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Jobisdone [24]
3 years ago
13

The Census Bureau estimates that in 2020, 15 of the 50 states will have a Spanish-speaking majority. What percent of the states

will have a Spanish-speaking majority?
A. 10%
B. 15%
C. 30%
D. 50%
Mathematics
1 answer:
Vinvika [58]3 years ago
4 0
The answer is:  [C]:  "30%" .
_____________________________________________________
Explanation:
_____________________________________________________
Note that: "%" ; or "percent" means "out of 100" ; or "divided by 100" .
______________________________________________________
Method 1) 

15/50 = ?/ 100 ;

Look at the denominators:

50 * (what value?) = 100 ? ;  → "100 ÷ 50 = 2" ;

→ 50 * 2 = 100 ;

So:  "15/50 = (15*2)/(50*2) = "30/100" ; which is "30%" .
___________________________________________
Method 2:
___________________________________________
 "15/50" = (15÷5) / (50÷5) = 3/10 ; 

3/10 = (3*10) / (10*10) = 30/100 ; which is:  "30%" .
__________________________________________
Method 3:  (slight variation of "Method 2" above):
___________________________________________
 "15/50" = (15÷5) / (50÷5) = 3/10 ; 

3/10 = 0.3 = 0.30 = (0.30 * 100) % =  " 30% " .
___________________________________________ 
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(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

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Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

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Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

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P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

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