What is the greatest integer a, such that a^2 + 3b is less than (2b)^2, assuming that b is 5? Please answer and have a nice day!
1 answer:
<h3>Answer: Largest value is a = 9</h3>
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Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
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Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get
So the greatest integer possible for 'a' is a = 9.
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Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work
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Answer:
the probability that we hit the bullseye at least 100 times is 0.0113
Step-by-step explanation:
Given the data in the question;
Binomial distribution
We find the probability of hitting the dart on the disk
⇒ Area of small disk / Area of bigger disk
⇒ πR₁² / πR₂²
given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1
so we substitute
⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04
Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.
so
X ~ Bin( 2000, 0.04 )
n = 2000
p = 0.04
np = 2000 × 0.04 = 80
Using central limit theorem;
X ~ N( np, np( 1 - p ) )
we substitute
X ~ N( 80, 80( 1 - 0.04 ) )
X ~ N( 80, 80( 0.96 ) )
X ~ N( 80, 76.8 )
So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;
we covert to standard normal variable
⇒ P( X ≥ )
⇒ P( X ≥ 2.28217 )
From standard normal distribution table
P( X ≥ 2.28217 ) = 0.0113
Therefore, the probability that we hit the bullseye at least 100 times is 0.0113