When two people are in a race, what do you need to know to determine who is the fastest? (Select all that apply.)

Use the terms below to complete the sequence of the levels of organization in a multicellular organism.

Valentin [98]

Cells -> Tissues -> Organs -> Organs systems -> Organisms

6 0

3 years ago

A person whose blood tests show neither protein A nor protein B present has blood type: 1.O 2.A 3. AB URGENTTTTTTT IN A QUIZ

Llana [10]

1, O because if either shows present it's O.

8 0

3 years ago

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0

3 years ago

Allen Dexter, a 19-year-old college student, was rock climbing when he fell 30 feet to the ground. Paramedics arriving at the sc

a_sh-v [17]

Answer:

Explanation:

A cervical collar was used to avoid movement and prevent further damage incase he has a fracture.

light is focused in one eye,the pupil will constrict as a respond to the light flashed. It is repeated Again and observed.

The vitals were normal until he had the accident as the patient reached hospital his vitals were abnormal he had achycardia, severe hypotension, increased temperature and shallow breathing which is an indicator of C5 injury. After the surgery and the patient was treated he stabilises and achieved his vitals normal ranges in day 4 of admission.

The X ray result shows he has fracture of the 5th cervical bone.

The blood test shows he has a decreased pH level and an alteration in

his breathing pattern shows he has respiratory acidosis

Cranial tongs was uses to immobilize the neck and prevent it from moving it also aid healing

After the operation it was revealed that he has a damage on both sides of of his 5th Cervical bone which has lead to loss of sensation and strength in his upper arms and lower .It leads to incontinence in urine.

6 0

3 years ago

Two atoms with an electronegativity difference of 2.1, will form a bond that is:

pogonyaev

Covalent because electrons are shared

4 0

3 years ago