Answer:
What's the question?
Step-by-step explanation:
342 is the product of q and 214
You should be able to set up a proportion between the sides something like (18+x)/x=(56+21)/56
Answer:
At 25 = 6.8612mm
At 50 years = 5.422mm
Step-by-step explanation:
Equation,
d = 2.115Logₑa + 13.669
d = diameter of the pupil
a = number of years
Note : Logₑa = In a (check logarithmic rule)
d = 2.115Ina + 13.669
1. At 25 years,
d = -2.115In25 + 13.669
d = -2.115 × 3.2188 + 13.669
d = -6.807762 + 13.669
d = 6.8612mm
At 25 years, the pupil shrinks by 6.86mm
2. At 50 years,
d = -2.1158In50 + 13.669
d = -2.1158 * 3.912 + 13.669
d = -8.2770 + 13.699
d = 5.422mm
At 50 years, the pupil shirks by 5.422mm
To save this question, I had to plug in the values into the equation.
Solving for Logₑa might be difficult, so instead I used Inx which is the same thing. Afterwards, i substituted in the values and solve the equation for each years.
// Input value is usernum.
// This code snippet sums 1 + 3 + 5 + ... + usernum
// The answer is stored in the variable summedvalue.
N = (int) (usernum+1)/2; // maximum number of integers to be summed
int *v = malloc(N*sizeof(int)); // allocate storage for array v
// Calculate the number of loop counts and assign array v..
count = 0;
k = 1;
while (1) {
if (k>usernum) { // do not extend v beyond usernum
break;
}
v(count) = k; // assign an odd integer to v, including usenum
count++;
k += 2; // k is an odd number
if k>usernum { // handle usernum as odd or even
k = usernum;
}
}
n = count; // the size of array v.
// Calculate the sum in a for loop
summedvalue = 0; // initialize summedvalue
for (i=0; i<=n; i++) {
summedvalue += v(i);
}
Answer:

Step-by-step explanation:
GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are
red candies,
blue,
green,
brown,
orange, and
yellow. Three candies are pulled from the bag in succession, without replacement.
TO FIND: What is the probability that the first two candies drawn are orange and the third is green.
SOLUTION:
Total candies in the bag 
Probability that first ball is orange, 

Probability that second ball is orange, 

Probability that third ball is green, 

Now, probability that first two balls are orange and third is green is




Hence, probability that first two balls are orange and third is green is 