add up all the numbers together. Then divide by how many numbers there are.

So use the chart and do what I said :)

2:

To find the mean: Add up all the numbers together. Then divide by how many numbers there are.

To find the median: Put all the numbers in least to greatest. Then if there is a odd amount of numbers. It would be the number in the middle.

To find the mode: The mode is the number that occurs the most.

So use the chart and do what I said :)

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I really need help i’m going to fail so answer if you can ! <3

Volgvan

Answer:

110

Step-by-step explanation:

you place 30 in the place of X, and it becomes 4×30-10. in PEMDAS, multiplication goes first and 4×30=120. 120-10=110

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3 years ago

Erin is a dog groomer. She charges a $6.00 base fee plus an extra $0.50 per pound. If x represents the weight of the dog in poun

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3 years ago

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Find two consecutive odd integers whose sum is 96

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3 years ago

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A 2ft-by-2ft square is divided into smaller squares and portions are shaded. What is the area of the shaded portion?

TiliK225 [7]

Answer:

1.5 ft^2

Step-by-step explanation:

Since the length is divided into 4 parts,

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3 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 120 customer orders to f

Aliun [14]

Answer:

a. P(X = 0) = 0.02586

b. $\mathbf{P(X \leq 2 ) =0.2879}$

c. $\mathbf{P(X \leq 5 ) =0.8387}$

Step-by-step explanation:

From the given information:

a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}$

$P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}$

P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰

P(X = 0) = 0.02586

b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]$

$P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]$ $P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]$

$P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]$

$\mathbf{P(X \leq 2 ) =0.2879}$

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?

$P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]$

$P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]$ $P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]$

$\mathbf{P(X \leq 5 ) =0.8387}$

5 0

3 years ago