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Montano1993 [528]
3 years ago
5

Look at the 95% confidence interval and say whether the following statement is true or false. ""This interval describes the pric

e of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."" Be sure to explain your answer.
Mathematics
1 answer:
zheka24 [161]3 years ago
8 0

Answer:

False

Step-by-step explanation:

Confidence intervals provide a range for a population parameter at a given significance level. The parameter can be mean, standard deviation etc.

In this example population is the prices of the rents of all the unfurnished one-bedroom apartments in the Boston area

significance level is 95%. Thus, the chance being the true population parameter in the given interval is 95%.

But, "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." statement is false because the population parameter is missing. Confidence interval may describe population mean for example but it does not describe the <em>whole</em> population.

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yaroslaw [1]

The value of distance between the station and the city in terms of miles which train and car travelled at interval of 2 hours is 129.6 miles.

<h3>What is the rate of speed?</h3>

The rate of speed is the rate at which the total distance is travelled in the time taken. Rate of speed can be given as,

r=d/t

Here, (d) is the distance travelled by the object and (t) is time taken but the object to cover that distance.

The train traveled 1/3 of the distance at 30 mph and the remaining distance at 40 mph. Let <em>t</em> is the time the train has taken to travel and x is the distance it travelled. Thus,

t=\dfrac{\dfrac{1}{3}x}{30}+\dfrac{\dfrac{2}{3}x}{40}\\t=\dfrac{x}{90}+\dfrac{x}{60}\\t=\dfrac{2x+3x}{180}\\t=\dfrac{5x}{180}\\t=\dfrac{x}{36}

After two hour a car left the same station traveled the first 3 hours at 35 mph and the remaining distance at 51 mph. Let <em>t</em> is the time the car has taken to travel and x is the distance it travelled. Thus,

t+2=3+\dfrac{(x-3\times35)}{51}

As t is same, thus put the value of t in this equation,

\dfrac{x}{36}=3-2+\dfrac{(x-3\times35)}{51}\\\dfrac{x}{36}=3-2+\dfrac{(x-3\times35)}{51}\\\dfrac{x}{36}=1+\dfrac{(x-105)}{51}\\\dfrac{x}{36}=\dfrac{(51+x-105)}{51}\\51x=36x-54\times36\\51x-36x=1944\\x=\dfrac{1944}{15}\\x=129.6

Thus, the value of distance between the station and the city in terms of miles which train and car travelled at interval of 2 hours is 129.6 miles.

Learn more about the rate of speed here:

brainly.com/question/359790

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2 years ago
What is the solution to equation? <br><br> 4x+5=2x-3
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4x - 2x = - 3 - 5
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3 years ago
9/16 divided by 9 plz help thx for everyone that has helped me
Verizon [17]
0.5625 will be the answer to 9 divided by 16
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3 years ago
Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six
baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.  

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

P(A|d) =\frac{P(A \cap d)}{P(d)}

P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}

P(A|d)= 0.3

Hence, the probability that a part was manufactured on machine A is 0.3

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3 years ago
Ezra enjoys gardening.
svlad2 [7]

Answer:

Ezra can plant 8 sunflowers

Step-by-step explanation:

0.7S + 0.5L <u><</u> 11

0.7S + 0.5(10) <u><</u> 11

0.7S  + 5 <u><</u> 11

0.7S <u><</u> 6

6 ÷ 0.7 = 8.5

7 0
3 years ago
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