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AfilCa [17]
3 years ago
15

What is the value of y in the equation 2(3y-4)=10?

Mathematics
2 answers:
Marta_Voda [28]3 years ago
8 0
2(3y-4)=10 \\ 6y-8=10 \\ 6y=18 \\ y= \frac{18}{6} \\ \boxed {y=3}
USPshnik [31]3 years ago
3 0
6y-8=10

6y=18

y=3
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The number of grams of carbohydrates contained in 5–ounce of randomly selected meals "with/with no" potatoes (fries) in McDonald
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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)

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41                 30                          11                              121

25                 38                        -13                              169

32                39                        -7                                49

29                 10                          19                             361

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34                 55                       -21                              441

24                29                         -5                              25

27                 27                         0                               0

<u>29                 31                         -2                              4                     </u>

<u> ∑                                                 -64                           2100               </u>

  1. 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]

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sd= √175.410 = 13.244

t = d`/sd/√n= - 4.923/13.244/√13

t= - 4.923/3.67344

t= -1.340

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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".

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