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3 years ago

The polar coordinates of a point are given. find the rectangular coordinates of this point.(5, 3pi/4

1 answer:

3 0

X would be r*cos theta and y would be r*sin theta.

Here, x is 5cos 135 deg = 5(-1/sqrt(3))
and y is 5* sin 135 deg = 5*(1/sqrt(3))

Write this point in the form (x,y), using the above values for x and y.

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Which decimal is between 85 and V86?

Romashka-Z-Leto [24]

$\sf \large \underbrace{ \underline{Understanding \: the \: Question}}$

This can be considered as a very easy Question where you have to find the value of √85 and √86 upto 2 decimal places then we could find it easily.

Let's start!

Value of √85=9.21

Value of √86=9.27

Now it is obvious that 9.25 lies between √85 and √86.

$\sf \large \underbrace{ \underline{Additional \: information}}$

What are rational no.?

Any pair of numbers which is in the form of p/q where p and q are integers and p and q are co-prime is called rational number.

4 0

3 years ago

Read 2 more answers

Find the angles y and x

Katen [24]

Answer:

x = 67

y = 23.5

Step-by-step explanation:

Sum of angles in a triangle is supplementary,

x + 90 + 23 = 180

x + 113 = 180

x = 180 - 113

x = 67

y = ?

180 - 67 / 2

= 133 / 2

= 66.5

y + 66.5 = 90

y = 90 - 66.5

y = 23.5

3 0

3 years ago

What is 600,319,287,914 + 297,997,000,026?

geniusboy [140]

898316287940

hope this helps, back to my hole i go

6 0

3 years ago

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Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120