So with this, I will be using the substitution method. With the first equation, substitute (y+3) into the x variable and solve for y:
Next, now that we have the value of y, substitute it into either equation to solve for x:
<u>And this is how you get your final answer (5,2).</u>
You do 3 x 8 / 100 to work out 3/8 as a decimal. So its 0.24. 0.38 is bigger
To solve this, since you know that h(x) is 11 and that it also equals -4x+3, you set them equal to one another which would look like this: 11=-4x+3.
Then, to solve for x, which is what I am assuming the question is asking, you would subtract 3 from both sides to isolate -4x, which would result in this:
-4x=8
Now, to solve for x, divide both sides by -4, and you get your answer which is x=-2
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
Answer:
x= -7-11
x=-18
Step-by-step explanation:
