Firstly, arrange the data.
19 23 26 28 28 30 31 35 38 40 43 46
a) 5 number summary
-min value = 19
- Q1= (26+28/2) = 27
- Q2/ Median = (30+31/2)= 30.5
- Q3= (38+40/2)= 39
- Max value- 46
b) range= max value-min value
∴46-19= 21
c)interquartile range = Q3-Q1
= 39-27
=12
Responder:
15 m e 75 m
Explicação passo a passo:
Deixar :
Largura = x
Comprimento = 5x
Perímetro = 180
Já o terreno tem uma forma retangular; usamos a relação;
Perímetro = 2 (comprimento + largura)
180 = 2 (5x + x)
180 = 2 (6x)
180 = 12x
180/12 = x
15 = x
Largura = x = 15 metros
Comprimento = 5x = 15 * 5 = 75 metros
Medida de largura e comprimento, respectivamente:
15 m e 75 m
The answer will probably be -0.665
Answer:
The mean is 40.35 and the standard deviation is 0.13.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean 
and standard deviation 
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
For a certain type of olivine assembly, the silicon dioxide (SiO2) content (in weight percent) in a randomly chosen rock has mean 40.35 and standard deviation 0.4.
Sample of 10:
By the Central Limit Theorem, the mean is 40.35, and the standard deviation is 
The mean is 40.35 and the standard deviation is 0.13.
Answer:
The distance when x = 6km is D = 498 KM
Step-by-step explanation:
Here, given the formula for the distance, we want to find the distance D given the value of x to be 6km
What we do simply here is to substitute the value 6km as x in the equation
The equation is
D = 4x^3 - 10x^2 -6
Substituting x = 6, we have
D = 4(6^3) -10(6^2) -6
D = 864-360-6
D = 498 km
