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ExtremeBDS [4]
3 years ago
10

One glass of milk provides 305mg of calcium. Maggie drinks one glass of milk every morning. How many grams of calcium does Maggi

e get from her milk in 14 days?
Mathematics
2 answers:
Komok [63]3 years ago
8 0
She drinks 4.27 grams on calcium in 14 days at one glass of milk containing 14 mg per day
Alik [6]3 years ago
3 0

Answer:

4.27grams

Step-by-step explanation:

The initial information we have can be represented in one table:

Glasses of milk        Calcium

           1               ⇒    305mg

and we are told that Maggie drinks one gass of milk every morning.

that means that in 14 days she has drunk 14 glasses of milk,  I will represent  the amount of calcium in 14 glasses of milk by x :

Glasses of milk        Calcium

           1               ⇒    305mg

          14               ⇒    x

We solve this problem by multiplying the cross quantities in the table (305 and 14):

x= (305mg)(14)

x = 4270mg

but we need the answer in grams not in miligrams.

So we make the conversion, since 100mg = 1g

4270mg = 4.27g

The answer is 4.27grams

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I wanna say 1/3 because there 3 even numbers on a dice

Step-by-step explanation:

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Step-by-step explanation:

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2.8

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7 0
3 years ago
Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro
Elanso [62]

Answer:

<em>Choose the first alternative</em>

\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}

Step-by-step explanation:

<u>Probabilities</u>

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions (N_s) and the total number of possible choices (N_t), i.e.

\displaystyle P=\frac{N_s}{N_t}

There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C} \text{ ways to do it}

The total number of possible choices is

_{2}^{20}\textrm{C} \text{ ways to do it}

The probability is then

\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}

Choose the first alternative

7 0
3 years ago
Read 2 more answers
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