To solve this problem, let us first assign some
variables. Let us say that:
x = pigs
y = chickens
z = ducks
From the problem statement, we can formulate the
following equations:
1. y + z = 30 --->
only chicken and ducks have feathers
2. 4 x + 2 y + 2 z = 120 --->
pig has 4 feet, while chicken and duck has 2 each
3. 2 x + 2 y + 2 z = 90 --->
each animal has 2 eyes only
Rewriting equation 1 in terms of y:
y = 30 – z
Plugging this in equation 2:
4 x + 2 (30 – z) + 2 z = 120
4 x + 60 – 2z + 2z = 120
4 x = 120 – 60
4 x = 60
x = 15
From the given choices, only one choice has 15 pigs. Therefore
the answers are:
She has 15 pigs, 12 chickens, and 18 ducks.
Answer: A 1 to 4
Step-by-step explanation:
Answer:
8 feet
Step-by-step explanation:
Given that 96 sq. feet of tile is bought to retile the bathroom.
It means area of bathroom is 9
6 sq. feet
Width of bathroom be x
Given that the length of the bathroom is one and a half times the width
Length of bathroom in term sod width = 1 1/2* x = 3/2 x
we know area of rectangle is given by length * width
therefore area of bathroom =( x* 3/2 x) = 3/2 x^2
Also area of bathroom is 86 sq. feet, therefore
![3/2 x^2 = 96\\=> 3 x^2 = 96*2=192\\=> x^2 = 192/3 = 64\\=>\sqrt{x^2} = \sqrt{64} \\=> x = 8](https://tex.z-dn.net/?f=3%2F2%20x%5E2%20%3D%2096%5C%5C%3D%3E%203%20x%5E2%20%3D%2096%2A2%3D192%5C%5C%3D%3E%20%20x%5E2%20%3D%20192%2F3%20%3D%2064%5C%5C%3D%3E%5Csqrt%7Bx%5E2%7D%20%20%3D%20%5Csqrt%7B64%7D%20%5C%5C%3D%3E%20x%20%3D%208)
Thus, width of bathroom is 8 feet.
Answer:
0.3907
Step-by-step explanation:
We are given that 36% of adults questioned reported that their health was excellent.
Probability of good health = 0.36
Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.
Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.
i.e. ![P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)](https://tex.z-dn.net/?f=P%28x%5Cleq%203%29%3DP%28x%3D1%29%2B%7BP%28x%3D2%29%2BP%28x%3D3%29)
Formula :![P(x=r)=^nC_r p^r q ^ {n-r}](https://tex.z-dn.net/?f=P%28x%3Dr%29%3D%5EnC_r%20p%5Er%20q%20%5E%20%7Bn-r%7D)
p is the probability of success i.e. p = 0.36
q = probability of failure = 1- 0.36 = 0.64
n = 11
So, ![P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)](https://tex.z-dn.net/?f=P%28x%5Cleq%203%29%3DP%28x%3D1%29%2B%7BP%28x%3D2%29%2BP%28x%3D3%29)
![P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}](https://tex.z-dn.net/?f=P%28x%5Cleq%203%29%3D%5E%7B11%7DC_1%20%280.36%29%5E1%20%280.64%29%5E%7B11-1%7D%2B%5E%7B11%7DC_2%20%280.36%29%5E2%20%280.64%29%5E%7B11-2%7D%2B%5E%7B11%7DC_3%20%280.36%29%5E3%20%280.64%29%5E%7B11-3%7D)
![P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}](https://tex.z-dn.net/?f=P%28x%5Cleq%203%29%3D%5Cfrac%7B11%21%7D%7B1%21%2811-1%29%21%7D%20%280.36%29%5E1%20%280.64%29%5E%7B11-1%7D%2B%5Cfrac%7B11%21%7D%7B2%21%2811-2%29%21%7D%20%20%280.36%29%5E2%20%280.64%29%5E%7B11-2%7D%2B%5Cfrac%7B11%21%7D%7B3%21%2811-3%29%21%7D%20%280.36%29%5E3%20%280.64%29%5E%7B11-3%7D)
![P(x\leq 3)=0.390748](https://tex.z-dn.net/?f=P%28x%5Cleq%203%29%3D0.390748)
Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907
Answer:
we don't know your questions