Solution :
It is given that :
So,
The relative change is then,
This means that after 5 weeks, the revenue from the DVD sales in $563 with a rate of change of $70 per week and the increasing at a continuous rate of 12% per week.
The random sampling method is used.
What is random sampling?
Random sampling is a statistical method of selecting a representative sample of individuals from a larger population in a way that ensures every member of the population has an equal chance of being chosen. In other words, each individual in the population has an equal probability of being selected for the sample, and the selection process is not biased towards any particular subgroup or characteristic of the population.
The sampling method used in this scenario is simple random sampling. This is because the survey is being given to a randomly selected group of students from the entire student body, where every student has an equal chance of being selected. The administration used a random number generator to select 100 numbers at random, which is a common method for achieving simple random sampling. This method is useful because it helps to eliminate bias in the selection process and ensures that the sample is representative of the entire population.
Hence, the random sampling method is used.
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answer: 22
15/7.5 = x/11
cross multiply
7.5x = 165
divide by 7.5 on both sides to get the x alone
7.5x/7.5 = 165/7.5
x = 22
the tree is 22 feet
sorry if it’s wrong
Prime factorization expresses an integer in terms of multiples of prime numbers. The prime factorization of 20 using exponents can be written as 2²×5¹.
Factorization is expressing a mathematical quantity in terms of multiples of smaller units of similar quantities.
Prime factorization is when all those factors are prime numbers.
Thus, prime factorization expresses an integer in terms of multiples of prime numbers.
The prime factorization of 20 using exponents can be written as,
20 = 1 × 2 × 2 × 5
= 2² × 5¹
Hence, the prime factorization of 20 using exponents can be written as 2²×5¹.
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The mean is the sum of all values divided by the total number of values. Then:
Answer:
Step-by-step explanation:
From the information given:
a human brain weighs = 1 kg ; = 1000 grams
Number of cells = 10¹¹ cells
The density of water filled in each cell = 1 g/mL
From above;
the weight of each of the brain cell = total weight of the human brain/the number of cells
the weight of each of the brain cell = 1000/10¹¹
the weight of each of the brain cell = 1 × 10⁻⁸ grams
Now, to calculate the quantity of water in each cell; we have:
= the weight of each brain × density
=
For cube; we know that
1 mL = 1 cm³
Thus:
Recall that; the volume of a cube as well =
where;
x = length of each sides
∴
=
x = 0.0022 cm
Thus, the length of each side of the cell = 0.0022 cm
The surface area of a single cell = x²
The surface area of a single cell = (0.0022 cm)²
The surface area of a single cell = 4.84 × 10⁻⁶ cm²
Therefore, the total surface area of is:
=
=
= 50 m²
If the human brain's cells were cube-shaped and filled with water, each cell would be roughly 21.5 micrometers on a side. If these cells were spread out into a single-cell-thick layer, the total surface area for one side of the layer would be approximately 4.63 square meters.
To answer your question, the human brain has about 1011 cells, each filled with water. Given the total mass of the brain (about 1 kg) and the number of cells, we can calculate the volume of a single cell. The density of water is 1 g/mL or 1,000 kg/m³, so the volume of all the cells (entire brain) is 1 m³. Therefore, the volume of a single cell must be 1 m³/1011 cells, which is approximately 10-14 m³. For a cubical cell, the side length of the cube (a) would be the cube root of this volume, which is approximately 2.15 x 10-5 m or 21.5 micrometers.
To calculate the total surface area for one side of the cell layer, we multiply the area of a single cell by the total number of cells: (2.15 x 10-5)² m²/cell x 1011 cells = approximately 4.63 m².
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Answer:
Step-by-step explanation:
Corresponding heights of presidents and height of their main opponents form matched pairs.
The data for the test are the differences between the heights.
μd = the president's height minus their main opponent's height.
President's height. main opp diff
191. 166. 25
180. 179. 1
180. 168. 12
182. 183. - 1
197. 194. 3
180. 186. - 6
Sample mean, xd
= (25 + 1 + 12 - 1 + 3 + 6)/6 = 5.67
xd = 5.67
Standard deviation = √(summation(x - mean)²/n
n = 6
Summation(x - mean)² = (25 - 5.67)^2 + (1 - 5.67)^2 + (12 - 5.67)^2+ (- 1 - 5.67)^2 + (3 - 5.67)^2 + (- 6 - 5.67)^2 = 623.3334
Standard deviation = √(623.3334/6 sd = 10.19
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 6 - 1 = 5
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (5.67 - 0)/(10.19/√6)
t = 1.36
We would determine the probability value by using the t test calculator.
p = 0.12
Since alpha, 0.05 < than the p value, 0.12, then we would fail to reject the null hypothesis.
Therefore, at 5% significance level, we can conclude that for the population of heights for presidents and their main opponents, the differences have a mean greater than 0 cm.
The null hypothesis in this case would be that there is no average height advantage for presidents over their main opponents (µd ≤ 0), while the alternative hypothesis is that presidents are taller on average (µd > 0). A paired t-test with a significance level of 0.05 is usually employed in testing these hypotheses using the p-value and t-score.
In hypothesis testing, the goal is to determine the validity of a claim made. In this case, the claim is that the mean difference in height, where the difference is calculated as the president's height minus their main opponent's height, is greater than 0 cm. This represents the theory that taller presidential candidates have an advantage.
For setting up a null hypothesis and an alternative hypothesis, we consider the following parameters:
To test these hypotheses, we would typically use a one-sample t-test for paired differences with a significance level (alpha) of 0.05. A p-value less than this would allow us to reject the null hypothesis in favor of the alternative hypothesis that presidents are on average taller than their main opponents. Use of p-value and t-score is essential in conducting such a test.
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