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3 years ago

How many 1 3/5 inch books can fit on a 14 1/2 inch bookshelf

Mathematics

2 answers:

lbvjy [14]3 years ago

6 0

Answer:

9 1/16 or about 9 books

Step-by-step explanation:

divide 14.5 by 1.6 to get 9.0625 or 9 books will fit!

14.5÷1.6=9.625 9.625= about 9

Hope that was helpful.Thank you!!!

OLEGan [10]3 years ago

5 0

9 books will fit on the shelf

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The world's tallest unsupported flagpole is believed to be a 282-ft-tall steel pole located in Surrey, British Columbia. During

EleoNora [17]

Answer:

As shown in the attached figure we have

$tan(\alpha )=\frac{282}{359}\\\\\therefore \alpha =tan^{-1}(\frac{282}{359})\\\alpha =37.95^{o}$

Now we have $\alpha =\lambda$

Thus angle of depression is $37.95^{0}$

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3 years ago

What is 23/2 converted into a mixed numeral

Cerrena [4.2K]

Answer:

11 1/2

Step-by-step explanation:

22 divided by 2 is 11, and there is 1 left over. So that is 1/2. The mixed number is 11 1/2

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3 years ago

Add the two complex numbers 6 + 7i and 10 + 2i. Write your answer as a complex number in standard form, a + bi.

lina2011 [118]

1. The complex number 6 + 7i has

real part 6;
imaginary part 7.

2. The complex number 10 + 2i has

real part 10;
imaginary part 2.

3. When we add two complex numbers, we add real parts and imaginary parts separately:

(6+7i)+(10+2i)=(6+10)+(7+2)i=16+9i.

Answer: 16+9i

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3 years ago

Read 2 more answers

At a certain university, the average attendance at basketball games has been 3125. Due to the dismal showing of the team this ye

Neporo4naja [7]

Answer:

The test statistic needed to evaluate the claim is t = -1.08.

Step-by-step explanation:

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value of the mean, s is the standard deviation of the sample and n is the size of the sample.

At a certain university, the average attendance at basketball games has been 3125. The athletic director claims that the attendance is the same as last year.

This means that $\mu = 3125$

Due to the dismal showing of the team this year, the attendance for the first 9 games has averaged only 2915 with a standard deviation of 585.

This means that $n = 9, X = 2915, s = 585$

What is the test statistic needed to evaluate the claim?

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{2915 - 3125}{\frac{585}{\sqrt{9}}}$

$t = -1.08$

The test statistic needed to evaluate the claim is t = -1.08.

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2 years ago

(2.3 x 108)/(9.8 x 108)

QveST [7]

Answer: 9.45

Step-by-step explanation:

2.3 * 108 = 248.4

9.8 * 108 = 1,058.4

248.4 divided by 1,058.4 = 9.45

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3 years ago