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3 years ago

Write each decimal as a percent. 3. 0.35. 4. 0.812

Mathematics

2 answers:

ruslelena [56]3 years ago

5 0

Answer:

30%.34 81%.02

Step-by-step explanation:

Olenka [21]3 years ago

4 0

Answer:

3.=300%

0.35=35%

4.=400%

0.812=81.2%

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a large painting has a length of 18 inches and width of 12 inches. If each dimension is reduced by x inches to make the ratio of

KiRa [710]

Answer:

The value of x = 3.

Step-by-step explanation:

The ratio between length and width is given by:

$\frac{L}{W}=\frac{5}{3}$

Where:

L represents the new length
W represents the new width

Now, we know that each dimension is reduced by x inches, it means:

$L=18-x$

$W=12-x$

Then, we will have:

$\frac{18-x}{12-x}=\frac{5}{3}$

Finally, we just need to solve it for x.

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$x=3$

Therefore, the value of x = 3.

I hope it helps you!

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3 years ago

Steven has 24 t-shirts. One third of the t-shirts are white, one fourth of the t-shirts are blue ; the remainder of the t-shirts

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3 years ago

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It fo

Harman [31]

Answer:

(a) The 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b) The sample size required is 107.

Step-by-step explanation:

(a)

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

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

Given:

$\bar x=\$10,979\\s=\$1000\\n=20$

Compute the critical value of t for 90% confidence level as follows:

$t_{\alpha/2, (n-1)}=t_{0.10/2, (20-1)}=t_{0.05, 19}=1.729$

*Use a t-table.

Compute the 90% confidence interval for population mean as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

$=10979\pm 1.729\times \frac{1000}{\sqrt{20}}\\=10979\pm4.47\\ =(10974.53, 10983.47)$

Thus, the 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b)

The margin of error is provided as:

MOE = $250

The confidence level is, 99%.

The critical value of z for 99% confidence level is:

$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$

Compute the sample size as follows:

$MOE= z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times s}{MOE} ]^{2}$

$=[\frac{2.58\times 1000}{250}]^{2}$

$=106.5024\\\approx107$

Thus, the sample size required is 107.

4 0

3 years ago