FORTY WHOLE POINTS FOR U AND THANKS AND BRAINLIEST AND 5 STARS PLEASE I NEED THIS IM CONFUSED IDK SLOPE IM NOT GOOD AT MATH I NE
ED SOMEONE TO ANSWER THIS FOR MEEEEEEEEEEEE
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<em>Directed numbers</em> are numbers that have either a <u>positive</u> or <u>negative </u>sign, which can be shown on a <em>number line</em>. Therefore, point F is Fifteen-halves of line <em>segment</em> DE.
A <u>number line</u> is a system that can show the positions of <em>positive</em> or <em>negative</em> numbers. It has its <em>ends</em> ranging from <em>negative infinity</em> to <em>positive infinity</em>. Thus any <em>directed</em> number can be located on the line.
Directed numbers are numbers with either a <u>negative</u> or <u>positive </u>sign, which shows their direction with respect to the <em>number line.</em>
In the given question, the <u>distance</u> between points D and E is <em>9 units</em>. So that <em>dividing</em> 9 units in the ratio of 5 to 6, we have;
x 9 =
=
Therefore, the <em>location</em> of point F, which <u>partitions</u> the directed line segment from d to E into a 5:6 ratio is . Thus the<em> answer</em> is <u>Fifteen-halves.</u>
For further clarifications on a number line, visit: brainly.com/question/23379455
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The required probability is
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that,
Probability for first draw to be non-defective
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective
(printers after first slot = 10; total defective printers = 2)
Then, total probability