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3 years ago

Can a triangle with the angles of 0, 90 , and 90 be considered a triangle?

1 answer:

5 0

A "triangle" with the angles 0, 90, and 90 degrees cannot be considered a triangle since the two 90 degree angles make a right angle, and 0 degrees would mean a flat line. Rather it would resemble something like a straight U that is stretched sideways.

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Marco paid 36.89 for 8.6 gallons of gas. What is the price for 1 gallon of gas

pishuonlain [190]

____
I'm pretty sure you do this 36.89)8.6

5 0

4 years ago

Read 2 more answers

out of 1000 student 350 students passed the examand rest were failed find the empirical probability of failed student

torisob [31]

Answer:

out of 1000 student 350 students passed the examand rest were failed find the empirical probability of failed student

1000-350=650

8 0

3 years ago

In a survey, a group of students were asked their favorite sport. Eighteen students chose “other” sports, while 37.5% chose foot

LUCKY_DIMON [66]

18 students are 100%-37,5%-40%=22,5%

18 students - 22,5%
x students - 100%

22,5x=18×100
x=80

Answer: 80 students

4 0

3 years ago

Find the interquartile range of the following data set.

BigorU [14]

The interquartile range of 48, 26, 31, 50, 38, 40, 42, 34, 44, 36, is: D. 10.

<h3>What is the Interquartile Range of a Data Distribution?</h3>

The interquartile range of a data distribution is determined as: upper quartile (Q3) - lower quartile (Q1).

<h3>How to Find the Upper Quartile and Lower Quartile of a Data Distribution?</h3>

The upper quartile of a data distribution is the center of the second half of a data distribution while the lower quartile is the center of the first half of a data distribution.

Given the data, 48, 26, 31, 50, 38, 40, 42, 34, 44, 36:

Order the data set as, 26, 31, 34, 36, 38, 40, 42, 44, 48, 50

The first half of the data is: 26, 31, 34, 36, 38.

The center is 34.

Lower quartile (Q1) = 34.

The second half of the data is: 40, 42, 44, 48, 50. The center is 44.

Upper quartile (Q3) = 44.

Interquartile range = 44 - 34

Interquartile range = 10

Learn more about interquartile range on:

brainly.com/question/4102829

#SPJ1

6 0

2 years ago

Differentiate with respect to t

True [87]

Answer:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$

Step-by-step explanation:

We want to differentiate the equation:

$x^2-y^3+z^4=1$

With respect to <em>t</em>, where <em>x, y, </em>and <em>z</em> are functions of <em>t. </em>

<em />

So:

$\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]$

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$

8 0

3 years ago