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Kobotan [32]
3 years ago
9

the total cost for material and labor for performing a blood test is $38.52. if the lab adds 15 2/5% to this cost for profit, wh

at is the charge for the blood test to a patient?​
Mathematics
2 answers:
just olya [345]3 years ago
6 0

Answer:

$44.45

Step-by-step explanation:

Convert 15 2/5% to decimal, =0.154

38.52+(38.52x0.154) (do this to add the cost of the % increase)

≈$44.45.

HACTEHA [7]3 years ago
3 0

Answer:

$44.45

Step-by-step explanation:

Convert 15 2/5% to decimal, =0.154

38.52+(38.52x0.154) (do this to add the cost of the % increase)

≈ $44.45.

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nekit [7.7K]
You would find the probability of getting an answer less than 10 by counting all of the possible outcomes, or numbers that are less than 10 and compare them to the total number of possible outcomes, or the total number of numbers. You would write the ratio of predicted outcomes to total possibilities.

The answer would be 4/9.
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3 years ago
Belle bought 18 seeds in order to plant an herb garden for her grandma. Of the seeds she bought, 10 were parsley seeds.
Harrizon [31]

Answer:

  0.3431

Step-by-step explanation:

Here, it can work well to consider the seeds from the group of 18 that are NOT selected to be part of the group of 15 that are planted.

There are 18C3 = 816 ways to choose 3 seeds from 18 NOT to plant.

We are interested in the number of ways exactly one of the 10 parsley seeds can be chosen NOT to plant. For each of the 10C1 = 10 ways we can ignore exactly 1 parsley seed, there are also 8C2 = 28 ways to ignore two non-parsley seeds from the 8 that are non-parsley seeds.

That is, there are 10×28 = 280 ways to choose to ignore 1 parsley seed and 2 non-parsley seeds.

So, the probability of interest is 280/816 ≈ 0.3431.

___

The notation nCk is used to represent the expression n!/(k!(n-k)!), the number of ways k objects can be chosen from a group of n. It can be pronounced "n choose k".

6 0
3 years ago
(1) Cos AcosecA=cot A​
Mashutka [201]

Answer:

The Basic Identities are :

cosec(A) =  \frac{1}{ \sin(A) }

\tan(A)  =  \frac{ \sin(A) }{ \cos(A) }

\cot(A)  =  \frac{1}{ \tan(A) }  =  \frac{ \cos(A) }{ \sin(A) }

So for this question :

\cos(A)cosec(A) =  \cot(A)

l.s.h =  \cos(A)  \times  \frac{1}{ \sin(A) }

l.s.h =  \frac{ \cos(A) }{ \sin(A) }

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Let x = total games played

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7 0
3 years ago
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This equation could represent the relationship e=8.75h
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