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3 years ago

In what ways is the measurement 3.2 m like 3.20 m? How isit different?

Mathematics

1 answer:

stich3 [128]3 years ago

7 0

Answer:

3.20 is much more accurate than 3.2 .

Step-by-step explanation:

3.2 m is way different than 3.20 m , It is because

We can't measure any length accurately , and thus we are having different devices used to measure different lengths with different precision .

In the case of 3.2 , device used can accurately measure the length one place after the decimal. In this device we can't say the digit after 2 means that it can be 1,2,.... . Thus this device is less accurate as compared to second one.

In case of 3.20 , device can accurately measure the length two places after the decimal and in this device we can confidentially say that digit after two is 0. This device is more accurate than First one.

Thus, they are different because 3.20 is much more accurate than 3.2 .

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Step-by-step explanation:

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3 years ago

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3 years ago

Which number is relatively prime to 63? A. 114 B. 99 C. 87 D. 25

ElenaW [278]

Answer:

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Step-by-step explanation:

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4 years ago

A circle has an area of 75 square centimeters. Which answer is closest to the measure of its radius?

REY [17]

Answer:

5 √(3/π) centimeters .

Step-by-step explanation:

I think that's the exact answer

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3 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago

In what ways is the measurement 3.2 m like 3.20 m? How isit different?​

In what ways is the measurement 3.2 m like 3.20 m? How isit different?