The slope-intercept form: y = mx + b
Combine these two like terms to create an equivalent expression: 4z - (-3z)
1 answer:
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The total number of students is 45 + 48 +40 = 133 students
1/2 * 133 ------ 3/4 * 133
(66.5 ; 99.75)
So we can say that b<span>etween 65 and 100 students packed their lunch</span>
1/2 * 133 ------ 3/4 * 133
(66.5 ; 99.75)
So we can say that b<span>etween 65 and 100 students packed their lunch</span>
The area of triangle BMN is 124.7 square centimeter.
<h3>What are similar triangles?</h3>Similar triangles are triangles that their corresponding angles are equal and the ratio of their corresponding lengths are equal. They are often similar When at least Two angles are equal or The ratio of any corresponding lengths are same.
Analysis:
AB = AC = BC = 50cm
AH = AC/2 = 50/2 = 25cm
From Δ ABH, using Pythagoras theorem to find BH.
=
-
=
-
= 2500 - 625
= 1875
BH = = 43.3cm
ΔBMN and BHC are similar,
so, BM/BH = MN/HC
BM/43.3 = 12/25
25BM = 12 X 43.3
BM = 519.6/25 = 20.78cm
Area of ΔBMN
= 1/2(MN)(BM) = 1/2 x 12 x 20.78 = 124.7 square centimeter
Learn more about similar triangles: brainly.com/question/2644832
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(a) For any probability distribution, the total probability must be 1. That is, the area under the probability density curve must be equal to 1.
The empirical rule for normal distributions says
• approximately 68% of the distribution lies within 1 standard deviation of the mean
• approx. 95% lies within 2 s.d. of the mean
• approx. 99.7% lies within 3 s.d. of the mean
In this case, with mean 3500 and s.d. 470, this translates to
• Pr(3500 - 470 < X < 3500 + 470) = Pr(3030 < X < 3970) ≈ 0.68
• Pr(3500 - 2*470 < X < 3500 + 2*470) = Pr(2560 < X < 4440) ≈ 0.95
• Pr(3500 - 3*470 < X < 3500 + 3*470) = Pr(2090 < X < 4910) ≈ 0.997
Continuous probability distributions also have the property that
Pr(a < X < b) = Pr(a < X < c) + Pr(c < X < b)
if a < c < b.
Combining all these properties, we can find the probabilities for each of the 8 regions in the graph to be (from left to right)
• Pr(-∞ < X < 2090) ≈ (1 - 0.997)/2 ≈ 0.0015
• Pr(2090 < X < 2560) ≈ (1 - 0.95 - 2*0.0015)/2 ≈ 0.0235
• Pr(2560 < X < 3030) ≈ (1 - 0.68 - 2*0.0235 - 2*0.0015)/2 ≈ 0.135
• Pr(3030 < X < 3500) ≈ 0.68/2 ≈ 0.34
and since the distribution is symmetric about its mean, we already know the remaining probabilities,
• Pr(3500 < X < 3970) ≈ 0.34
• Pr(3970 < X < 4440) ≈ 0.135
• Pr(4440 < X< 4910) ≈ 0.0235
• Pr(4910 < X < ∞) ≈ 0.0015
(b) Per the rule, 99.7% of babies would weight between 2090 and 4910 grams.
(c) The proportion of babies weighing less than 3030 grams is the sum of the proportions of babies weighing less than 2090, between 2090 and 2560, and between 2560 and 3030 grams. So
Pr(X < 3030) = Pr(-∞ < X < 2090) + Pr(2090 < X < 2560) + P(2560 < X < 3030)
Pr(X < 3030) ≈ 0.0015 + 0.0235 + 0.135
Pr(X < 3030) ≈ 0.16 = 16%
(d) Similarly,
Pr(X > 2560) = Pr(2560 < X < 3030) + Pr(3030 < X < 3500) + … + Pr(4910 < X < ∞)
Pr(X > 2560) ≈ 0.135 + 0.34 + 0.34 + 0.135 + 0.0235 + 0.0015
Pr(X > 2560) ≈ 0.975 = 97.5%
