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3 years ago

Question 2 of 14 Which part of this graph shows a nonlinear relationship?

Mathematics

1 answer:

victus00 [196]3 years ago

6 0

Answer is 2!
All of the other parts show straight segments

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The price of a hat is $27.When it goes on sale, it wll be 1/3.How much would you save if you were to buy ti on sale

Bond [772]

Answer:

You would save $18

Step-by-step explanation:

(27/3)*2 = $18

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3 years ago

What’s 43% out of 200?

alukav5142 [94]

Answer:

200x43%=86

Step-by-step explanation:

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4 years ago

Pls do some research but can I bring a bearded dragon on a 12 hour car trip without any heath issues happening?

creativ13 [48]

Answer:

after doing a little bit of research. NO your bearded dragon will not be affected as long as you feed it and keep it on a level surface so the enclosure it's in does not tip at all other than that it will be perfectly fine.

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3 years ago

What is the sign of the product (3)(−25)(7)(−24)?

german

Answer:

- * - = +

then:

(3)(−25)(7)(−24)

= (3*-25)(7*-24)

= (-75)(-168)

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The sign is:

6 0

4 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