Wish I knew need the same answer
Answer:
Required Probability = 0.97062
Step-by-step explanation:
We are given that the weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4016 grams and a standard deviation of 532 grams.
Let X = weight of the newborn baby, so X ~ N(
)
The standard normal z distribution is given by;
Z =
~ N(0,1)
Now, probability that the weight will be less than 5026 grams = P(X < 5026)
P(X < 5026) = P(
<
) = P(Z < 1.89) = 0.97062
Therefore, the probability that the weight will be less than 5026 grams is 0.97062 .
Answer:
no Solution
Step-by-step explanation:
-12x-12y=4\\ 3x+3y=0
12x-12y=4
add 12y to both sides
12x-12y+12y=4+12y
divid both sides by -12
\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}
simplfy
x=-\frac{1+3y}{3}
\mathrm{Substitute\:}x=-\frac{1+3y}{3}
\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}
\begin{bmatrix}-1=0\end{bmatrix}
34 is larger than 3, so 34.653
Answer 4
4multiplyed by 4 is 16