Answer:
5907.2 grams
Step-by-step explanation:
16 x 28.4 x 13 = 5907.2grams
Answer:
Since the question is partial, i will assume 2 scenarios:
They need to raise 1000 income
They need to make 1000 as profit
If $1000 as income:
Each ticket costs $15, so tickets would bring them $1000 income. Fractional ticket is not possible, so rounding gives us 67 tickets as the answer.
If $1000 as profit:
Their cost of renting is $700. We know that .
So, . So, to raise $1700, we need tickets. Fractional ticket is not possible, so rounding gives us 114 tickets as the answer.
ANSWER:
If need to raise atleast $1000 as income, they need to sell 67 tickets.
If need to raise atleast $1000 as profit, they need to sell 114 tickets.
The most probable answer would be 114 tickets
PLZ MARK ME BRAINLIEST
Lets get all the information we need first.
A week has 7 days, you practice 30 min each day, so in a week you will practice a total time t of 30 min for 7 times, that is:
t = 7day(30 min/day)
t = 210 min
So in a week you will practice a total of 210 minutes.
The new equation after shifting will be:

Step-by-step explanation:
Function trnafomations upward and left are defined as:
Upward:
f(x) => f(x)+b where b is an integer
Left:
f(x) => f(x+b) where b is an integers
Given function is:

Shifting the function 5 units left

Shifting the function upward 2 units
So,

The new equation after shifting will be:

Keywords: Functions, shifting
Learn more about functions at:
#LearnwithBrainly
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.