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topjm [15]
3 years ago
15

The perimeter of a rectangular field is 344m.if the length of the field is 93m.what is it’s width?

Mathematics
2 answers:
Alisiya [41]3 years ago
6 0

Step-by-step explanation:

perimeter = 344

length = 93

perimeter of rectangle = 2( l+ b )

344 = 2 (93 + b)

344÷2 = 93 + b

172 = 93 + b

172 - 93 = b

79 = b

in other words width is 79

hope it helps

snow_tiger [21]3 years ago
4 0

Answer:

79m

Step-by-step explanation:

344-(93x2)=158

158/2=79m

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Step-by-step explanation:

<em>Step 1: Define significance level</em>

In this hypothesis testing problem, significance levels α is selected: 0.05, the associated z-value from Laplace table:

Φ(z) = α - 0.5 = 0.05 - 0.5 = -0.45

=> z = -1.645

<em>Step 2: Define null hypothesis (</em>H_{0}<em>) and alternative hypothesis (</em>H_{1}<em>)</em>

H_{0} : rate of flu infection p = 8.3% or 8.3/100 = 0.083

H_{1} : rate of flu infection p < 8.3% or 8.3/100 = 0.083

<em>Step 3: Apply the formula to check test statistic:</em>

K = \frac{f - p}{\sqrt{p(1 - p)} } * \sqrt{n}

with f is actual sampling percent, p is rate of flu infection of H_{0}, n is number of samples.

The null hypothesis will be rejected if K < z

<em>Step 4: Calculate the value of K and compare with </em>z

K = \frac{(\frac{3.5}{100})  - 0.083}{\sqrt{0.083(1 - 0.083)} } * \sqrt{200} = -2.46  

We have -2.46 < -1.645

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Step-by-step explanation:

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Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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}}.

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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This means that p = 0.6

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This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

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p-value of Z when X = 0.5. So

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

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