Answer:
I would love to help you but what am I helping with there is no question there other that "Help me plssssss"
Step-by-step explanation:
Solve the following using Substitution method
2x – 5y = -13
3x + 4y = 15
- To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
- Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.
- Add 5y to both sides of the equation.
- Multiply
times 5y - 13.
- Substitute
for x in the other equation, 3x + 4y = 15.
- Multiply 3 times .
- Add
to 4y.
- Add
to both sides of the equation.
- Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.
- Substitute 3 for y in
. Because the resulting equation contains only one variable, you can solve for x directly.
- Add
to 
by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.
- The system is now solved. The value of x & y will be 1 & 3 respectively.
Answer:
Step-by-step explanation:
Domain is the x values. The interval of x-values that the function encompasses are from 0 inclusive to 7 exclusive. In interval notation that is [0, 7). The range is the y values. The interval of y-values that the function encompasses are from what looks like -2 to 4. In interval notation that is [-2, 4]. The domain goes from the lowest x-value to the highest; the range goes from the lowest y-value to the highest.
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
21x -15 + 5 = 10x - 5
21x -5 = 10x
11x = 5
x = 5/11
