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Tcecarenko [31]
3 years ago
8

250% converted into a decimal

Mathematics
2 answers:
IgorC [24]3 years ago
6 0
2.5 is the answer in this question that u have just asked
ollegr [7]3 years ago
6 0
2.5
just move the decimal twice to the left, since we are dividing by 100
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Daniel can wash 200 dishes in 30 minutes about how long would it take him to wash 250​
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37.5 minutes

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30/200 = .15

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The weight of National Football League (NFL) players has increased steadily, gaining up to 1.5 lb. per year since 1942. Accordin
GuDViN [60]

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the weight of National Football League (NFL) players now.

The mean weight is, <em>μ</em> = 252.8 lb.

The standard deviation of the weights is, <em>σ</em> = 25 lb.

A random sample of <em>n</em> = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The sample of players selected is quite large, i.e. <em>n</em> = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)

Compute the probability that the sample mean weight will be more than 262 lb as follows:

P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

6 0
3 years ago
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