Right now the dirt bike is worth $500. Next year it will be worth 15% less, or 85% of the current value (because currently it's at 100%, so you would subtract the 15%). To find 85% of the dirt bike's value, you would take $500 and multiply by 85% (or 0.85). This would give you $425 after 1 year. You'd take $425 and multiply that by 85% for the next year, and continue until you reached 5 years of the value decreasing. Does this help?
10 (13) 15
35 (37) 40
45 (48) 50
60 (61) 65
90 (93) 95
Answer:
Step-by-step explanation:
A.
P=(219/800)×(218/799)≈0.0747≈0.075
B.
800-219=581
P=(581/800)×(580/799)≈0.527
c.
P( 1 think role models and one thinks not role models)+P(both thinks they are role models)=(219/800)×(581/799)+(219/800)×(218/799)=(219×581+219×218)/(800×799)
=[219(581+218)]/(800×799)
=(219×799)/(800×799)
=219/800
=0.27375
=0.274
The length and width of a rectangular field fully enclosed with 218 metes fencing are 63 meters and 46 meters respectively.
The perimeter of a rectangle is the sum of the whole four sides. Therefore, the perimeter of a rectangle is defined as follows:
where
l = length
w = width
perimeter = 218 meters
The length of the field is 17 meters longer than the width, w. Therefore, the length is defined as follows:
The length and the width can be calculated as follows:
218 = 2(l + w)
218 = 2(17 + w + w)
218 = 34 + 4w
218 - 34 = 4w
184 = 4w
divide both sides by 4
w = 184 / 4
w = 46 meters
length = 17 + 46
length = 63 meters
learn more on rectangle here: brainly.com/question/15989799?referrer=searchResul
// 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);
}
