The answer you're looking for is Panda because a panda has an organ system, the other choices are all microscopic organisms so they don't have an organ system
Hope this helped you out :)
Answer: Increase the depth of ventilation.
In high altitude, the oxygen pressure will be much lower that makes the gas exchange rate is lower and decrease the oxygen delivery to <span>the tissue.</span> To counter this, the mouse will increase its depth of ventilation so there is more air inside the lungs for every inspiration.
Answer:
Remain with the client and be silent
Explanation:
Bowel obstruction can be explained or described as a situation whereby the intestines, i.e, both small and large intestines are blocked, and thereby, preventing the normal movement of food, liquid or stool in the intestines.
The blockage could result or occur due to tumor, foreign substance or digestive problem.
The Miller-Abbott tube is the tube that was invented or developed to treat the obstruction that is present in both the small and the large intestines. It is used by passing it through the stomach into the intestine.
In this case, the bowel obstruction is caused by recurrent malignancy, that is recurrent of formation of tumor.
The best thing the nurse should do, is to remain silent and with the client, until the client is ready for explanation on the process, and that he/she is in position and capable of inserting the tube.
The answer is Integumentary System. it is the organ framework that shields the body from different sorts of harm, for example, loss of water or scraped spot from outside. The framework involves the skin and its limbs.
The integumentary system comprises of the skin, hair, nails, organs, and nerves. Its principal work is to go about as a hindrance to shield the body from the outside world. It additionally capacities to hold body liquids, ensure against ailment, dispense with squandering items, and direct body temperature.
Answer:
Explanation:
Bacterial count in stock- 1.85x10^6 cfu/ml
Dilution methods
Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.
Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.
Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.
Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml
So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.
If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.
