First you input the equation into the quadratic formula:
__________
x=<span><span><span>−<span>(<span>−2</span>)</span></span>±<span>√<span><span><span>(<span>−2</span>)</span>2</span>−<span><span>4<span>(1)</span></span><span>(5)</span></span></span></span></span>
</span> -----------------------------
2(1)
Next you simplify the formula:
___
x=<span><span>2±<span>√<span>−16
</span></span></span></span> ------------
2
This problem has no real solutions.
Answer:
Intervals = (1,064) , (1,036)
Step-by-step explanation:
Given:
Use 95% method
Mean = 1,050
Standard deviation = 7
Find:
Intervals.
Computation:
95% method.
⇒ Intervals = Mean ± 2(Standard deviation)
⇒ Intervals = 1,050 ± 2(7)
⇒Intervals = 1,050 ± 14
⇒ Intervals = (1,050 + 14) , (1,050 - 14)
⇒ Intervals = (1,064) , (1,036)
Answer:
center = 98.6, variability = 0.08
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation 
The center is the mean.
So 
The standard deviation of the sample of 50 adults is the variability, so

So the correct answer is:
center = 98.6, variability = 0.08

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

Now the zero product property tells us that there are two cases where this is true,

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of

, so

where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval

for

and

. More generally, if you think of

as a point on the unit circle, this occurs whenever

also completes a full revolution about the origin. This means for any integer

, the general solution in this case would be

and

.
Answer:
No. Remember, a right angle must have a 90 degree angle. We can find the lengths with the Pythagorean Theorem.
Step-by-step explanation:
Given the length 7, 10, and 12, we can assume that 12 is the hypotenuse (it is the longest length).
- we can use 7 and 10 interchangeably.
Fill in the equation, 
where c = 12, and a or b = 7 or 10.
To indicate if the given lengths would form a right angle, we can only input 7 or 10, not both.
Therefore,
or 
==> 49 + b^2 = 144 ==> <u>b= </u>
<u> ==> </u><u>9.746</u>
b= 9.7, not 10.
==> 100 + b^2 = 144 ==> <u>b = </u>
<u> ==> </u><u>6.633 </u>
b= 6.6, not 7.
Therefore, the lengths 7, 10, and 12, does NOT make a right triangle.
Hope this helps!