6x3x = 18x and 6x2=12
answer:18x+12
Answer:
In both cases, we have similar figures.
This means that the shape of the figures is the same, but the size is different:
PQST is similar to STNR and to NRPQ
VUYZ is similar to YZWX and to VUWX
this means that, for example, in problem 18, the ratio between ST and NR must be the same as the ratio between PQ and ST. This happens because the measure increases by the same scale factor.
With this in mind, we can solve the problem:
18)
ST = 7.5
NR = 5.5
Then the quotient ST/NR is:
ST/NR = 7.5/5.5
And, as we said above:
PQ/ST = ST/NR
PQ/7.5 = 7.5/5.5
PQ = (7.5/5.5)*7.5 = 10.23
19) Here we should have:
YZ/VU = WX/YZ
Then:
22.9/35 = WX/22.9
(22.9/35)*22.9 = WX = 14.98
5( y + 2/5) = -13
Distribute the 5 by multiplying the 5 by each term inside the set of parentheses:
5y + 2 = -13
Subtract 2 from both sides:
5y = -15
Divide both sides by 5:
Y = 3
The answer is y = 3
Answer:
Probability = 0.8444
Step-by-step explanation:
To solve this problem, we need to obtain the z-value which corresponds to the probability that the mean of a sample of 84 computers would differ from the population mean by less than 1.39 months
Let x' denote the mean life of the computer.
Mean = 88 months
Variance = 81
Thus, standard deviation = √81 = 9
Thus, we obtain;
P(|x - μ| < 1.39) = P(-1.39 > x < 1.39)
For x' = - 1.39,
z = -1.39/(9/√84)
z = -1.39/0.982
z = -1.42
For x' = 1.39,
z = 1.39/(9/√84)
z = 1.39/0.982
z = 1.42
Thus, P(|x - μ| < 1.39) =
P(−1.42 < z < 1.42)
From the Z-distribution tables i attached, we obtain ;
0.92220 - 0.07780 = 0.8444