How do I solve this problem
1 answer:
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a) I have attached an image of a probability tree diagram.
b) 0.02332 (please round to where you need)
The probability that Terri chooses a blue pen is 2/7. The probability that Terri chooses a red pen is 4/7. The probability that Terri chooses neither a red nor blue pen is 1/7.
2/7 x 4/7 x 1/7 = 0.02332
Hope this helps! :)
The first problem:
y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴
The correct options are 2 , 7
Solution of the system of equations is
and
==================================
The second problem:
The general equation of the hyperbole is
For the given equation:
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36 ⇒⇒⇒ b = √36 = 6
∴ the slope of the <span><span>asymptotes = </span>b/a and -b/a
</span><span>
</span><span>b/a = 6/15 = 2/5
</span><span>
</span><span>-b/a = -6/15 = -2/5
</span><span>
</span><span>∴ m = 2/5 and m = -2/5
</span>
y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴
The correct options are 2 , 7
Solution of the system of equations is
and
==================================
The second problem:
The general equation of the hyperbole is
For the given equation:
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36 ⇒⇒⇒ b = √36 = 6
∴ the slope of the <span><span>asymptotes = </span>b/a and -b/a
</span><span>
</span><span>b/a = 6/15 = 2/5
</span><span>
</span><span>-b/a = -6/15 = -2/5
</span><span>
</span><span>∴ m = 2/5 and m = -2/5
</span>