Answer:
y=−7−(√161/4), −7+(√161)/4
Decimal form- y=1.42214438, -4.92214438
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
First, suppose that nothing has changed, and possibility p is still 0.56. It's our null hypothesis. Now, we've got Bernoulli distribution, but 30 is big enough to consider Gaussian distribution instead.
It has mean μ= np = 30×0.56=16.8
standard deviation s = √npq
sqrt(30×0.56×(1-0.56)) = 2.71
So 21 is (21-16.8)/2.71 = 1.5494 standard deviations above the mean. So the level increased with a ˜ 0.005 level of significance, and there is sufficient evidence.
What exactly are you trying to find but I tried to do the problem and I got 3
Answer: 18x^4 - 5x^3 + x^2 - 4
Step-by-step explanation:
We need to add like terms.
-10 and 6 are like terms
-10 + 6 = - 4
-3x^3 and -2^3 are like terms
-3x^3 + -2^3 = -5x^3
There are no other like terms.
18x^4 - 5x^3 + x^2 - 4
