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shtirl [24]
3 years ago
6

Solve for 34 is equal to 17/x

Mathematics
2 answers:
Margarita [4]3 years ago
8 0

Answer:

x = 0.5

Step-by-step explanation:

<u>Step 1:  Convert words into an expression</u>

34 is equal to 17/x

34 = 17/x

<u>Step 2:  Multiply both sides by x</u>

34 * x = 17/x * x

34x = 17

<u>Step 3:  Divide both sides by 34</u>

34x / 34 = 17 / 34

<em>x = 0.5</em>

Answer:  x = 0.5

Sophie [7]3 years ago
5 0

Answer:

1/2

Step-by-step explanation:

34=17/x

x=17/34=1/2

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A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of
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Answer:

Ix = Iy = \frac{ρπR^{4} }{16}

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Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

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4 years ago
The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

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