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Aneli [31]
3 years ago
7

What is s²t-10 if s=-8 and t=

\frac{3}{4}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
FrozenT [24]3 years ago
7 0

Hi ;-)

\text{s}=-8 \ \text{and} \ \text{t}=\frac{3}{4}\\\\\text{s}^2\text{t}-10=(-8)^2\cdot\frac{3}{4}-10=\\\\=64\cdot\frac{3}{4}-10=16\cdot3-10=48-10=\boxed{38}

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Help plz i have a f in this class
Mnenie [13.5K]

Answer:

For number 3 the answer is -4

4 0
2 years ago
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2x-(10+4x)=3x-7 what is the answer
ZanzabumX [31]
X=-3/5

    3
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 X=-3/5 would be the answer
6 0
3 years ago
Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample
fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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.

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}}

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

Z = \frac{0.38 - 0.37}{0.0322}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

Z = \frac{0.26 - 0.37}{0.0322}

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

5 0
3 years ago
How much interest is earned on $300 at 6% for 6 years?
miskamm [114]

Answer:

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Step-by-step explanation:

First, let's convert 6% to a decimal.

We get 0.06 because we multiplied (or divided by 100) it by 1/100.

Now, let's do 0.06 * 300 to get interest on one year. This equals to 18$

Let's now do 18 * 6 to get

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3 years ago
What is the degree of 6g^2h^3k, is it a monomial
Igoryamba

Answer:

6

Step-by-step explanation:

7 0
3 years ago
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