Answer:
(C) 2√15
Step-by-step explanation:
Recognize that all the triangles are right triangles, so are similar to each other. In these similar triangles, the ratio of the short side to the long side is the same for all.
... CB/CA = CT/CB
... CB² = CA·CT = 10·6 = 60 . . . . . . . . . . multiply by CA·CB; substitute values
... CB = √60 = 2√15 . . . . . . . take the square root; simplify
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<em>Comment on this solution</em>
The altitude to the hypotenuse of a right triangle (CB in this case) divides the hypotenuse into lengths such that the altitude is their geometric mean. That is ...
... CB = √(AC·CT) . . . . as above
This is true for any right triangle — another fact of geometry to put in your list of geometry facts.
Cos(x) = sin(90 - x)
cos(53) = sin(90 - 53)
cos(53) = sin(37)
If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.
Calculation for the Amount of IV
It is given that,
1L IV contains 60meq of kcal
⇒ 1000 mL of IV contains 60meq of kcal
⇒ 1 mL of IV will contain 60 / 1000 meq of kcal
As per the question,
The amount of IV infused in the patient = 400 mL
⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)
= 4 × 6
= 24 meq of kcal
Hence, the patient receives 24meq of kcal amount of IV.
Learn more about amount here:
brainly.com/question/4567186
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Answer:
7
Step-by-step explanation:
Multiply the first collum by 7. It's basically unit rates
We should first calculate the average number of checks he wrote
per day. To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463. This will be λ in our Poisson distribution. Our formula is
. We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.
To find P(X≥1), we find 1-P(X<1). Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0). Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.<em />
