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4 years ago

Evaluate the following expression if x = 23, y = 0.6 and z = -9:2(x - y)

Mathematics

1 answer:

MAVERICK [17]4 years ago

7 0

Answer:

201.6 im not sure but if u put this: and meant to put a * multiplacation simble then this should be right.

Step-by-step explanation:

x=23 y=0.6 z=-9 (x-y) = ( 23-0.6 ) = 22.4*9 = 201.6

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zaharov [31]

Answer:

2x² - 14x + 24

Step-by-step explanation:

Use the FOIL method (First, Outside, Inside, Last).

First, multiply the first term of both parenthesis:

2x * x = 2x²

Next, multiply the outside terms of both parenthesis:

2x * -4 = -8x

Then, multiply the inside terms of both parenthesis:

-6 * x = -6x

Finally, multiply the last terms of both parenthesis:

-6 * -4 = 24

Combine like terms:

2x² - 8x - 6x + 24

2x² (-8x - 6x) + 24

2x² (-14x) + 24

2x² - 14x + 24 is your answer.

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stiks02 [169]

Answer:

Mean = 30516.67

Standard deviation, s = 3996.55

P(x < 27000) = 0.0011518

Step-by-step explanation:

Given the data:

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Mean, xbar = Σx / n = 366200 /12 = 30516.67

Standard deviation, s = [√Σ(x - xbar) / n-1]

Using calculator, s = 3996.55

The ZSCORE = (x - mean) / s/√n

Zscore = (27000 - 30516.67) / (3996.55/√12)

Zscore = - 3516.67 / 1153.7046

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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viva [34]

Answer:

The critical value of z for 99% confidence interval is 2.5760.

The 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

Step-by-step explanation:

Let X = number of lightning strikes on each day.

A random sample of n = 23 days is selected to observe the number of lightning strikes on each day.

The random variable X has a sample mean of, $\bar x=8.1\ mn$ and the population standard deviation, $\sigma=0.51\ mn$ .

The (1 - α)% confidence interval for population mean μ is:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

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*Use a z-table.

The critical value of z for 99% confidence interval is 2.5760.

Compute the 99% confidence interval for population mean number of lightning strike as follows:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=8.1\pm2.5760\times\frac{0.51}{\sqrt{23}}\\=8.1\pm0.2738\\=(7.8262, 8.3738)\\\approx(7.83, 8.37)$

Thus, the 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

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