Answer:
48 hats and 104 shirts
Step-by-step explanation:
These are the equations you build from the problem:
h + s = 152
8.50h + 12s = 1656
This is how I solved them:
s= 152-h
8.5h + 12(152-h) = 1656
8.5h + 1824 - 12h = 1656
Solve for h
h= 48
Put this into first equation (h +s = 152) to get s
The expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
Solution:
Given expressions are 15t – 2t and 2t – 15t.
To determine 15t – 2t is equivalent to 2t – 15t or not.
Substitute t = 2 in above two expressions.
15t – 2t = 15(2) – 2(2)
= 30 – 4
= 26
2t – 15t = 2(2) – 15(2)
= 4 – 30
= –26
The values of the expressions are different when t = 2.
So, 15t – 2t is not equivalent to 2t – 15t.
Hence the expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
<span>Given that triangle
NLM is reflected over the line segment as shown, forming triangle ABC.
When a point is refrected across a line, the relative distance form the point to the line of refrection is preserved. That is the distance from the point to the line of refrection is equal to the distance of the image to the line of refrection.
Thus, from the figure, it can be seen the point B is of the same distance to the line of refrection as point M, so is point A to point L and point C to point N.
Thus, </span><span>ΔNLM is similar to </span><span><span>ΔCAB
Therefore, the</span> congruency statement that is correct is ΔNLM ≅ ΔCAB</span>
Part A: [<span>P + (A + G) - M
</span>Part B: [0.75 + (0.25 + 0.30) - 0.20] = 1.1
Answer:
(A) 0.15625
(B) 0.1875
(C) Can't be computed
Step-by-step explanation:
We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.
Let X = Amount of time taken by student to complete a statistics quiz
So, X ~ U(32 , 64)
The PDF of uniform distribution is given by;
f(X) = 
, a < X < b where a = 32 and b = 64
The CDF of Uniform distribution is P(X <= x) = 
(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)
P(X > 59) = 1 - P(X <= 59) = 1 - = 1 - 
= 
= 0.15625
(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)
P(X <= 43) = 
= 
= 0.34375
P(X < 37) = 
= 
= 0.15625
P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875
(C) Probability that student complete the quiz in exactly 44.74 minutes
= P(X = 44.74)
The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.
