Answer:
42
Step-by-step explanation:
Let total no. of trees be 'X'
Lemon are (1/6) of X = X/6
Orange are (1/2) of X = X/2
Difference is 14
X/2 - X/6 = 14
Lcm is 6
(3X-X)/6 = 14
2X = 14×6
X = 42
Answer:
Suppose that a data set has a minimum value of 18 and a max of 83 and that you want 5 classes. Explain how to find the class width for this frequency table. What happens if you mistakenly use a class width of 13 instead of 14?
Step-by-step explanation:
Answer:
the number of times in a month the train must be used, so that the total monthly cost without the pass is the same as the total monthly cost with the pass, is b. 24 times
Step-by-step explanation:
in normal purchase, train ticket (A) = $2.00
using frequent pass,
frequent pass (P) = $18
train ticket using frequent pass (B) = $1.25
Now, let assume the number of times in a month the train must be used = M
so,
A x M = P + (B x M)
$2.00 x M = $18 + ($1.25 x M)
($2.00 x M) - ($1.25 x M) = $18
M x ($2.00 - $1.25) = $18
M = $18 : $0.75
M = 24
Thus, the number of times in a month the train must be used is 24 times
Answer:
0.79
Step-by-step explanation:
Here,
Let X be the event that the flights depart on time
Let Y be the event that flights arrive on time
So,
X∩Y will denote the event that the flights departing on time also arrive on time.
Let P be the probability
P(X∩Y)=0.65
And
P(X)=0.82
We have to find P((Y│X)
We know that
P((Y│X)=P(X∩Y)/P(X) )
=0.65/0.82
=0.79
So the probability that a flight that departs on schedule also arrives on schedule is: 0.79
9514 1404 393
Answer:
3) x = 9
4) x = 3
Step-by-step explanation:
3) The two short segments are indicated as having a sum equal to the long segment.
(x +2) +(-5 +x) = 15
2x = 18 . . . . . . . . . . . . add 3
x = 9 . . . . . . . . . divide by 2
(This makes the segments be 9+2 = 11 and -5+9 = 4, which total 15.)
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4) Same deal.
3x +3 = 4x
3 = x . . . . . . . . subtract 3x
(This makes the segments be 3(3) = 9 and 4(3) = 12, where 9+3=12.)
