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Likurg_2 [28]
2 years ago
7

The solution of the following equation as 0.15 ( 10t - 9= 0.75 (4t - 3)​

Mathematics
1 answer:
Alenkinab [10]2 years ago
4 0

Answer:

\huge\boxed{\sf t = 0.6}

Step-by-step explanation:

\sf 0.15(10t-9) = 0.75(4t-3)\\\\Resolving \ Parenthesis\\\\1.5t-1.35 = 3t - 2.25\\\\Combining \ like \ terms\\\\-1.35 + 2.25 = 3t-1.5t\\\\0.9 = 1.5t\\\\Dividing \ both \ sides \ by \ 1.5\\\\0.9/1.5 = t\\\\0.6 = t\\\\OR\\\\\bold{t = 0.6}

Hope this helped!

<h2>~AnonymousHelper1807</h2>
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In an exam 35 out of 125 students failed. What percentage passed the exam?​
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Answer:

28%

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Hope this helps you out! : )

4 0
2 years ago
Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

Z = \frac{46 - 40}{5.66}

Z = 1.06

Z = 1.06 has a pvalue of 0.8554

X = 34

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

Z = \frac{34 - 40}{5.66}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0
2 years ago
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