4x-3=5x-21 what does x=

Suppose a deck of cards contains 13 cards:

ira [324]

Answer:

$P(G_1) = \frac{5}{13}$

$P(G_2) = \frac{1}{3}$

$P(X \ge 1) = \frac{25}{39}$

Step-by-step explanation:

Given

$G = 5$

$R = 4$

$B = 4$

$n = 13$

Solving (a): $P(G_1)$

This is calculated as:

$P(G_1) = \frac{G}{n}$

$P(G_1) = \frac{5}{13}$

Solving (b): $P(G_2)$

This is calculated as:

$P(G_2) = \frac{G - 1}{n - 1}$ -- this is so because the selection is without replacement

$P(G_2) = \frac{5 - 1}{13 - 1}$

$P(G_2) = \frac{4}{12}$

$P(G_2) = \frac{1}{3}$

Solving (c): $P(X \ge 1)$

Using the complement rule, we have:

$P(X \ge 1) = 1 - P(X = 0)$

To calculate $P(X = 0)$ , we have:

$G = 5$ --- Green

$G' = 8$ ---- Not green

The probability that both selections are not green is:

$P(X = 0) = P(G'_1) * P(G'_2)$

So, we have:

$P(X = 0) = \frac{G'}{n} * \frac{G'-1}{n-1}$

$P(X = 0) = \frac{8}{13} * \frac{8-1}{13-1}$

$P(X = 0) = \frac{8}{13} * \frac{7}{12}$

Simplify

$P(X = 0) = \frac{2}{13} * \frac{7}{3}$

$P(X = 0) = \frac{14}{39}$

Recall that:

$P(X \ge 1) = 1 - P(X = 0)$

$P(X \ge 1) = 1 - \frac{14}{39}$

Take LCM

$P(X \ge 1) = \frac{39 -14}{39}$

$P(X \ge 1) = \frac{25}{39}$

7 0

2 years ago

A number is called squarefree if it is not divisible by the square of a positive integer greater than one. find the number of sq

german

Count the number of positive integers less than 100 that do not contain any perfect square factors greater than 1.

Possible perfect squares are the squares of integers 2-9.
In fact, only squares of primes need be considered, since for example, 6^2=36 actually contains factors 2^2 and 3^2.
Tabulate the number (in [ ])of integers containing factors of
2^2=4: 4,8,12,16,...96 [24]
3^2=9: 9,18,....99 [11]
5^2=25: 25,50,75 [3]
7^2=49: 49,98 [2]

So the total number of integers from 1 to 99
N=24+11+3+2=40
=>
Number of positive square-free integers below 100 = 99-40 = 59

3 0

2 years ago

Extracting roots of 5x²+15=15

IgorLugansk [536]

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7 0

2 years ago

A 2003 survey showed that 14 out of 250 Americans surveyed had suffered some kind of identity theft in the past 12 months. What

ivann1987 [24]

Answer:

The lower confidence limit of the 95% confidence interval for the population proportion of Americans who were victims of identity theft is 0.0275.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.