6.7 10 squared positive 8
Sh(2x) = (e^2x + e^-2x)/2
<span>Thus the integral becomes </span>
<span>Int[e^3x*(e^2x + e^-2x)/2] = Int[(e^5x + e^x)/2] </span>
<span>= e^5x/10 + e^x/2 + C
</span>=(1/10)(e^5x) + (1/2)(e^x) + C
Answer: the mode is 67.2
Step-by-step explanation:
Given that;
Data Frequency
30 - 34 1
35 - 39 0
40 - 44 3
45 - 49 7
50 - 54 5
55 - 59 10
60 - 64 10
65 - 69 21
70 - 74 12
Mode = ?
we know that mode is the number that has the highest number of appearance of frequency, so in this case, the data group that has the highest frequency (21) is 65 - 69
Lower class boundary of the modal group; L = 65
Frequency of the group before the modal group; Fm-1 = 10
Frequency of the modal group; Fm = 21
Frequency of the group after the modal group; Fm+1 = 12
Group width; G = 4
Now using the formula
Mode = L + [ (Fm - Fm-1) / ( (Fm - Fm-1) + (Fm - Fm+1) ) ] × W
so we substitute
Mode = 65 + [ (21 - 10) / ( (21 - 10) + (21 - 12) ) ] × 4
= 65 + [ 11 / 20] × 4
= 65 + 2.2
= 67.2
Therefore the mode is 67.2
X2+5x-150= <span><span> x = 10 </span><span> x = -15
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