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xz_007 [3.2K]
3 years ago
14

The weather forecast states that the temperature will be lower than –5°F tonight. Let t represent the temperature in degrees Fah

renheit.
Which inequality models this situation?

A.
t < –5 and t may be a negative number

B.
t < –5 and t must be a positive number

C.
t > –5 and t may be a negative number

D.
t > –5 and t must be a positive number
Mathematics
2 answers:
Deffense [45]3 years ago
7 0
Hi There!

<span>t < –5 and t may be a negative number!</span>
Tcecarenko [31]3 years ago
7 0
Hey there,
 
It would be A

This <span> < is a bracket 
</span>
Bracket means something could be (greater then or less than)
In this case it showing the definition or the out come! 

Hope this helped!  

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Jung wanted to find out which band was most popular among the high school students. Where should he conduct his survey to get th
emmasim [6.3K]

Answer:

I'd say A

Step-by-step explanation:

Reason: Because B and C are both having bands there and the cafeteria has the students popular opinion where there isn't band influence.

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X/-4 + 14&gt;20<br><br> someone help a girl out
MrRa [10]

Answer:

x < -20

Step-by-step explanation:

x / -4  + 14 >20

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x / -4 +14-14 > 20 -14

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Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste
Volgvan

Answer:

a) The calculated value   t =  5.903 > 2.572 at 0.01 level of significance

Null hypothesis is rejected at 0.01 level of significance

There is a true mean value is less than 10

b) p - value  0.00001

The p - value 0.00001  < 0.01

The result is significant at p<0.01

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given mean of the population = 10

Sample size 'n'= 35

Mean of the sample x⁻ = 14.44

Standard deviation of the sample 's' = 4.45

Level of significance ∝ = 0.01

<u><em>Step(ii)</em></u>:-

<em>Null Hypothesis H₀</em> : μ >10

<em>Alternative Hypothesis</em> : H₁ : μ < 10

Test statistic

                 t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }

                t = \frac{14.44-10}{\frac{4.45}{\sqrt{35} } }

               t =  5.903

Degrees of freedom = n-1 = 35-1 = 34

tabulated value t₀.₀₁,₃₄  = 2.572

The calculated value   t =  5.903 > 2.572 at 0.01 level of significance

Null hypothesis is rejected at 0.01 level of significance

There is a true mean value is less than 10

<u><em>p- value</em></u> :

p - value  0.00001

The p - value 0.00001  < 0.01

The result is significant at p<0.01

8 0
3 years ago
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