Answer:
a) 90.695 lb
b) 85.305 lb
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
(a) The 65th percentile
X when Z has a pvalue of 0.65. So X when Z = 0.385.
(b) The 35th percentile
X when Z has a pvalue of 0.35. So X when Z = -0.385.
Answer:
![-7ab\sqrt[3]{3ab^2}](https://tex.z-dn.net/?f=-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D)
Step-by-step explanation:
Remove perfect cubes from under the radical and combine like terms.
![2ab\sqrt[3]{192ab^2}-5\sqrt[3]{81a^4b^5}=2ab\sqrt[3]{4^3\cdot 3ab^2}-5\sqrt[3]{(3ab)^3\cdot 3ab^2}\\\\=(8ab -15ab)\sqrt[3]{3ab^2}=\boxed{-7ab\sqrt[3]{3ab^2} }](https://tex.z-dn.net/?f=2ab%5Csqrt%5B3%5D%7B192ab%5E2%7D-5%5Csqrt%5B3%5D%7B81a%5E4b%5E5%7D%3D2ab%5Csqrt%5B3%5D%7B4%5E3%5Ccdot%203ab%5E2%7D-5%5Csqrt%5B3%5D%7B%283ab%29%5E3%5Ccdot%203ab%5E2%7D%5C%5C%5C%5C%3D%288ab%20-15ab%29%5Csqrt%5B3%5D%7B3ab%5E2%7D%3D%5Cboxed%7B-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D%20%7D)
Answer:
80%
Step-by-step explanation:
What I did was just play with the numbers. But first, you'd multiply 6.00 time 3 because it's for the three people. Then just multiply 18 with .8 and bingo! The answer you get is 14.40, so the discount would be 80%
Hope this helps!
Answer:
35
Step-by-step explanation:
Here we see 5 black keys for every 7 white keys.
So the ratio is 5:7
If we need 49 white keys, find the amount we scale the original ratio by:
49/7 = 7
So we are scaling by a factor of 7.
The number of black keys would be 5 * the scale of 7. = 35
So there should be 35 black keys.
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
