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kap26 [50]
3 years ago
14

A length of glass tubing is 10 cm long. What is its length in inches to the nearest inch?

Mathematics
2 answers:
Lelechka [254]3 years ago
6 0
3.93701 inches exactly or to the nearest inch is 4 inches
lapo4ka [179]3 years ago
3 0

Answer:

4\ in

Step-by-step explanation:

we know that

1\ in =2.54\ cm

so

By proportion

Convert 10\ cm to inches

\frac{1}{2.54}\frac{in}{cm}=\frac{x}{10}\frac{in}{cm}\\ \\x=10/2.54\\ \\x=3.94\ in

Round to the nearest inch

3.94\ in=4\ in

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In a sample of eight whitefish caught in Yellowknife Bay, the mean arsenic concentration in the liver was 0.32 mg/kg, with a sta
Advocard [28]

Answer:

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 7

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.3246

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.3646\frac{0.06}{\sqrt{8}} = 0.0502

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg

The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

8 0
3 years ago
20 PTSS !!!<br> PLEASE HELPP !!
butalik [34]

Because it is extremely hard to find the area of this figure all together, it would be in our best interest to split this figure up into three different pieces: the two horizontal rectangles, and the verticle rectangle. We can find the area of all three and add them up. Be aware that there are two different ways that you can break this figure up, As shown in the attachments. I will be using the first image (the one with the tall horizontal rectangles, NOT the almost-squares).

So, we see that we have enough information to solve for the area of the left-most rectangle. Area = lw. 10 x 4 = 40, so the area is 40. Next, we have to notice, that the horizontal rectangles are also the same, so both of the areas of the two horizontal rectangles are 40.

Now, we can find the middle rectangle. We know that the length of the entire thing is 18, but it is taken up by 8 (4+4) of the horizontal triangles, so 18-8=10, so the length Is 10. We also know that the height of the horizontal rectangles is 10, so 10-3=7. Our dimensions for the rectangle are 10x7 or 70 square units. If we add them all together, 40+40+70=150.

The area is 150 square units

8 0
3 years ago
The diagram shows a right-angled triangle.
crimeas [40]

Answer:

Maybe try 23-9. Or maybe the answer is 23, hopefully this helped.

Step-by-step explanation:

8 0
3 years ago
Sphere with a volume of 1767.1 . what is the radius?
Llana [10]

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=1767.1 \end{cases}\implies 1767.1=\cfrac{4\pi r^3}{3}\implies 5301.3=4\pi r^3 \\\\\\ \cfrac{5301.3}{4\pi }=r^3\implies \sqrt[3]{\cfrac{5301.3}{4\pi }}=r\implies 7.4999\approx r

3 0
3 years ago
X - (2x + 1) = 8 - (3x + 2)
vlada-n [284]

Answer:

x=7/2

Step-by-step explanation:

7 0
3 years ago
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