Answer:
Step-by-step explanation:
v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h
vE = 20sin36º + 10 = 21.76 mi/h
vN = 20cos36° = 16.18 mi/h
v = √(vE2 + vN2) = √(21.762 + 16.182) mi/h = 27.12 mi/h
θ = tan-1(vN/vE) = tan-1(16.18/21.76) = 36.6º north of east
Answer:
one day 2 babies were born in 1 hour home babie was 0.28LBS less heavier that the baby that was 13.44LBS whats the difference between the babies?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given is a paired data which consist of temperatures (X in mm) and growth
We have to find the linear correlation i.e. the measure of association between these two variables.
x y xy x^2 y^2
62 36 2232 3844 1296
76 39 2964 5776 1521
50 50 2500 2500 2500
51 13 663 2601 169
71 33 2343 5041 1089
46 33 1518 2116 1089
51 17 867 2601 289
44 6 264 1936 36
79 16 1264 6241 256
Mean 58.88888889 27 1623.888889 3628.444444 916.1111111
cov 33.88888889
std dev x 13.43916333 14.50861813
sx *sy
r 0.195529176
Hence we find that correlation coefficient 0.1955.
Answer:
m∠QPM=43°
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠NPQ=m∠MPN+m∠MPQ
we have
m∠NPQ=(9x-25)°
m∠MPN=(4x+12)°
m∠MPQ=(3x-5)°
substitute the given values and solve for x
(9x-25)°=(4x+12)°+(3x-5)°
(9x-25)°=(7x+7)°
9x-7x=25+7
2x=32
x=16
Find the measure of angle QPM
Remember that
m∠QPM=m∠MPQ
m∠MPQ=(3x-5)°
substitute the value of x
m∠MPQ=(3(16)-5)=43°
therefore
m∠QPM=43°
