Answer:
Y=7
Step-by-step explanation:
(3y+5)= -4 -8y +9 + 77
-5 -5
3y= -9 - 8y + 9 + 77
+9 -9
3y= -8y + 77
+8y +8y
11y = 77
11y/11 77/11
y = 7
1. subtract 5 from both sides
2. add -9 and +9 and you get 0
3. add 8y to both sides
4. divide both sides by 11
5. you get y=7
Answer:
You can draw three rectangles:
1- 28 * 1.
2- 14*2
3- 7*4
Step-by-step explanation:
We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.
28 = 1*2*2*7
We can organize them as follows:
R1: 28 = 1*28
R2: 28 = 2*14
R3: 28 = 4*7
Finally, we can conclude that there are only three possibilities
1- 28 by 1.
2- 14 by 2
3- 7 by4
The perimeters will be:
Perimeter 1 = 2x1 + 2x28 = 58
Perimeter 2 = 2x2 + 2x14 = 32
Perimeter 3 = 2x4+2x7 = 22
Answer:
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
Step-by-step explanation:
Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.
Let five years ago be group I X and as of now be group II Y
(Two tailed test at 5% level of significance)
Group I Group II combined p
n 270 300 570
favor 120 140 260
p 0.4444 0.4667 0.4561
Std error for differene = 
p difference = -0.0223
Z statistic = p diff/std error = -1.066
p value =0.2864
Since p value >0.05, we accept null hypothesis.
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.
B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4.
(13−10)! 3!
C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =
10! 3!
9!1! 1!2!
10! 3!
8!2! 2!1!
10! 3!
7!3! 3!0!
= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =
= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =
= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285
1·2·3
<span>possible choices.</span><span>
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