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nydimaria [60]
3 years ago
8

On-the-Go Phone Company has two monthly plans for their customers.The EZ Pay Plan costs $0.15 per minute. The 40 to Go Plan cost

s $40 per month plus $0.05 per minute. Write an expression that represents the monthly bill for x minutes on EZ Pay Plan.
Mathematics
2 answers:
natulia [17]3 years ago
4 0
For one moth it would be....
30 + 40 + 0.05
$57.00 In total but im not sure if this is correct so if you could make sure please.
alexandr1967 [171]3 years ago
4 0

Answer: Expression that represents the monthly bill for x minutes on EZ pay plan is 0.15x

Step-by-step explanation:

Since we have given that

Cost per minute through EZ Pay plan = $0.15

Cost of Go plan per month = $30

Cost per minute in Go plan = $0.05

so, Expression for EZ pay plan is given by

0.15x

Expression for Go plan is given by

40+0.05x

Hence, expression that represents the monthly bill for x minutes on EZ pay plan is 0.15x

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Answer:

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Step-by-step explanation:

convert each mixed number into a improper fraction

27/11  - 117/11

= -90/11

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Which expression shows the result of applying the distributive property to  12(5y+4) ?
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<span>12(5y+4)
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2 years ago
An NBA fan named Mark claims that there are more fouls called on his team 1 point
inn [45]

Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

df = n-1=34-1=33

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

Since is a one side test the p value would be:  

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

7 0
3 years ago
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