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3 years ago

What is the scientific notation for 0.000639

Mathematics

1 answer:

daser333 [38]3 years ago

4 0

Answer:

6.39*10^-4

Step-by-step explanation:

according to the form that is used to convert usual into scientific notation it be like 6.39*10^-4

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32.8 + # = 100 What is #?

Marrrta [24]

Answer:

67.2

Step-by-step explanation:

You take 100 and minus 32.8

it is the same as 10minus 8 is 2 so we right the 2. then 9-2 is 7. and 9-3 is 6

99.(1)0

32.8

---------------

67.2

5 0

2 years ago

A particular computer takes 14 minutes to download a 56-minute TV show. How long will it take the computer to download a 2.5-h

charle [14.2K]

Answer:

240 minutes

Step-by-step explanation:

step 1 - 1:56÷14=4

step 2 - 2:4×60=240

4 0

3 years ago

Suppose y= 2x - 2.25 Find x if y = 10

sukhopar [10]

Y = 2x - 2.25 y = 10
10 = 2x - 2.25
10 + 2.25 = 2x
12.25 = 2x
6.125 = x

hope this helps, God bless!

5 0

2 years ago

An assembly line produces 20 items per hour. If the line runs 24 hours per day, 7 days week, how many items are made by this mac

Margaret [11]

20x24x30 = 14,400
14,400 items

7 0

3 years ago

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

1 year ago