190 miles every 4 days. How many miles he could bike in 6 days
Answers
Answer:
285 miles
Step-by-step explanation:
2 days = 95 miles, 4 days = 190. 95+190 = 285
He could bike 285 miles in 6 days.
Answer:
285
Step-by-step explanation:
Sorry can't explain
Related Questions
Does anyone know the answer
Twelve of the 20 students in Ms. Sarno’s homeroom are female. Each month, Ms. Sarno randomly selects a student to act as “teacher’s helper.”Which simulation best models the selection of female “teacher’s helpers” for January, February, March, and April?A spinner, divided into 5 congruent sectors with three of the sectors representing female students, is spun 12 times.A coin, with heads up representing female students, is flipped 4 times.A bag of 15 marbles, with 9 marbles representing female students, has a marble drawn 4 times with replacement.A number cube, with odd numbers representing female students, is rolled 12 times.
Express Fx and Fy in terms of the length of the vector F and the angle θ, with the components separated by a comma.
the sum of the interior angles of a polygon is 540. determine and state the number of degrees in one interior angle of the polygon.
B. $33,175.00
C. $35,140.00
D. $35,162.50
Answers
6% of that is 1,987.5.
1,987.5+33,125+50=
$35,162.50
Answers
Ax- 3x = -(3a+b)
x(A-3) = - (3a+b)
x = - (3a+b) / (A-3)
ax+b=3(x)+3(−a)
Multiply 3 by x to get 3x.
ax+b=3x+3(−a)
Move −1.
ax+b=−1⋅3a
Multiply −1 by 3 to get −3.
ax+b=−3a
Replace back in to larger expression.
ax+b=3x−3a
Since 3x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 3x from both sides.
ax+b−3x=−3a
Factor out the GCF of x from each term in the polynomial.
x(a)+x(−3)=−b−3a
Factor out the GCF of x from ax−3x.
x(a−3)=−b−3a
Divide each term in the equation by (a−3).
x=−3a+b/a−3
Answer:
x=−3a+b/a−3
Answers
Answer:
10.98
Step-by-step explanation:
the answer to 9.1×8.31 is 10.98
Answers
Answer:
Step-by-step explanation:
r = 9.3 ft
h = y ft
Volume = 3000 ft
πr²h = 3000
3.14 * 9.3 *9.3 * y = 3000
y = 11.05 ft
A) 0.26
B) 0.42
C) 0.75
D) 0.91
E) 1.00
2. If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is ?
A) 0.147
B) 0.203
C) 0.300
D) 0.375
E) 0.750
Answers
p = P(not defective) = 1 - P(defective) = 1 - 0.2 = 0.8
P(x) = nCr p^x q^(n-x)
P(x ≥ 4) = P(4) + P(5) + P(6) = 6C4 * (0.8)^4 * (0.2)^2 + 6C5 * (0.8)^5 * 0.2 + (0.8)^6 = 15 * 0.4096 * 0.04 + 6 * 0.32768 * 0.2 + 0.262144 = 0.24576 + 0.393216 + 0.262144 = 0.90112
Option D is the correct answer.
2.) p = P(defective) = 10/50 = 0.2
q = P(not defective) = 1 - P(defective) = 1 - 0.2 = 0.8
P(x) = nCr p^x q^(n-x)
P(x = 3) = 8C3 * (0.2)^3 * (0.8)^5 = 8C3 * 0.008 * 0.32768 = 56 * 0.008 * 0.32768 = 0.1468
Option A is the correct answer.
Answers
d = 3, the common difference
a(n) = a₁ + d(n-1)
a₇ = -8 + 3(7-1)
a₇ = -8 + 3(6)
a₇ = -8 + 18
a₇ = 10
To check:
a₁ = -8
a₂ = -8 + 3 = -5
a₃ = -5 + 3 = -2
a₄ = -2 + 3 = 1
a₅ = 1 + 3 = 4
a₆ = 4 + 3 = 7
a₇ = 7 + 3 = 10