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(A) Since the sum of all probabilities in a probability distribution is $1$,we have:$\sum_{x=0}^7 P(X=x) = 1$$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^

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$\frac{19}{100}$

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(A) Since the sum of all probabilities in a probability distribution is $1$,we have:$\sum_{x=0}^7 P(X=x) = 1$$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$$10k^2 + 9k - 1 = 0$$(10k - 1)(k + 1) = 0$Since $k \geq 0$,we have $k = \frac{1}{10}$.Now,we need to find $P(X \geq 6)$:$P(X \geq 6) = P(X=6) + P(X=7)$$P(X \geq 6) = 2k^2 + (7k^2 + k) = 9k^2 + k$Substituting $k = \frac{1}{10}$:$P(X \geq 6) = 9\left(\frac{1}{10}\right)^2 + \frac{1}{10} = \frac{9}{100} + \frac{10}{100} = \frac{19}{100}$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

If a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

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If a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:$X$$0$$1$$2$$3$$4$$5$$6$$7$$P(X)$$0$$k$$2k$$2k$$3k$$k^2$$2k^2$$7k^2 + k$