Please help there are a few questions but I will give brainliest to first person who answers all of them! Thank you!!!!!!!!!!!!!
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You can do this using synthetic division, which is the easiest way. If x - 2 = 0, then x = 2. That 2 will go outside the "box" and the leading coefficients of the terms in the polynomial will go inside the "box". 2 (1 -3 -10 24). Bring down the first number, the 1. Multiply that 1 by the 2 to get 2. Put that 2 up under the -3 and add to get -1. Multiply that -1 by the 2 to get -2. Put that =-2 up under the -10 and add to get -12. Multiply that -12 by the 2 to get -24. Put the -24 up under the 24 and add to get 0. That means that x - 2 is a factor of the polynomial. What's left, the bolded numbers, are the coefficients of a new polynomial that is one degree less than the polynomial you started with. In other words, when we divide your polynomial by x-2, you get 
.
Check the picture below.
notice that the triangle ADH, since the segment AL is an angle bisector, meaning it cuts the angle A in two equal halves, then the triangle ADH is only using half of A.
notice that the triangle ADH, since the segment AL is an angle bisector, meaning it cuts the angle A in two equal halves, then the triangle ADH is only using half of A.
well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
You're looking for a value 
such that
Because the distribution is symmetric, the value of in either case will be the same.
Now, because the distribution is continuous, you have that
The mean for the standard normal distribution is
, and because the distribution is symmetric about its mean, it follows that 
.
You can consult a score table to find the corresponding score for this probability. It turns out to be 
.
Because the distribution is symmetric, the value of
Now, because the distribution is continuous, you have that
The mean for the standard normal distribution is
You can consult a