Answer:
P(2): 1/5
P(4): 1/5
P(odd number): 3/5
P(whole number): 5/5
P(6): 0/5
P(2 or 3): 2/5
Step-by-step explanation:
There are 5 <em>equal </em>sections in this circle. So, the probability to land in each section is 1/5.
The odd numbers are 1, 3, and 5. Since each section is 1/5, you add
1/5 + 1/5 + 1/5 = 3/5. That is the probability that you will land in any odd number.
Because all the numbers listed are whole numbers, no matter where the spinner lands it will be a whole number. So, the probability is 5/5 for whole numbers.
Since 6 is not a section, it's probability will be 0/5 (or you can just put 0).
"Or" means you add the two probabilities. Add the probability of landing on 3 (which is 1/5) to the probability of landing on 2 (which is also 1/5). So, you get 2/5.
Im not entirely sure but, if you're on plato answer D is correct
model 1 has a random pattern and is fit for the data
6(4)^2 + 2
= 6(16) + 2
= 96 + 2
= 98
If (x^2 -10) is one of the factors, that can be further factored into:
(x - sqrt(10) ) * (x+sqrt(10)) =0
making 2 of the 4 solutions equal:
3.1623 and -3.1623
I then used an algebraic long division calculator
http://calculus-calculator.com/longdivision/
to calculate:
<span>x^4 + 5x^3 ‒ x^2 ‒ 50x ‒ 90 divided by x^2 -10 which equals
</span>x^2 + 5x + 9
Using the quadratic formula, the roots of that equation are:
x = -5 + sqrt (-11) / 2
and
x = -5 - sqrt (-11) / 2
Both of those roots are not real.
I tried using online graphing calculators for x^4+5x^3-x^2-50x-90=0 but none worked.
2. For this equation,
<span>3x^2 ‒ 8x + k = 0
I used my OWN quadratic formula calculator
http://www.1728.org/quadratc.htm
and found that real roots no longer exist after "k" is greater than 5.3
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