Answer:
The solution is g = 4
Step-by-step explanation:
* 6(-2g - 1) = -(13g + 2)
- We need to solve it to find the value of g
- Let us simplify each side and then solve the equation
∵ 6(-2g - 1) = (6)(-2g) - (6)(1)
∴ 6(-2g - 1) = -12g - 6
∴ The simplify of 6(-2g - 1) is -12g - 6 ⇒ (1)
∵ -(13g + 2) = -13g - 2
∴ The simplify of -(13g + 2) is -13g - 2 ⇒ (2)
- Equate (1) and (2)
∴ -12g - 6 = -13g - 2
- Add 13g to both sides
∴ g - 6 = - 2
- Add 6 to both sides
∴ g = 4
* The solution is g = 4
You can do this by converting this into a ratio or a fraction
Ratio:
1 lb for 4 guinea pigs
1:4
If he was 8 lbs for x guinea pigsit would be
1:4
8:x
Multiply the first ration until the one is equal to 8, then whatever the four turns into is x
Fraction: (easier way)
1/4 (1lb for 4 guinea pigs)
8/x (8lb for x guinea pigs)
1/4 = 8/x
Multiply one until it is equal to 8
Correspond that to the 4
1*8 = 8
4*8 = 32
32 guinea pigs
Yes! The lengths of each side must be less than the sum of the other two lengths. It looks like this:
4+4>7
7+4>4
4+7>4
Answer:
Answer:
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
Step-by-step explanation:
Potatoes : No Potatoes : Difference Difference (d)²
(Potatoes- No Potatoes)
29 41 -12 144
25 41 -16 256
17 37 -20 400
36 29 -7 49
41 30 11 121
25 38 -13 169
32 39 -7 49
29 10 19 361
38 29 9 81
34 55 -21 441
24 29 -5 25
27 27 0 0
<u>29 31 -2 4 </u>
<u> ∑ -64 2100 </u>
- We state our null and alternative hypotheses as
H0 : μd= 0 and Ha: μd≠0
2. The significance level alpha is set at α = 0.01
3. The test statistic under H0 is
t= d`/sd/√n
which has t distribution with n-1 degrees of freedom.
4. The critical region is t > t (0.005,12) = 3.055
5. Computations
d`= ∑d/n = -64/ 13= -4.923
sd²= ∑(di-d`)²/ n-1 = 1/n01 [ ∑di² - (∑di)²/n]
= 1/12 [2100- ( -4.923)] = 175.410
sd= √175.410 = 13.244
t = d`/sd/√n= - 4.923/13.244/√13
t= - 4.923/3.67344
t= -1.340
6. Conclusion :
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
