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Stella [2.4K]
2 years ago
10

An oddly-shaped water bottle holds 30.5 L of water. If the water bottle is 2/3 full, how many L are in the water

Mathematics
1 answer:
Sveta_85 [38]2 years ago
5 0

Answer: <em>2/3 of 30.5 is 20.3333333333</em>

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How to find the scale factor of two triangles
BabaBlast [244]

Answer:

Here's how to do it.  

Step-by-step explanation:

The two triangles must be similar, that is, corresponding angles must be equal, as in the diagram below.

1. Find a corresponding pair of sides

Assume that ∆ABC is the original and ∆DEF is its scaled image.

Then, AB and DE are corresponding sides, with lengths of 8 and 12 units , respectively.

2. Set up a ratio

Ratio = side length in image/side length in original = 12/8

3. Simplify the ratio

The simplified ratio is the scale factor (SF).

SF = 12/8 = 3/2 = 1.5

The scale factor is 1.5.

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3 years ago
The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minus.
AnnyKZ [126]

Answer:

Step-by-step explanation:

Given that the  time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

P(completing exam before 1 hour)

= P(less than an hour) = P(X<60)

=P(Z<\frac{60-70}{10} =-1)

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3 years ago
A consequence can be defined as:
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3 years ago
Read 2 more answers
How many half-lives of radon-222 have passed in 11.46 days? If 5.2 × 10−8 g of radon-222 remain in a sealed box after 11.46 days
Svetach [21]

Answer:

(i) Approximately 3 half lifes

(ii) 4.21\times 10^{-7}\text{ g}

Step-by-step explanation:

(i) ∵ The half life of Radon-222 is approximately 3.8 days,

So, the number of half life in 11.46 days = \frac{11.46}{3.8} ≈ 3

(ii) Since, the half life formula is,

N=N_0 (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

Where,

N_0 = initial quantity,

t = number of periods

t_{\frac{1}{2}} = half life of the quantity,

Given,

N = 5.2\times 10^{-8}\text{ g}

t = 11.46 days,

t_{\frac{1}{2}} = 3.8\text{ days}

\implies 5.2\times 10^{-8}=N_0 (\frac{1}{2})^\frac{11.46}{3.8}

\implies N_0=2^{\frac{11.46}{3.8}}\times 5.2\times 10^{-8}\approx 4.21\times 10^{-7}\text{ g}

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4 years ago
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