Please help, i will give brainliest to whos first

Mathematics

1 answer:

Wittaler [7]3 years ago

6 0

Thanks with all of my heart!

ik im not first but have a lovely day! u are really kind! never forget that!

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A) 6 square meters<br> B) 16 square meters <br> C) 28 square meters <br> D) 49 square meters

coldgirl [10]

C 28 square meters smokejumper

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3 years ago

For every pound a company spends on advertising, it spends £0.43 on its website. Express the amount spent on advertising to its

Wittaler [7]

Answer:

£1 : £0.43

Step-by-step explanation:

So to make the ratio, you need the amount of money spent on advertising and the money spent on its website

Website money = £0.43

Advertising money = £1

£1 : £0.43

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3 years ago

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strojnjashka [21]

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Step-by-step explanation:

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3 years ago

A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of

Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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