So
a. so 100-1/10 of 100 or 100-10=90
then second dose minus 10% or 90-9=81
third dose 81 minus 10%=81-8.1=72.9
fourth dose 72.9 minus 10% or 72.9-7.29=65.61
so after 4th dose only 65.61% are left
b. so to get it to 5% you do alot and find that it takes 27 tries to get the percent down to 5.2% and 28 doses to get it to 4.7%
Answer:
784
Step-by-step explanation:
caculator. of course.
Answer:
Step-by-step explanation:
a) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as
Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}
where
p1 = sample proportion of population 1
p2 = sample proportion of population 2
n1 = number of samples in population 1,
n2 = number of samples in population 2,
From the information given
p1 = 0.77
1 - p1 = 1 - 0.77 = 0.23
n1 = 58
p2 = 0.67
1 - p2 = 1 - 0.67 = 0.33
n2 = 70
Standard error = √{(0.77 - 0.67)/[(0.77)(0.23)/58) + (0.67)(0.33)/70}
= √0.1/(0.0031 + 0.0032)
= √1/0.0063
= 12.6
the standard error of the distribution of differences in sample proportions is 12.6
b) the sample sizes are large enough for the Central Limit Theorem to apply because it is greater than 30
Answer:
8
Step-by-step explanation:
8*8=64
Mark me brainliest if I helped;D
Answer:
Step-by-step explanation:
17 : i 12x^2(4x−3) ,
ii : 5x(x−3y) ,
iii: 5xy^2z(3x^2−5z^2)
18) since the two polynomial are equal then :
2x^3+ax^2+3x-5 = x^3+x^2-2x+a same remainder x-2 then x=2
2(2)^3+a(2)^2+3(2)-5=(2)^3+(2)^2-2(2)+a solve for a
-3a=9
a=-3
18 ) x^3+y^3 -125+15xy
x^3+y^3-(5)^3-3(x)(y)(-5) then factorize x^3+y^3
(x+y)(x^2-xy+y^2)- (5)(5)^2-3(x)(y)(-5) common factor x+y-5
(x+y-5)(x^2-xy+y^2-5^2-3xy given x+y=5
5-5( x^2-xy+y^2-25-3xy) = 0
the value of the expression equal to zero
