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2 years ago

A guy bikes 15 miles in 1 hour, then rests for an hour. Then he bikes 25 in 2 hours. What was his average speed for the trip?

1 answer:

4 0

He went 15 miles/hour in the first sequence, and then 12.5 miles/hour (25 miles/2 hours) in the second sequence. If you take into account the hour he rested between segments: 15 miles/hour, 0 miles/hour, and 12.5 miles/hour. So (15+0+12.5)/3 = 9.17 miles/hour is his average. However, if you are not taking into account the whole trip, and only the time he was biking: (15+12.5)/2 = 13.75 miles/hour

You might be interested in

Give role of biochips and biofertilizers?

Stels [109]

Answer:

Biochips The biochip is used to simultaneously analyze a panel of related tests in a single sample, producing a patient profile. The patient profile can be used in disease screening, diagnosis, monitoring disease progression or monitoring treatment.

Biofertlizers: Biofertilizers add nutrients through the natural processes of nitrogen fixation, solubilizing phosphorus, and stimulating plant growth through the synthesis of growth-promoting substances. The microorganisms in biofertilizers restore the soil's natural nutrient cycle and build soil organic matter.

Explanation:

3 0

2 years ago

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent as

ASHA 777 [7]

Answer:

(a) AABBCC × aabbcc->AaBbCc → 1

(b) AABbCc × AaBbCc->AAbbCC → 1/32

(d) aaBbCC × AABbcc->AaBbCc → 1/2

Explanation:

In such cases we calculate the probability of each allelic pair separately.

(a) AABBCC × aabbcc -> AaBbCc

Aa Aa Aa Aa Bb Bb Bb Bb Cc Cc Cc Cc

↓ ↓ ↓

4/4 = 1 4/4 = 1 4/4 = 1

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 1 x 1 = 1.

(b) AABbCc × AaBbCc -> AA bb CC

AA AA Aa Aa BB Bb Bb bb CC Cc Cc cc

↓ ↓ ↓

2/4 1/4 1/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.

In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.

So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.

AA Aa Aa aa BB Bb Bb bb CC Cc Cc cc

↓ ↓ ↓

2/4 2/4 2/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.

So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.

(d) aaBbCC × AABbcc->AaBbCc

Aa Aa Aa Aa BB Bb Bb bb Cc Cc Cc Cc

↓ ↓ ↓

4/4 = 1 2/4 4/4 = 1

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.

5 0

3 years ago

According to this food web, which of the following would be considered secondary consumers?

mamaluj [8]

It is the grasshopper because it is

4 0

3 years ago

Which of the three balls pictured below have gravitational potential energy.

kumpel [21]

Answer:

i believe the basketball falling from the net

Explanation:

5 0

2 years ago

Although most solutes that enter or exit the cell are relatively small (inorganic ions, sugars, or amino acids), occasionally th

TEA [102]

Complete question:

Although most solutes that enter or exit the cell are relatively small (inorganic ions, sugars, or amino acids), occasionally the movement of larger molecules into or out of the cell is required. For example, certain cells secrete proteins, such as pancreatic cells that secrete insulin. Other cells take up food particles or even entire cells, such as lymphocytes in your immune system. Transport of such large materials into and out of the cytosol cannot be accomplished by transport proteins in the plasma membrane. The transport of large molecules is accomplished by bulk transport via exocytosis and endocytosis. Like other forms of active transport, bulk transport requires an input of energy. Before beginning this tutorial, watch the segment of the Membrane Transport animation on exocytosis and endocytosis. Observe the role that vesicles play in the transport of materials into and out of the cell.

Part A

Sort the phrases into the appropriate bins depending on whether they describe exocytosis, endocytosis, or both.

a. decreases the surface area of the plasma membrane

b. requires fusion of vesicles with the plasma membrane

c. forms vesicles from inward folding of the plasma membrane

d. transported substances never physically cross the plasma membrane

e. increases the surface area of the plasma membrane

f. requires cellular energy

g. secretes large molecules out of the cell

Answer:

a. --> ENDOCYTOSIS

b. -->EXOCYTOSIS

c.--> ENDOCYTOSIS.

d. --> BOTH

e. --> EXOCYTOSIS

f. --> BOTH

g. --> EXOCYTOSIS

Explanation:

To answer the question, first, we need to know the meaning of endocytosis and exocytosis.

Endocytosis and exocytosis process refers to molecules transport mediated by vesicles. It occurs from and into the cell and needs energy to happen because it is active transport.

Cells can export many substances in vesicles or vacuoles that originate in specific organelles like, for instance, the Golgi complex. Organelles form new vesicles or vacuoles with substances destined for exportation and send them forward to the membrane. When they get there, they fuse to the cell membrane, releasing all their content to the extracellular space. When fussing, the vesicle membrane increases the size of the cellular membrane. This process is <u>exocytosis</u>, and its direction is always from the inside of the cell to the outside.

But there is also another vesicle-mediated transport that occurs from the exterior to the cell interior, the <u>endocytosis process. </u>It happens when the cell imports material from the extracellular space through the invagination of the membrane, producing a new vesicle that surrounds the incoming material. The invagination ends in strangulation of the membrane, releasing the new vesicle in the cytoplasm. This process decreases the size of the cellular membrane because it uses part of it to engulf the material.

Now that we know these concepts, we can sort the phrases into the appropriate bins.

a. decreases the surface area of the plasma membrane. ENDOCYTOSIS, because the new vesicles are formed using the cell membrane

b. requires fusion of vesicles with the plasma membrane. EXOCYTOSIS, because vesicle needs to fuse to the cell membrane to release the carrying material

c. forms vesicles from inward folding of the plasma membrane. ENDOCYTOSIS. Refers to the invagination process of the cell membrane. It only occurs during endocytosis

d. transported substances never physically cross the plasma membrane. BOTH. Substances are too big or too many to be physically transported through the cell membrane

e. increases the surface area of the plasma membrane. EXOCYTOSIS. Occurs when the vesicle fuses the cell membrane to release the content to the exterior.

f. requires cellular energy. BOTH of them, because they are active transport.

g. secretes large molecules out of the cell. EXOCYTOSIS. The direction is from the interior of the cell to the exterior. These are exportation molecules.

8 0

3 years ago