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Alexeev081
4 hours ago
7

What is the range of the function y = 2 e Superscript x Baseline minus 1

Mathematics
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"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight
AnnZ [12381]

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

</pfor>
6 0
3 months ago
A store sold a case of scented candles for $17.85 that had been marked up 110%. What was the original price?
Zina [12379]
1.1x = 17.85 x ≈ 16.227272727272727... Rounded to $16.23
5 0
2 months ago
Read 2 more answers
Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over a
lawyer [12517]

Answer:

a) The average of the time interval is 45 minutes.

b) There is a 27.78% chance that the time gap between two successive defective light bulbs will lie between 10 and 35 minutes.

c) The standard deviation for the time duration is 25.98 minutes.

d) There is an 11.11% likelihood that the time interval between two successive defective light bulbs will be under 10 minutes.

Step-by-step explanation:

A uniform probability situation occurs when all outcomes have an equal chance of happening.

In this context, we identify a lower and upper limit for the distribution known as 'a' and 'b', respectively.

The likelihood of finding a value X that is less than x is determined by this formula.

P(X < x) = \frac{x - a}{b-a}

The likelihood of obtaining a value X that lies between c and d is expressed as:

P(c \leq X \leq d) = \frac{d-c}{b-a}

The average of the uniform distribution is:

M = \frac{a+b}{2}

The standard deviation of the uniform distribution can be calculated as:

S = \sqrt{\frac{(b-a)^{2}}{12}}

The uniform distribution covers an interval from 0 to 90 minutes.

This indicates that a = 0, b = 90

a) What is the mean of the time interval?

M = \frac{a+b}{2} = \frac{0 + 90}{2} = 45

The mean of the time interval is 45 minutes.

b) What is the probability that the time gap between two consecutive defective light bulbs will be between 10 and 35 minutes?

P(10 \leq X \leq 35) = \frac{35-10}{90-0} = 0.2778

There is a 27.78% chance that the time interval between two successive defective light bulbs falls between 10 and 35 minutes.

c) What is the standard deviation of the time interval?

S = \sqrt{\frac{(90-0)^{2}}{12}} = 25.98

The standard deviation of the time interval is 25.98 minutes.

d) What is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes?

P(X < 10) = \frac{10 - 0}{90 - 0} = 0.1111

There is an 11.11% likelihood that the time duration between two consecutive defective light bulbs will be shorter than 10 minutes.

8 0
3 months ago
What is the greatest common divisor of $1665,$ $4554,$ $1683,$ and $4050$?
Inessa [12570]
9 is the highest common factor
7 0
2 months ago
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