A bag contains 8 black, 5 white, and 7 pink identical bottles. Two bottles are picked at random one after the…

Question

A bag contains 8 black, 5 white, and 7 pink identical bottles. Two bottles are picked at random one after the other with replacement. Find the probabilities that they are all pink, the first is white

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to questions 11, 12, 13, and 36.

Question 11:
A bag contains 8 black, 5 white, and 7 pink identical bottles.
Total number of bottles $= 8 + 5 + 7 = 20$ .
Two bottles are picked at random one after the other with replacement.

a) They are all pink?

Calculate the probability of picking a pink bottle.
$P(\text{pink}) = \frac{\text{Number of pink bottles}}{\text{Total bottles}} = \frac{7}{20}$
Since the picking is with replacement, the probability of picking two pink bottles is the product of their individual probabilities.
$P(\text{pink and pink}) = P(\text{pink}) \times P(\text{pink}) = \frac{7}{20} \times \frac{7}{20} = \frac{49}{400}$
The probability that they are all pink is $\boxed{\frac{49}{400}}$ .

b) The first is white and the second is white?
Calculate the probability of picking a white bottle.
$P(\text{white}) = \frac{\text{Number of white bottles}}{\text{Total bottles}} = \frac{5}{20} = \frac{1}{4}$
Since the picking is with replacement, the probability of picking two white bottles is the product of their individual probabilities.
$P(\text{white and white}) = P(\text{white}) \times P(\text{white}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
The probability that the first is white and the second is white is $\boxed{\frac{1}{16}}$ .

c) The first is pink and the second is white?
Calculate the probability of picking a pink bottle first.
$P(\text{first pink}) = \frac{7}{20}$
Calculate the probability of picking a white bottle second (with replacement, so the total number of bottles and the number of white bottles remain the same).
$P(\text{second white}) = \frac{5}{20} = \frac{1}{4}$
Multiply the probabilities.
$P(\text{first pink and second white}) = P(\text{first pink}) \times P(\text{second white}) = \frac{7}{20} \times \frac{1}{4} = \frac{7}{80}$
The probability that the first is pink and the second is white is $\boxed{\frac{7}{80}}$ .

Question 12:
A motorist travels from town R to town S on a bearing of $065^\circ$ for 54 km. He then travels from town S to town Q on a bearing of $155^\circ$ for 80 km.

i) The distance between town R and town Q, correct to one decimal place.
Draw a diagram representing the journey.
Let $N_R$ be the North line at R and $N_S$ be the North line at S.
The bearing of S from R is $065^\circ$ . So, the angle between $N_R$ and RS is $65^\circ$ .
The line SR makes an angle of $65^\circ$ with the South line at S (alternate interior angles).
The angle from the North line at S, clockwise to SR, is $180^\circ + 65^\circ = 245^\circ$ .
The bearing of Q from S is $155^\circ$ . This is the angle from the North line at S, clockwise to SQ.
Calculate the interior angle $\angle RSQ$ of the triangle RSQ.
$\angle RSQ = 245^\circ - 155^\circ = 90^\circ$
So, triangle RSQ is a right-angled triangle with the right angle at S.
Use the Pythagorean theorem to find the distance RQ.
$RQ^2 = RS^2 + SQ^2$ $RQ^2 = (54 \text{ km})^2 + (80 \text{ km})^2$ $RQ^2 = 2916 + 6400$ $RQ^2 = 9316$ $RQ = \sqrt{9316} \approx 96.519 \text{ km}$
Correct to one decimal place, the distance between town R and town Q is $\boxed{96.5 \text{ km}}$ .

ii) The bearing of town Q from town R.
Find the angle $\angle SRQ$ in the right-angled triangle RSQ.
$\tan(\angle SRQ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{SQ}{RS} = \frac{80}{54}$ $\angle SRQ = \arctan\left(\frac{80}{54}\right) \approx \arctan(1.4815) \approx 55.99^\circ$
The bearing of Q from R is the angle from the North line at R, clockwise to RQ.
This is the initial bearing of S from R plus $\angle SRQ$ .
$\text{Bearing of Q from R} = 065^\circ + \angle SRQ$ $\text{Bearing of Q from R} = 65^\circ + 55.99^\circ = 120.99^\circ$
Correct to one decimal place, the bearing of town Q from town R is $\boxed{121.0^\circ}$ .

Question 13:
Points A, B, and C have position vectors: $\vec{OA} = (2, 3)$ , $\vec{OB} = (6, -1)$ , and $\vec{OC} = (1, 5)$ .

a) Find the vectors $\vec{AB}$ and $\vec{AC}$ .
Calculate $\vec{AB}$ .
$\vec{AB} = \vec{OB} - \vec{OA} = (6, -1) - (2, 3) = (6-2, -1-3) = (4, -4)$
Calculate $\vec{AC}$ .
$\vec{AC} = \vec{OC} - \vec{OA} = (1, 5) - (2, 3) = (1-2, 5-3) = (-1, 2)$
The vectors are $\boxed{\vec{AB} = (4, -4)}$ and $\boxed{\vec{AC} = (-1, 2)}$ .

b) Determine the position vector of point D such that $\vec{AD} = \vec{AB} + \vec{AC}$ .
Calculate $\vec{AD}$ .
$\vec{AD} = \vec{AB} + \vec{AC} = (4, -4) + (-1, 2) = (4-1, -4+2) = (3, -2)$
Find the position vector of D, $\vec{OD}$ .
We know $\vec{AD} = \vec{OD} - \vec{OA}$ , so $\vec{OD} = \vec{OA} + \vec{AD}$ .
$\vec{OD} = (2, 3) + (3, -2) = (2+3, 3-2) = (5, 1)$
The position vector of point D is $\boxed{(5, 1)}$ .

c) Show whether points A, B, C are collinear.
For points A, B, C to be collinear, vector $\vec{AB}$ must be a scalar multiple of vector $\vec{AC}$ (i.e., $\vec{AB} = k \vec{AC}$ for some scalar $k$ ).
We have $\vec{AB} = (4, -4)$ and $\vec{AC} = (-1, 2)$ .
Check if there is a scalar $k$ such that $(4, -4) = k(-1, 2)$ .
From the x-components: $4 = k(-1) \implies k = -4$ .
From the y-components: $-4 = k(2) \implies k = -2$ .
Since the value of $k$ is not consistent ( $-4 \neq -2$ ), $\vec{AB}$ is not a scalar multiple of $\vec{AC}$ .
Therefore, points A, B, C are $\boxed{\text{not collinear}}$ .

Question 36:
Solve the equation: $\frac{7x}{2} + \frac{2}{3x} = 34$ .
Find a common denominator for the fractions, which is $6x$.
Multiply the entire equation by $6x$ to eliminate the denominators.
$6x \left( \frac{7x}{2} \right) + 6x \left( \frac{2}{3x} \right) = 6x (34)$
Simplify the terms.
$3(7x^2) + 2(2) = 204x$ $21x^2 + 4 = 204x$
Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$ .
$21x^2 - 204x + 4 = 0$
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $x$ .
Here, $a=21$ , $b=-204$ , $c=4$ .
$x = \frac{-(-204) \pm \sqrt{(-204)^2 - 4(21)(4)}}{2(21)}$ $x = \frac{204 \pm \sqrt{41616 - 336}}{42}$ $x = \frac{204 \pm \sqrt{41280}}{42}$
Calculate the square root and simplify.
$\sqrt{41280} \approx 203.1743$ $x_1 = \frac{204 - 203.1743}{42} = \frac{0.8257}{42} \approx 0.01966$ $x_2 = \frac{204 + 203.1743}{42} = \frac{407.1743}{42} \approx 9.6946$

A bag contains 8 black, 5 white, and 7 pink identical bottles. Two bottles are picked at random one after the other with replacement. Find the probabilities that they are all pink, the first is white

Step-by-step solution

Calculate the probability of picking a pink bottle.

Since the picking is with replacement, the probability of picking two pink bottles is the product of their individual probabilities.

Calculate the probability of picking a white bottle.

Since the picking is with replacement, the probability of picking two white bottles is the product of their individual probabilities.

Calculate the probability of picking a pink bottle first.

Calculate the probability of picking a white bottle second (with replacement, so the total number of bottles and the number of white bottles remain the same).

Multiply the probabilities.

Draw a diagram representing the journey.

Calculate the interior angle $\angle RSQ$ of the triangle RSQ.

Use the Pythagorean theorem to find the distance RQ.

Find the angle $\angle SRQ$ in the right-angled triangle RSQ.

The bearing of Q from R is the angle from the North line at R, clockwise to RQ.

Calculate $\vec{AB}$ .

Calculate $\vec{AC}$ .

Calculate $\vec{AD}$ .

Find the position vector of D, $\vec{OD}$ .

For points A, B, C to be collinear, vector $\vec{AB}$ must be a scalar multiple of vector $\vec{AC}$ (i.e., $\vec{AB} = k \vec{AC}$ for some scalar $k$ ).

Check if there is a scalar $k$ such that $(4, -4) = k(-1, 2)$ .

Find a common denominator for the fractions, which is $6x$.

Simplify the terms.

Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$ .

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $x$ .

Calculate the square root and simplify.

Still stuck on this one?

More Mathematics Questions

A bag contains 8 black, 5 white, and 7 pink identical bottles. Two bottles are picked at random one after the other with replacement. Find the probabilities that they are all pink, the first is white

Step-by-step solution

Calculate the probability of picking a pink bottle.

Since the picking is with replacement, the probability of picking two pink bottles is the product of their individual probabilities.

Calculate the probability of picking a white bottle.

Since the picking is with replacement, the probability of picking two white bottles is the product of their individual probabilities.

Calculate the probability of picking a pink bottle first.

Calculate the probability of picking a white bottle second (with replacement, so the total number of bottles and the number of white bottles remain the same).

Multiply the probabilities.

Draw a diagram representing the journey.

Calculate the interior angle ∠RSQ\angle RSQ∠RSQ of the triangle RSQ.

Use the Pythagorean theorem to find the distance RQ.

Find the angle ∠SRQ\angle SRQ∠SRQ in the right-angled triangle RSQ.

The bearing of Q from R is the angle from the North line at R, clockwise to RQ.

Calculate AB⃗\vec{AB}AB.

Calculate AC⃗\vec{AC}AC.

Calculate AD⃗\vec{AD}AD.

Find the position vector of D, OD⃗\vec{OD}OD.

For points A, B, C to be collinear, vector AB⃗\vec{AB}AB must be a scalar multiple of vector AC⃗\vec{AC}AC (i.e., AB⃗=kAC⃗\vec{AB} = k \vec{AC}AB=kAC for some scalar kkk).

Check if there is a scalar kkk such that (4,−4)=k(−1,2)(4, -4) = k(-1, 2)(4,−4)=k(−1,2).

Find a common denominator for the fractions, which is $6x$.

Simplify the terms.

Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

Use the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​ to solve for xxx.

Calculate the square root and simplify.

Still stuck on this one?

More Mathematics Questions

Calculate the interior angle $\angle RSQ$ of the triangle RSQ.

Find the angle $\angle SRQ$ in the right-angled triangle RSQ.

Calculate $\vec{AB}$ .

Calculate $\vec{AC}$ .

Calculate $\vec{AD}$ .

Find the position vector of D, $\vec{OD}$ .

For points A, B, C to be collinear, vector $\vec{AB}$ must be a scalar multiple of vector $\vec{AC}$ (i.e., $\vec{AB} = k \vec{AC}$ for some scalar $k$ ).

Check if there is a scalar $k$ such that $(4, -4) = k(-1, 2)$ .

Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$ .

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $x$ .