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3 years ago

Im confused on this question on how to solve it

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

Answer:

9/4

Step-by-step explanation:

see image

2/3 is 2 × 1/3.

We can separate the exponent into 1/3 power raised to the 2nd power. 1/3 power is the same as cube root.

27 and 8 are both perfect cubes. 3^3 and 2^3.

Thats how you go from 27/8 to 3/2.

Then square it.

see image.

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The ratio of girls to boys is 2 to 5 if they're 80 girls how many boys are there?

I am Lyosha [343]

2/5=80/?
2÷80=40
5×40=200
there is 200 boys
80:200

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3 years ago

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4 years ago

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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

Which statement is true about angles 3 and 6

vlada-n [284]

Answer:

They are vertical angles.

Step-by-step explanation:

8 0

4 years ago

Im confused on this question on how to solve it​

Im confused on this question on how to solve it