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Sliva [168]
3 years ago
7

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (

c) below. TInterval ​(13.046,22.15) x overbarxequals=17.598 Sxequals=16.01712719 nequals=50 a. Express the confidence interval in the format that uses the​ "less than" symbol. Given that the original listed data use one decimal​ place, round the confidence interval limits accordingly.
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
4 0

Answer:

13.05MbpS < x < 22.15 Mbps

Step-by-step explanation:

Confidence interval CI can be expressed in the form;

Lower bound < x < upper bound

Given;

CI = (13.046, 22.15)

Lower bound = 13.046 = 13.05 Mbps (to two s.f.)

Upper bound = 22.15 Mbps

So, the Confidence interval is given as;

13.05 MbpS < x < 22.15 Mbps

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Find the area of a regular pentagon with the apothem of 24.3 cm and a side length of 35.3
steposvetlana [31]

Answer: A=2144\ cm^2

Step-by-step explanation:

For this exercise you need  to use the following formula for calculate the area of regular polygon:

A=\frac{s^2n}{4tan(\frac{180}{n})}

Where is "s" the length of any side, "n" is the number of sides.

Int this case you know that it is regular pentagon, which means that it has five sides. Then you can identify that the values of "n" and "s":

s=35.3cm\\n=5

 Therefore, substituitng values into the formula, you get that the area f the pentagon is:

A=\frac{(35.3cm)^2(5)}{4tan(\frac{180}{5})}\\\\A=2144\ cm^2

8 0
3 years ago
The students at Southern Junior High School are divided into four homerooms. Benjamin just moved into the school district and wi
Klio2033 [76]

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
The batteries from a certain manufacturer have a mean lifetime of 850 hours, with a standard deviation of 70 hours, assuming tha
matrenka [14]

To answer (a), we will need to find the Z value for 710 hours and 990 hours.

\begin{gathered} For\text{ 710} \\ Z\text{ = }\frac{x-\mu}{\sigma} \\ \text{    = }\frac{710-850}{70} \\ \text{    = -2} \\ p\text{ =0.0228 } \\ For\text{ 990:} \\ Z\text{ = }\frac{990-850}{70} \\ \text{ =2} \\ p\text{ = 0.9772} \end{gathered}\begin{gathered} To\text{ find P \lparen–2 \le Z \le 2\rparen} \\ =\text{ 0.9772 -0.0228} \\ =\text{ 0.9544} \\ =95.44\% \end{gathered}

B) 68% of data lies between one standard deviation of the mean.

850 +70 = 920

850 - 70 = 780

5 0
1 year ago
HELP ME ASAP NOW PLEASE!
Aleksandr [31]
1760
13+25+10/150=.32
.32x5500=1760
8 0
3 years ago
The weights of a certain brand of candies are normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.
aleksandr82 [10.1K]

Answer:

The probability is 0.508 = 50.8%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g.

This means that \mu = 0.8544, \sigma = 0.0525

If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g.

This is 1 subtracted by the pvalue of Z when X = 0.8535. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.8535 - 0.8544}{0.0525}

Z = -0.02

Z = -0.02 has a pvalue of 0.492

1 - 0.492 = 0.508

The probability is 0.508 = 50.8%.

8 0
3 years ago
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