Answer:
3.14 x 10^-7
Step-by-step explanation:
move the decimal behind the first significant figure and count the amount of spaces from there to the original decimal place
Let the width = x
=>length = 2x-1
91= x(2x-1)=2x^2-x
2x^2-x-91=0
=(2x+13)(x-7)
=> x=-13/2=> no
=>x= 7 cm
We want to find the probability that the two students chosen for the duet are boys. We will find that the probability that both students chosen for the duet are boys is 0.458
If we assume that the selection is totally random, then all the students have the same<em> </em><em>probability </em><em>of being chosen.</em>
This means that, for the first place in the duet, the probability of randomly selecting a boy is equal to the quotient between the number of boys and the total number of students, this is:
P = 11/16
For the second member of the duet we compute the probability in the same way, but this time there is one student less and one boy less (because one was already selected).
Q = 10/15
The joint probability (so both of these events happen together) is just the product of the individual probabilities, this will give:
Probability = P*Q = (11/16)*(10/15) = 0.458
So the probability that both students chosen for the duet are boys is 0.458
If you want to learn more, you can read:
brainly.com/question/1349408
Answer:
c
Step-by-step explanation:
you divide each side by four to get a=2.5 b=2.5 and c=3
We are given compound inequality 4p + 1 > −7 or 6p + 3 < 33.
Let us solve each of the inequality one by one.
4p + 1 > −7
Subtracting 1 from both sides, we get
4p + 1-1 > -7-1
4p > -8
Dividing both sides by 4, we get
p > -2. <em> (Shading right side for greater than sign)</em>
Solving 6p + 3 < 33.
Subtracting 3 from both sides, we get
6p + 3-3 < 33-3.
6p < 30
Dividing both sides by 6, we get
p < 5. <em> (Shading left for less than sign)</em>
<em>We have less than and greater than symbols in both inequalities, therefore we would have open circles(dots) on -2 and 5.</em>
And because we have "OR" composite inequality.
<em>So, we would take the combination of both shaded portion.</em>
<h3>Therefore, correct option is C.number line with shading everywhere.. </h3>
