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zheka24 [161]
2 years ago
12

Problem of the Week for April 24, 2017<br /><br /><br />

Mathematics
1 answer:
Lesechka [4]2 years ago
4 0
4/8 gallon chocolate sauce (1/2)
2/8 gallon strawberry sauce (1/4)
2/8 gallon caramel sauce (1/4)

at the end I put the simplified version if thats required
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Nelson Mandela was a South African politician known for fighting racism and inequality. He was president of South Africa until h
user100 [1]
One possible inequality would be

1999-x<0.

If you subtract any year after he stepped down from 1999, the year he stepped down, the result will be a negative number.
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2 years ago
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The zoo needed to raise $25,000. Zoo officials planned a dinner/dance event and charged $300 per couple. What is the fewest numb
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84 the least hope that helps

3 0
2 years ago
QUESTION THREE (30 MARKS) 3.1 The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard
kiruha [24]

Using the normal distribution and the central limit theorem, it is found that  there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

----------------------------------

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

----------------------------------

  • Mean of 700g means that \mu = 700
  • Standard deviation of 21g means that \sigma = 21
  • Sample of 64, thus n = 64
  • <u>For the sampling distribution of the sample mean</u>, the standard deviation is of s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625

The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{695 - 700}{2.625}

Z = -1.905

Z = -1.905 has a p-value of 0.0284.

0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

A similar problem is given at brainly.com/question/22934264

7 0
2 years ago
11/12​−6/1​q+6/5​q−3/1​, Combine like terms to create an equivalent expression.
Nikolay [14]

Answer: -\frac{25}{12} -\frac{24q}{5}

Step-by-step explanation:

\frac{11}{12} -\frac{6}{1} q+\frac{6}{5}q-\frac{3}{1}

\frac{11}{12} -6q+ \frac{6q}{5} -\frac{3}{1}

\frac{11}{12}-6q \frac{6q}{5} -3

(\frac{11}{12} -3)+(-6q+\frac{6q}{5} )

-\frac{25}{12}- \frac{24q}{5}

Simplify three times, collect like terms and simplify again, how to collect like terms see their differences and then put them into groups, for example I put all the numbers without variables on one side and all the numbers with variables on another.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

- Cutiepatutie ☺❀❤

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Solnce55 [7]
First off, let's find the 1st term's value, and the 30th term's value,

\bf a1=3(1)+2\implies a1=5\qquad \qquad \quad   a30=3(30)+2\implies a30=92\\\\&#10;-------------------------------\\\\&#10;~~~~~~~\textit{ Sum of an arithmetic sequence}\\\\&#10;S_n=\cfrac{n(a1+an)}{2}~ &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a1=\textit{first term's value}\\&#10;----------\\&#10;a1=5\\&#10;a30=92\\&#10;n=30&#10;\end{cases} \implies S_{30}=\cfrac{30(5+92)}{2}&#10;\\\\\\&#10;S_{30}=15(97)
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