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cupoosta [38]
3 years ago
14

Number of Text Messages a Day A random sample of n = 755 US cell phone users age 18 and older in May 2011 found that the average

number of text messages sent or received per day is 41.5 messages,32 with standard error about 6.1. State the population and parameter of interest. Use the information from the sample to give the best estimate of the population parameter. Find and interpret a 95% confidence interval for the mean number of text messages.
Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

The 95% confidence interval is  (29.54 - 53.46}

Step-by-step explanation:

given data:

\hat X = 41.5

Se = 6.1

n = 755

a) best estimate  \hat x = 41.5

b) at 95% confidence interval

\alpha = 1- 0.95 = 0.05

\alpha /2 = 0.025

z_{\alpha/2} = z_{0.025} = 1.96

at 95% confidence interval for [/tex]\mu[/tex]

\hat x \pm z_{\alpha /2} \times Se

41.5 \pm 1.96\times 6.1

41.5 \pm 11.96

The 95% confidence interval is  (29.54 - 53.46}

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Answer:

C 47.12 L

Step-by-step explanation:

1 gallon = 3.8 liters

12.4 gallons = x

Cross Multiply

= 12.4 × 3.8

47.12 liters

Option C is correct

5 0
3 years ago
Angle W and angle X are congruent. If their sum is 121 degrees, what is the measure of angle X?
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Angle X = 60.5 degrees
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IceJOKER [234]

Answer:

Step-by-step explanation:

instead of (2x - 60 ) there should be (60 - 2x) because since they are parallel lines corresponding sides must be equal . It is contradictory because the size of the corresponding angles is different.

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2 years ago
What is the solution to the following system of equations? x+2y=2 x-2y=-2
Tcecarenko [31]
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7 0
3 years ago
From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the a
padilas [110]

Answer:

See explanation

Step-by-step explanation:

Given:

50 liter container

40% concentration of acid

x liters are removed and replaced with 100% acid

(A)

Acid in 50-liter container =  40% of 50 liters

                                          = (40/100) * 50

                                          = 0.4 * 50

                                          = 20

So there are 20 liters amount of acid in 50 liter container.

Now x liters are removed and replaced with 100% acid

The acid in x-liter

40% of x liters = (40/100) * x  = 0.4 * x

When x liters are removed then amount of acid that remained in container:

20 - 40% of x liters = 20 -  (40/100) * x  = 20 - 0.4 * x

This can also be written as:

(50-x) (40/100) =  (40*50/100) - (40*x/100) = 2000/100 - 40x/100 =  20 - 2x/5

Since x liters replaced with 100% acid so this becomes:

20 - 40% of x liters + 100% of x liters

= 20 -  (40/100) * x + 100/100 * x

= 20 - (40/100) x + (100/100)x

= 20 +  (100-40/100)x

= 20 + (60/100)x

= 20 + 0.6 x    

This can also be written as:

20 - 2x/5 + x = 20 + 3x/5 =  20 + 0.6 x

Hence the amount of acid in the final mixture as a function of x:

f(x) = 20 + 0.6 x

(B)

The value of x can not be greater than 50 liters and can not be lesser than 0 liters.  

so, range is [0, 50] where both 0 and 50 are inclusive.  

(C)

Since final mixture is 50% acid So

Acid = 50% of 50 litres =

        = (50/100) * 50

        = 1/2 * 50

         = 50/2

        = 25

Put 25 in the computed function of x

20 + 0.6 x = 25

0.6x = 25 - 20

0.6x = 5

x = 5/0.6

x = 8.333333

x = 8.3 liters

3 0
3 years ago
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