Answer: X=40
Step-by-step explanation:
The question is incorrect
the correct question is
A local citizen wants to fence a rectangular community garden. The length of the garden should be at least 110 ft,and the distance around should be no more than 380 ft. Write a system of inequality that model the possible dimensions of he garden. Graph the system to show all possible solutionslet
x---------------> t<span>he length of the garden
</span>y---------------> the wide of the garden
we know that
x>=110
2x+2y <=380---------------> x+y <= 190
Part A) <span>Write a system of inequality that model the possible dimensions of he garden
</span>
the answer part A) is
x>=110
x+y <= 190
Part B) <span>Graph the system to show all possible solutions
using a graph tool
see the attached figure
the solution is the triangle show in the figure
</span><span>the possible solutions of y (wide) would be between 0 and 80 ft
</span>the possible solutions of x (length) would be between 110 ft and 190 ft
Answer:
9 total
2/9 - 5/9 -2/9 so 3 i guess...
Step-by-step explanation:
The probability that the first two electric toothbrushes sold are defective is 0.016.
The probability of an event, say E occurring is:
![P(E)=\frac{n(E)}{N}](https://tex.z-dn.net/?f=P%28E%29%3D%5Cfrac%7Bn%28E%29%7D%7BN%7D)
Here,
n (E) = favorable outcomes
N = total number of outcomes
Let X = the number of defective electric toothbrushes sold.
The number of electric toothbrushes that were delivered to a store is n = 20.
The number of defective electric toothbrushes is x = 3.
The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:
![(\left {{20} \atop {2}} \right. )](https://tex.z-dn.net/?f=%28%5Cleft%20%7B%7B20%7D%20%5Catop%20%7B2%7D%7D%20%5Cright.%20%29)
![=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190](https://tex.z-dn.net/?f=%3D%5Cfrac%7B20%21%7D%7B2%21%2820-2%29%21%7D%20%3D%5Cfrac%7B20%21%7D%7B2%2118%21%7D%20%3D%5Cfrac%7B20%2A19%2A18%21%7D%7B2%21%2A18%21%7D%20%3D%20190)
The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:
![(\left {{3} \atop {2}} \right. )](https://tex.z-dn.net/?f=%28%5Cleft%20%7B%7B3%7D%20%5Catop%20%7B2%7D%7D%20%5Cright.%20%29)
![=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =3](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3%21%7D%7B2%21%283-2%29%21%7D%20%3D%5Cfrac%7B3%21%7D%7B2%211%21%7D%20%3D%5Cfrac%7B3%2A2%21%7D%7B1%212%21%7D%20%3D3)
Compute the probability that the first two electric toothbrushes sold are defective as follows:
P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes
![\frac{3}{190}\\ =0.01579\\=0.016](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B190%7D%5C%5C%20%3D0.01579%5C%5C%3D0.016)
Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.
Learn more about probability here brainly.com/question/27474070
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