Answer:
m greater than 44
Step-by-step explanation:
The equal payments will add up to $160 which is greater than 44 and the most reasonable answer.
In a rhombus, all sides are equal:
5x = 15
and x = 15/5 = 3
(you could have done the same thing with the other side: 5x = 4x +3)
Answer:
A. 4(x + 8) = 56
Step-by-step explanation:
the correct equation would be 4(x + 8) = 56
because we need to find the number of hours on Saturday, since there Sheldon gets paid $4 per hour, we must subtract the amount Sheldon made on Sunday by the total he made and then divide that by 4 to get the number of hours on Saturday
$4 = number of dollars per hour
$56 = total on Saturday and Sunday
8 hours = Number of hours babysat on Sunday
x = hours spent on Saturday
4(x + 8) = 56
Step 1: Simplify both sides of the equation.
4(x + 8) = 56
(4)(x) + (4)(8) = 56(Distribute)
4x + 32 = 56
Step 2: Subtract 32 from both sides.
4x + 32 − 32 = 56 − 32
4x = 24
Step 3: Divide both sides by 4.
4x/4 = 24/4
x = 6
To check if this is correct, we add the amount Sheldon earned on Saturday to the amount he earned on Sunday, which is
6(4) + 8(4) = 56
24 + 32 = 56
56 = 56
or you could substitute the answer in the problem
4(6 + 8) = 56
Distribute
24 + 32 = 56
Simplify
56 = 56
Answer:If we're solving for t my work is below:
t = 4 s - 2
If we're solving for s my work is below:
s = t + 2/4
Step-by-step explanation:
Answer: 0.171887
Step-by-step explanation:
Given that S0 and S1 are binary symbol of equal probabilty;
P(S0) = P(S1) = 0.5
This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.
P(Sn =X) = nCx p^x (1-p)^(n-1)
Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)
Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),...(1111111110) for Sn= 1 is the sequence for 1 error.
10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5
P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7
= 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10
=0.000977 + 0.00977 + 0.04395 + 0.11719
= 0.171887
