FobelloE
2021-03-06
Answered

Kareem has 216 role-playing game cards. His goal is to collect all 15 sets of cards. There are 72 cards in a set. How many more cards does Kareem need to reach his goal?

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firmablogF

Answered 2021-03-07
Author has **92** answers

Since there are 72 cards in a set, then Kareem needs to collect a total of $15\times 72=1080$ cards. Since Kareem currently has 216 role-playing game cards, subtract 1080 by 216 to find how many more he needs:

$1080-216=864$

asked 2021-08-11

A ball is tossed upward from the ground. Its height in feet above ground after t seconds is given by the function $h\left(t\right)=-16{t}^{2}+24t$ . Find the maximum height of the ball and the number of seconds it took for the ball to reach the maximum height.

asked 2021-01-04

Write in words how to read each of the following out loud.

a. $\{x\in {R}^{\prime}\mid 0<x<1\}$

b. $\{x\in R\mid x\le 0{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}x\Rightarrow 1\}$

c. $\{n\in Z\mid n\text{}is\text{}a\text{}factor\text{}of\text{}6\}$

d. $\{n\in Z\cdot \mid n\text{}is\text{}a\text{}factor\text{}of\text{}6\}$

asked 2022-07-07

Bayes' Theorem Application Question

I'm reading Networks by Mark Newman and I'm confused about one of his applications of Bayes' theorem. He begins with the following equation:

$P(A,x,y\mid \text{data})=\frac{P(\text{data}\mid A,x,y)P(A)P(x)P(y)}{P(\text{data})}$

He then assumes the prior probabilities P(x) and P(y) are uniform and therefore constants. He then states that we cannot assume the same for P(A) and introduces the following as its prior probability $P(A|p)$(where p is another probability on which A depends)

From this he updates the formula above as follows:

$P(A,x,y,p\mid \text{data})=\frac{P(\text{data}\mid A,x,y)P(A\mid p)P(p)P(x)P(y)}{P(\text{data})}$

Why doesn't P(data|A,x,y) include p (and why did he introduce p into the posterior probability)?

I'm reading Networks by Mark Newman and I'm confused about one of his applications of Bayes' theorem. He begins with the following equation:

$P(A,x,y\mid \text{data})=\frac{P(\text{data}\mid A,x,y)P(A)P(x)P(y)}{P(\text{data})}$

He then assumes the prior probabilities P(x) and P(y) are uniform and therefore constants. He then states that we cannot assume the same for P(A) and introduces the following as its prior probability $P(A|p)$(where p is another probability on which A depends)

From this he updates the formula above as follows:

$P(A,x,y,p\mid \text{data})=\frac{P(\text{data}\mid A,x,y)P(A\mid p)P(p)P(x)P(y)}{P(\text{data})}$

Why doesn't P(data|A,x,y) include p (and why did he introduce p into the posterior probability)?

asked 2021-10-26

Simplify the expression

$\frac{(2x-3)}{5}-\frac{(2x-4)}{4}=2$

asked 2021-08-10

The time taken to finish 26.2 miles.

Given: 6 miles$=25$ minutes.

Given: 6 miles

asked 2022-05-24

How do you solve $-11>-\frac{c}{11}$ ?

asked 2022-07-25

Estimate each quotient:

121.6 divided by 43.5

69.1 divided by 10.7

38.9 divided by 13.1

435.8 divided by 88.6

52.7 divided by 9.2

75.6 divided by 15.3

121.6 divided by 43.5

69.1 divided by 10.7

38.9 divided by 13.1

435.8 divided by 88.6

52.7 divided by 9.2

75.6 divided by 15.3