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Tanzania [10]
3 years ago
6

The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 998 and a standard

deviation of 202 (based on date from the College Board). If a college includes a minimum score of 1025 among its requirements, what percentage of females do not satisfy that requirement?
Mathematics
1 answer:
WARRIOR [948]3 years ago
3 0

Answer: There is 64.8% of females do not satisfy that requirement.

Step-by-step explanation:

Since we have given that

Mean = 998

Standard deviation = 202

We need to find the percentage of females do not satisfy that requirement.

If a college includes a minimum score of 1025 among its requirements,

So, it becomes,

P(X

Hence, there is 64.8% of females do not satisfy that requirement.

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When factored completely, the expression 3x2 - 9x + 6 is equivalent to
Lunna [17]
Let's see what the options look like when we multiply the expressions in brackets:

(first, i multiply both parts of the second bracked by the first part of the first bracket, and then the same with the second part of the first bracket:
<span>(1) (3x - 3)(x - 2))

3x2 +6x -3x +6// this is not correct

(2) (3x + 3)(x - 2) </span>


3x2-6x+3x-6//this is not correct

(3)

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8 0
3 years ago
Read 2 more answers
X/5=2/3=5/y<br><br> 2/3 <br> 4/9<br> 1
Mademuasel [1]
\frac{x}{5}= \frac{2}{3}  = \frac{5}{y}
Break the equation into parts;
\frac{x}{5} = \frac{2}{3} ;  \frac{2}{3}= \frac{5}{y}
Solving for x;
\frac{x}{5} = \frac{2}{3}
x=\frac{10}{3}
Solving for y;
\frac{2}{3} = \frac{5}{y} &#10;y= \frac{3*5}{2} &#10;y= \frac{15}{2}
\frac{x}{y} = \frac{10/3}{15/2}
\frac{x}{y} = \frac{4}{15}
5 0
3 years ago
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