Graph the solutions of the inequality on a number line.
1 answer:
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Answer:
The probability of using one or the other is 36%
Step-by-step explanation:
For solving this problem it is easy if we see it in a ven diagram, for this first we are going to name the initial conditions with some variables:
Probability of passing Professor Jones math class = 64% =0,64
P(J) = 0.64
Probabiliry of passing Professor Smith's physics class = 32% =0.32
P(S) = 0.32
Probability of passing both is = 30% = 0.30
P(JnS) = 0.30 (Is is an intersection so it is in the middle of the ven diagram
We need to know which is the probability of pasing one or the other for this we need to take out the probability of passing both for this we have to add the probability of passing Professor Jones math class with the probabiliry of passing Professor Smith's physics class and substract the probability of passing both for each one:
P(JuS) = (P(J) - P(JnS)) + (P(S) - P(JnS)) = (0.64 - 0.30) + (0.32 - 0.30) = 0.34 + 0.02 = 0.36 = 36%
If you check the ven diagram you can see that if we add all what is in red we will have the probability of passing Professor Jones math class and if we add all what is in blue we wiill have the probability of passing Professor Smith's physics class, and if we add just what is in each corner we will get the same value that is the probabilty of passsing one or the other.
Answer:
Janelle will make <u>62 baskets</u> on 14th day.
Step-by-step explanation:
Given,
Number of goals on 1st day = 10
We need to find the number of goals on 14th day.
Solution,
Since Janelle made 10 goals on 1st day and want to make 4 more basket from the day before.
So we can say that;
and
And also
here = first term
d = common difference
n = number of terms
Now we apply the formula of A.P.
now substituting the given values, we get;
Hence Janelle will make <u>62 basket</u> on 14th day.