<h2>(-7) x (-3) =<u> 21</u></h2>
.... Hope this will help....
5 5/8 as an improper fraction
(8*5 +5) /8 = 45/8
45/8 divided by 1/4 =
copy dot flip
45/8 * 4/1 = 45/2
22 1/2
She has 22 1/2 servings
Answer:
<h2>SEE BELOW</h2>
Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>
<h3>let's solve:</h3>
vertex:(h,k)
therefore
vertex:(-1,4)
axis of symmetry:x=h
therefore
axis of symmetry:x=-1
- to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0
vertex form of quadratic equation:
therefore
- y=a(x-(-1))²+4
- y=a(x+1)²+4
it's to notice that we don't know what a is
therefore we have to figure it out
the graph crosses y-asix at (0,3) coordinates
so,
3=a(0+1)²+4
simplify parentheses:

simplify exponent:

therefore

our vertex form of quadratic equation is
let's simplify it to standard form
simplify square:

simplify parentheses:

simplify addition:

therefore our answer is D)y=-x²-2x+3
the domain of the function

and the range of the function is

zeroes of the function:




factor out x and -1 respectively:

group:

therefore

Answer:
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.
The probability of k online retail orders that turn out to be fraudulent in the sample is:

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:
![P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483](https://tex.z-dn.net/?f=P%28x%5Cgeq2%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%5D%5C%5C%5C%5C%5C%5CP%28x%3D0%29%3D%5Cdbinom%7B20%7D%7B0%7D%5Ccdot0.08%5E%7B0%7D%5Ccdot0.92%5E%7B20%7D%3D1%5Ccdot1%5Ccdot0.189%3D0.189%5C%5C%5C%5C%5C%5CP%28x%3D1%29%3D%5Cdbinom%7B20%7D%7B1%7D%5Ccdot0.08%5E%7B1%7D%5Ccdot0.92%5E%7B19%7D%3D20%5Ccdot0.08%5Ccdot0.205%3D0.328%5C%5C%5C%5C%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-%5B0.189%2B0.328%5D%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-0.517%3D0.483)
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.