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NISA [10]
4 years ago
10

You have a box that is a good size for your tape collection. two rows of tapes will fit in the box. the box is 10 inches wide. e

ach tape is ⅝ inches wide. how many tapes will fit in the box?
Mathematics
1 answer:
jeyben [28]4 years ago
3 0

Solution:

Two rows of tapes will fit in the box.

Box wide = 10 inches

Each tape wide = 5/8 inches

10/ (5/8) = (10 * 8)/5

= 16

There are two in each row. So, 32 tapes will fit in the box.


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What word matches the definition “the largest number that divides exactly into two or more numbers”
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Greatest or greatest common factor

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(7 + 7i)(2 − 2i)
Ostrovityanka [42]

The complex number  -7i into trigonometric form is 7 (cos (90) + sin (90) i) and  3 + 3i in trigonometric form is 4.2426 (cos (45) + sin (45) i)

<h3>What is a complex number?</h3>

It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.

We have a complex number shown in the picture:

-7i(3 + 3i)

= -7i

In trigonometric form:

z = 7 (cos (90) + sin (90) i)

= 3 + 3i

z = 4.2426 (cos (45) + sin (45) i)

\rm 7\:\left(cos\:\left(90\right)\:+\:sin\:\left(90\right)\:i\right)4.2426\:\left(cos\:\left(45\right)\:+\:sin\:\left(45\right)\:i\right)

\rm =7\left(\cos \left(\dfrac{\pi }{2}\right)+\sin \left(\dfrac{\pi }{2}\right)i\right)\cdot \:4.2426\left(\cos \left(\dfrac{\pi }{4}\right)+\sin \left(\dfrac{\pi }{4}\right)i\right)

\rm 7\cdot \dfrac{21213}{5000}e^{i\dfrac{\pi }{2}}e^{i\dfrac{\pi }{4}}

\rm =\dfrac{148491\left(-1\right)^{\dfrac{3}{4}}}{5000}

=21-21i

After converting into the exponential form:

\rm =\dfrac{148491\left(-1\right)^{\dfrac{3}{4}}}{5000}

From part (b) and part (c) both results are the same.

Thus, the complex number  -7i into trigonometric form is 7 (cos (90) + sin (90) i) and  3 + 3i in trigonometric form is 4.2426 (cos (45) + sin (45) i)

Learn more about the complex number here:

brainly.com/question/10251853

#SPJ1

3 0
2 years ago
7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value
mina [271]

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]±t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{n} }]

t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365

[1403±2.365*\frac{27}{\sqrt{7} }]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!

4 0
3 years ago
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