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elixir [45]
2 years ago
8

How do we convert kilogram to gram​

Mathematics
1 answer:
Sloan [31]2 years ago
5 0

<u>1kg=1000g</u>

7kg*1000=7000g

10kg*1000=10000g

15g*1000=15000g

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Divide <br> 8 2/5 ÷ (-2 1/5)
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8 2/5 ÷(-2 1/5) = -3.81818181818

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3 years ago
Please help brainiest for correct answer please i need this done quick
babunello [35]

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Its the third answer

Step-by-step explanation:

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2 years ago
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A bag contains balls in a radio of 3 5 4 if there are twelve balls how many balls are in the bag!!
Elena-2011 [213]
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It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i
USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 55, \pi = \frac{24}{55} = 0.4364

93% confidence level

So \alpha = 0.07, z is the value of Z that has a pvalue of 1 - \frac{0.07}{2} = 0.965, so Z = 1.81.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0
3 years ago
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If this helped, give me brainliest :)
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2 years ago
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