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OLEGan [10]
3 years ago
9

John was building a wall out of bricks the weight of the wall was 543.75

Mathematics
2 answers:
Ilya [14]3 years ago
8 0

Answer:

John was building a wall out of bricks. The weight of the wall was 543.75 pounds. Each brick weighed 4.35 pounds. How many bricks did John use?

Step-by-step explanation:

You have to divide.

adell [148]3 years ago
3 0
Is there a picture?
To go with this question?
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John is about to roll a six-sided, fair number cube 60 times. He wants to predict how many times the cube will land on an even n
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Even numbers
so each dice has numbers
1,2,3,4,5,6
the eve numbers are
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odd are others
1,3,5
so 1/2 are odd

therefor 1/2 of 60 is how many times it would get even
1/2 of 60=1/2 times 60=30

No not good predicition
my prediction would be 30 tiemes

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18kg is the correct answer

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Desert, tundra, ice, and mountains make up 24% of Earth's surface. Which fraction is equivalent to 24%?
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Answer:

6/25

Step-by-step explanation:

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Which graph shows a negative rate of change for the interval 0 to 2 on the x-axis?
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According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif
suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

P(0.79

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3 years ago
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