Answer:
1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle
Step-by-step explanation:
Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is
V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³
the mass in that volume would be m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)
The density of an alpha particle is ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be
ρ= m/V
since both should be equal ρ=ρa , then
ρa= m/V =N*L/V → N =ρa*V/L
replacing values
N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg = 1.544*10⁹ Linebackers
N=1.544*10⁹ Linebackers
The data given on the table about the ten participants in the health study shows that B. The post-run mean and median are the same value.
<h3>What is the post run mean?</h3>
This can be found as:
= (87 + 95 + 96 + 101 + 110 + 104 + 100 + 101 + 90 + 96) / 10
= 98
<h3>What is the post run median?</h3>
First order the pulses from smallest to largest:
= 87, 90, 95, 96, 96, 100, 101, 101, 104, 110
The two digits in the middle are:
96 and 100
The median will be:
= (96 + 100) / 10
= 98
In conclusion, option B is correct.
Find out more on the median at brainly.com/question/396397.
Answer:
The margin of error for the 94% confidence interval is 0.6154.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean is:
The margin of error of this interval is:
The critical value of <em>z</em> for 94% confidence level is, <em>z</em> = 1.88.
Compute the margin of error for the 94% confidence interval as follows:
Thus, the margin of error for the 94% confidence interval is 0.6154.
