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Minchanka [31]
2 years ago
9

The cost for making a telephone call from Vero beach to Miami is 37 cents for The first 3 minutes and 9 cents per each additiona

l minute. There is a 10% discount for call placed after 10pm. what is the cost of a 10 minute telephone call at placed at 11pm? Round your answer to the nearsest cent.
Mathematics
2 answers:
Evgen [1.6K]2 years ago
7 0

Answer:90cents

Cost of a  10 minute call before 10pm:

37 cents + 9cents*7minutes = 37 cents+63 cents =100 cents

10% of 100cents:

100*10/100= 10 cents

Cost of  the call after 10 pm (i.e with a 10% discount)

100-10= 90 cents

Vinvika [58]2 years ago
4 0

Answer:

  $0.90

Step-by-step explanation:

A 10-minute call exceeds the 3-minute initial period by 7 minutes. The initial period charge is 37¢. The additional minute charge is 7·9¢ = 63¢. Then the regular charge for that call is ...

  37¢ +63¢ = 100¢ = $1.00

The 10% discount reduces the charge by ...

  10% × $1.00 = $0.10

so the final cost of the 10-minute call is ...

  $1.00 -0.10 = $0.90 . . . . . cost of the call

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Consider the table showing the given, predicted and residual values for a data set. A 4-column table with 4 rows. The first colu
Alecsey [184]

Answer:

Step-by-step explanation: I believe the answer would be (4, 0.1).

When plotting a residual, use the x-value, in this case 4, and the residual value as the y. (4, 0.1) The 4 is the x value, and the 0.1 replaces the y value. In the table the column headers will show you what the x, y, and residuals are. Just disregard the y-value and "predicted" and "given" columns, they are not needed when plotting the residual. I really hope this helps, and I hope I explained it well!

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2 years ago
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image Alan, Benton, Claire, and Dita divided the cost of throwing a party as shown in the circle graph above. If Dita spent $ 12
sdas [7]

Let total amount spent =x

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35% of x = 0.35x

So amount spent by Dita = 0.35x

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We are given that Benton spent $12 more than Claire,

so making equation we have ,

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5 0
2 years ago
Select the functions that have identical graphs.
shtirl [24]

Answer:

c. 1 and 3

Step-by-step explanation:

To quickly solve this problem, we can use a graphing tool or a calculator to plot each equation.

Please see the attached image below, to find more information about the graph s

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A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious
Dominik [7]

Answer:

Part1) z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191    

p_v =2*P(Z>2.191)=0.0284

Readjustment is needed

Part 2)  z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

p_v =2*P(Z

No readjustment is needed

Step-by-step explanation:

Data given and notation

Part 1    

\bar X=16.32 represent the sample mean    

\sigma=0.8 represent the population standard deviation    

n=30 sample size    

\mu_o =16 represent the value that we want to test    

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

z would represent the statistic (variable of interest)    

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

State the null and alternative hypotheses.    

We need to apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  \mu =16    

Alternative hypothesis :\mu \neq 16    

Compute the test statistic  

The statistic for this case is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191    

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Since is a two tailed test the p value would be:    

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Using the value of \alpha =0.05 we see that pv and we have enough evidence to reject the null hypothesis.

Readjustment is needed

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Part 2

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We can replace in formula (1) the info given like this:    

z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

Give the appropriate conclusion for the test  

Since is a two tailed test the p value would be:    

p_v =2*P(Z

Using the value of \alpha =0.05 we see that pv>\alpha and we have enough evidence to FAIL to reject the null hypothesis.

No readjustment is needed

The rejection zone is given by:

(-\infty ;-1.96) , (1.96;\infty)  

What action would you recommend ?

Review the procedure.

Do you reach the same conclusion ?

No we got different conclusions

5 0
2 years ago
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