83 POINTS !!! I NEED HELP WITH THIS PLZZ DONT SCAM ME I NEED HELP WITH THIS!!! PLZZ HELP ME OUT

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

6 0

Question 1

Lets convert these to decimals first:

2 1/2 = 2.5

1/4 = 0.25

Now divide 2.5 by 0.25 to find how many scoops Zack uses.

If you divide correctly, you get 10.

Now multiply this by two becuase Zack has TWO bird feeders.

So Zach uses 20 scoops to fill both bird feeders.

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Question 2

Let x be the variable for the scoops Zack needs for both bird feeders

x = 2 $\frac{1}{2}$ / $\frac{1}{4}$ x 2

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Question 3

x = 2 $\frac{1}{2}$ / $\frac{1}{4}$ x 2

2 $\frac{1}{2}$ divided by $\frac{1}{4}$ = 10

x = 10 x2

10 multiplied by 2 is 20

So...

x = 20

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The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$