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olchik [2.2K]
2 years ago
12

What is the difference of the polynomials? (x2 - 6x + 2) - (3x? – 7x-8)

Mathematics
1 answer:
bagirrra123 [75]2 years ago
4 0
The difference is 10.
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Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0
Deffense [45]
\bf sin(x)cos(x)=0\implies 
\begin{cases}
sin(x)=0\\
\measuredangle x=sin^{-1}(0)\\
\measuredangle x=0\ ,\ \pi \\
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cos(x)=0\\
\measuredangle x=cos^{-1}(0)\\
\measuredangle x=\frac{\pi }{2}\ ,\ \frac{3\pi }{2}
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Rewrite the equation by completing the square.
ser-zykov [4K]

Answer:

4 × 2 + 20×

Step-by-step explanation:

please your questions are mixed up

4 0
3 years ago
An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

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 question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
3 years ago
Elsie is going to flip a coin and spin a spinner. The coin has a heads and a tails. The spinner has four equal parts that are ye
Vedmedyk [2.9K]

Answer:

1/6

Step-by-step explanation:

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2 years ago
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Wat is 10,000 more than 6,310
ziro4ka [17]
To find 10,000 more than 6,310, you add; so the answer is 16,310
3 0
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