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Nady [450]
2 years ago
9

Hey guys who needs gf I am lonely​

Biology
2 answers:
zhenek [66]2 years ago
5 0

an echo when the same with us and we can use the form of the cardboard box of light and plural form of the cardboard box of light and plural form of the cardboard box of light and plural form of the cardboard box of light and plural form of the cardboard box of light and plural form of the cardboard box of light and plural form

horsena [70]2 years ago
5 0

Answer:

You're asking for yourself to get banned.

No

Explanation:

#Nova

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The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (hg) and a st
Damm [24]

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 = \frac{X-u}{s} = \frac{100-152}{18} = -2.89

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.

7 0
3 years ago
Some environmental factors are studied by the use of:
Tresset [83]
Space satellites
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6 0
3 years ago
Is nutrition a result of cell division?
Stella [2.4K]
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


4 0
3 years ago
Transcribe the following section of DNA 3’ G C C A T T A G 5’
rjkz [21]

Answer:

theres 2 answers actually

Explanation:

DNA TO RNA :

5’ CGG UAA UC 3’

DNA TO DNA :

5’ CGG ATT TC

3 0
3 years ago
Read 2 more answers
For two autosomal dominant genes, what is the probability that a cross between two heterozygotes (i.e. AaBb x AaBb) will produce
oksian1 [2.3K]

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.

7 0
3 years ago
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