Answer:
<u><em>They each had </em></u><u><em>1/20 of the original meal</em></u>
Step-by-step explanation:
<u><em>To begin, </em></u><u><em>there are 4 people</em></u><u><em>. Her siblings, and herself. She is </em></u><u><em>splitting the 1/5 leftover</em></u><u><em> to feed them. To do this, we simply </em></u><u><em>take 1/5 and put it into a fraction to get 0.2</em></u><u><em>. Next, we can </em></u><u><em>put 0.2 / 4 into a calculator</em></u><u><em> to get how much each person can get.</em></u>
<u><em>0.2 / 4 = 0.05.</em></u>
<u><em>Now, we need to </em></u><u><em>turn it into a fraction</em></u><u><em> to find out how much they had in fraction form of the original.</em></u>
<u><em>0.05 = 1/20</em></u>
<u><em>They each had </em></u><u><em>1/20 of the original meal</em></u>
Answer:
d-"5"_""__"_'uvyctxycuvycycycyv
Answer:
y = 5
Step-by-step explanation:
Calculate the slope m using the slope formula
m = 
with (x₁, y₁ ) = (7, 5) and (x₂, y₂ ) = (- 9, 5)
m =
=
= 0
This means the line is horizontal and parallel to the x- axis with equation
y = c
where c is the value of the y- coordinates the line passes through.
The line passes through (7, 5) and (- 9, 5) with y- coordinates 5, thus
y = 5 ← equation of line
The aides of a right triangle will always satisfy the Pythagorean relationship ...
a² + b²= c²
We can take each of these and try it out:
(1) ... 10² + 24² = 100 + 576 = 676
√676 = <em>26 Yes</em>.
(2) ... 12² + 18² = 144 + 324 = 468
√468 = 21.6, not 20. <em>No.</em>
(3) ... 15² + 26² = 225 + 676 = 901
√901 = 30.017 Awfully close. As an engineer, I'll buy this one. <em>Yes.</em>
(4) ... 40² + 50² = 1,600 + 2,500 = 4,100
√4,100 = 64.03, not 80.<em> No.</em>
Answer:
No, because all the expected frequencies must be higher than 5.
Step-by-step explanation:
The chi-square test might be performed in those situations in which we need to know if a series of observations adjust or not to a theoretical function, such as the normal, Poisson, or binomial.
The chi-square test does not require the number of files to coincide with the number of columns in the table. It does not establish any restriction about the number of modalities per variable. However, the expected frequencies or counts should not be less than five.
In the exposed example, this unique rule is not accomplished, so the chi-square test can not be performed.