The y intercept is 0
Y = mx + b
m = 7, there is no b
I don’t really know I just need to answer two questions to ask mine sorry
Answer:
95% confidence interval for the variance of the bowling ball weight is between a lower limit of 13.328 pounds and an upper limit of 14.672 pounds.
Step-by-step explanation:
Confidence interval is given as weight +/- margin of error (E)
weight = 14 pounds
sample sd = 0.94 pound
n = 10
degree of freedom = n - 1 = 10 - 1 = 9
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value (t) corresponding to 9 degrees of freedom and 5% significance level is 2.262
E = t × sample sd/√n = 2.262×0.94/√10 = 0.672 pounds
Lower limit of weight = weight - E = 14 - 0.672 = 13.328 pounds
Upper limit of weight = weight + E = 14 + 0.672 = 14.672 pounds.
95% confidence interval is (13.328, 14.672)
FORMULA:
- Volume of hemisphere = 2/3πr³
ANSWER:
We are given a hemisphere of radius 3.1inches.
So, Volume = 2/3 × 3.14 × (3.1)³
- 187.08/3
- 62.36in³ or 62.4in³ rounded.
The correct answer is: [B]: "40 yd² " .
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First, find the area of the triangle:
The formula of the area of a triangle, "A":
A = (1/2) * b * h ;
in which: " A = area (in units 'squared') ; in our case, " yd² " ;
" b = base length" = 6 yd.
" h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
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→ A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ;
= " 24 yd² " .
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Now, find the area, "A", of the square:
The formula for the area, "A" of a square:
A = s² ;
in which: "A = area (in "units squared") ; in our case, " yd² " ;
"s = side length (since a 'square' has all FOUR (4) equal side lengths);
A = s² = (4 yd)² = 4² * yd² = "16 yd² "
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Now, we add the areas of BOTH the triangle AND the square:
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→ " 24 yd² + 16 yd² " ;
to get: " 40 yd² " ; which is: Answer choice: [B]: " 40 yd² " .
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