Answer:
there should be a pic or a diagram
Step-by-step explanation:
Answer:
The degrees of freedom associated with the critical value is 25.
Step-by-step explanation:
The number of values in the final calculation of a statistic that are free to vary is referred to as the degrees of freedom. That is, it is the number of independent ways by which a dynamic system can move, without disrupting any constraint imposed on it.
The degrees of freedom for the t-distribution is obtained by substituting the values of n1 and n2 in the degrees of freedom formula.
Degrees of freedom, df = n1+n2−2
= 15+12−2=27−2=25
Therefore, the degrees of freedom associated with the critical value is 25.
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
Answer:
The coach should start recruiting players with weight 269.55 pounds or more.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 225 pounds
Standard Deviation, σ = 43 pounds
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.15
Calculation the value from standard normal z table, we have,
Thus, the coach should start recruiting players with weight 269.55 pounds or more.
