Given: The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (Hg) and a standard deviation of 18 mm of Hg.
To find: P(100 < 152)
Method: Calculation of Z-Score followed by the probability or area of the bell curve at X = 100.
Solution:
Mean u = 152, std s = 18
Z score = 
The value of P(100<152) is calculated by looking at the value of Z in the Z score for the standard normal distribution given in the image.
P(Z=-2.89) = 0.0019
The P(Z = -2.89) corresponds to the area in the left tail of the bell curve.
Thus the probability of 100 mm Hg blood pressure is 0.0019.
Space satellites
Environmental
TritionalshiftsofEscherichiacoliB/rtorichermediahavebeenanalyzedinsynchronouslygrowingandexponential-phasepopulations.Earlyperturbationsinthetimingofcelldivisionwereobserved.Attheslowgrowth,divisionpro-gressedatarateequaltoorlessthanthepreshiftrateforabout1h.Atintermediategrowth,bothdelaysandaccelerationindivisionwereobserved.Theextentoftheperturbationdependedupontheageofthecellsatthetimeoftheshiftandthecompositionofthepreshiftandpostshiftmedia.TheperturbationwasdifferentinthetwosubstrainsofE.coliB/r I got this from http://jb.asm.org/content/136/2/631.full.pdf hopefully it helps you
Answer:
theres 2 answers actually
Explanation:
DNA TO RNA :
5’ CGG UAA UC 3’
DNA TO DNA :
5’ CGG ATT TC
Answer:
1 in 16
Explanation:
When two heterozygotes (AaBb x AaBb) for two autosomal genes are crossed the expected probability for the offspring is 9 A-B-, 3 A-bb, 3 aaB- and 1 aabb. In other words, out of sixteen offspring, 9 are expected to be dominant on both genes (either homozygous AA or BB or heterozygous Aa or Bb), 3 are expected to be dominant on the A gene but recessive on the b gene, 3 are expected to be recessive on the a gene but dominant on the A gene, and only 1 is expected to be recessive on both genes aabb.