Answer:
parents
Step-by-step explanation:
ask them for it
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
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Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x
<h3><u>Solution:</u></h3>
Given, There are "x" number of students at helms.
The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x
Number of students in next year = number of students in present year + increased number of students.
Thus there will be 1.078x students in next year
<span>1220
Subtracting the lower boundary of 1492 grams from the mean of 3234 gives you 1742 grams below the mean. Dividing 1742 by the standard deviation of 871 gives you 2 standard deviations below the curve. Now doing the same with the upper limit of 4976 grams also gives you 2 standard deviations above the mean (4976-3234)/871 = 2
So you now look for what percentage of the population lies within 2 standard deviations of the mean. Standard lookup tables will indicate that 95.4499736% of the population will be within 2Ď of the mean. So multiply 1278 by 0.954499736 giving 1219.851. Then round to the nearest whole number and you have an estimated 1220 babies that weigh between 1492 grams and 4976 grams.</span>
