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garri49 [273]
3 years ago
13

The temperature at a mountain base camp was −3 degrees Celsius on Monday. Tuesday morning, the temperature was 2 degrees Celsius

lower than it was on Monday. By Tuesday evening, the temperature was 4 degrees Celsius lower than it was that morning. What was the temperature at the base camp Tuesday evening? -9 degrees Celsius −6 degrees Celsius −5 degrees Celsius −1 degrees Celsius
Mathematics
2 answers:
AnnZ [28]3 years ago
5 0

-9 degrees Celsius

because when the temp is negative it and it keeps going down, then the numbers basically get added and have a negative sign in front of it.

-3 +-2 +-4 = -9

gogolik [260]3 years ago
5 0

It would be, negative nine.

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Read 2 more answers
The number of grams of carbohydrates contained in 5–ounce of randomly selected meals "with/with no" potatoes (fries) in McDonald
Ede4ka [16]

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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>

  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]

    = 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".

7 0
3 years ago
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