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Margaret [11]
3 years ago
13

Factor 49a– 14a +1. 49a– 14a + 1 =

Mathematics
1 answer:
Advocard [28]3 years ago
6 0
(7a+1)^2 hope this helps
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It’s around 450 euros right now - to figure this out you multiply the number of pounds by the current exchange rate ( which you can look up)
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What percent of 40 is 120?​
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the correct answer is 33.33%

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Two sides of a triangle are 7 and 11 the length of the third side x is expressed
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The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ... No; The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: No; 9.

8 0
3 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame
lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 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 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 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.

In this problem, we have that:

\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

Z = \frac{207.9 - 208}{0.1678}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0
3 years ago
Graph the first six terms of a sequence where a1=5 and r=1.25
emmainna [20.7K]
The sequence above is geometric progression. 
The nth term of such sequence is given by;

Tn = ar∧(n-1),
Where a⇒first term and
            r⇒common ratio

So, 1st term = 5×1.25∧(1-1) = 5×1.25∧0 =5 
      2nd term = 5×1.25∧(2-1) = 5×1.25 = 6.25
      3rd term = 5×1.25∧(3-1) = 5×1.25² = 7.8125
       4th term = 5×1.25∧(4-1) =5×1.25³ = 9.765625
       5th term = 5×1.25∧(5-1) = 5×1.25∧4 = 12.20703125
       6th term = 5×1.25∧(6-1) = 5×1.25∧5 = 15.25878909
7 0
3 years ago
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