Ask question

Ask question

All categories

2 years ago

5

(a) A company that makes crayons is trying to decide which colors to include in a promotional mini-box of crayons. The company c

an choose the mini-box colors from its collection of colors. How many mini-boxes are possible?

1 answer:

Novosadov [1.4K]2 years ago

8 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The number of color are possible $\left n} \atop }} \right. C _r = 1215450$

Step-by-step explanation:

From the question we are told that

The number of colors is r = 4

The sample size n = 75

The number of color that are possible can be mathematically evaluated as

$\left n} \atop }} \right. C _r = \frac{n !}{ (n-r)! r!}$

substituting values

$\left n} \atop }} \right. C _r = \frac{75 !}{ (75-4)! 4!}$

$\left n} \atop }} \right. C _r = \frac{75 !}{ (71)! 4 !}$

$\left n} \atop }} \right. C _r = \frac{75 *74 * 73* 72 * 71!}{ (71)! (4*3*2 *1)}$

$\left n} \atop }} \right. C _r = 1215450$

You might be interested in

Amanda is placing an order for running shoes and leather boots for her footwear boutique. She needs a total of 48 pairs of shoes

Basile [38]

Constraint 1:

Let the total number of running shoes be = R

Let the total number of leather boots be = L

As the given number of total shoes are 48,

The equations becomes,

R + L = 48............(1)

Constraint 2:

As running shoes are twice the leather boots, equation becomes,

R = 2L..............(2)

Putting the value of R from equation(2) in equation (1)

$2L+ L=48$

$3L=48$

$L=16$

Now putting the value of L in equation(2)

R= 2L

R = $2*16$

R=32

Hence, Amanda needs 16 pairs of leather boots and 32 pairs of running shoes.

4 0

2 years ago

A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 1.5 feet, 30 feet a

Nimfa-mama [501]

Answer:

x₁ > x₂

Step-by-step explanation:

Both actions imply a parable trajectories, since both are projectile shot cases.

Let´s call x₁ maximum distance in the first case

The maximum height is just in the middle of the curve, therefore x₁ the maximum horizontal distance is equal to 60 feet.

In the second case, the parable curve is modeled by:

y = x₂*( 0.08 - 0.002x₂) or y = 0.08*x₂ - 0.002*x₂²

A second degree equation, solving for x₂ and dismissing the value x₂ = 0

we get:

y = 0 ⇒ x₂*( 0.08 - 0.002x₂) = 0 x₂ = 0

And 0,08 - 0.002*x₂ = 0

- 0.002*x₂ = - 0.08

x₂ = 0.08/0.002

x₂ = 40 f

Then x₁ > x₂

3 0

2 years ago

A website randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. Det

Varvara68 [4.7K]

Answer:

a) P = 0.039

b) The expected number of days is 10 days.

Step-by-step explanation:

The most appropiate distribution to use in this case is the geometric distribution, in order to calculate the probability of a success after k failure trials.

The probability of success, as each of the 10 products are assumed to have fair probabilities, is:

$p=1/10=0.1$

Then, the probability that our product is not selected any given day is:

$q=1-p=1-0.1=0.9$

a) The probability that exactly this product is selected exactly 10 days from now is the probability that is not selected (probbility q) for the next 9 days and selected (probability p) at the 10th day:

$P=q^9p^1=0.9^9\cdot0.1=0.3874\cdot0.1=0.039$

b) The expected number of days is calculated as:

$E(X)=\dfrac{1}{p}=\dfrac{1}{0.1}=10$

6 0

2 years ago

Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a differen

Anna007 [38]

Answer:

5 miles.

Step-by-step explanation:

Consider the question: Jinghua hiked 4 1/2 miles through the woods in 2 1/4 hours. She hiked the return trip at the same average rate but by a different route taking 2 1/2 hours. How many miles did Jinghua hike on the return trip ?

First of all, we will find Jinghua's speed using given information as:

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Convert mixed fractions into improper fractions:

$4\frac{1}{2}\Rightarrow \frac{9}{2}$

$2\frac{1}{4}\Rightarrow \frac{9}{4}$

$\text{Jinghua's speed}=\frac{\frac{9}{2}\text{ Miles}}{\frac{9}{4}\text{ Hours}}$

Using property $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$ :

$\text{Jinghua's speed}=\frac{9*4\text{ Miles}}{9*2\text{ Hours}}$

$\text{Jinghua's speed}=\frac{2\text{ Miles}}{\text{ Hour}}$

We know that distance is equal to the product of speed and time.

$\text{Distance}=\text{Speed}\times\text{Time}$

Since we have been given that Jingua hiked the return trip at the same average rate, so distance covered by her on return trip would be speed (2 miles her hour) times given time (2 1/2 hours).

$\text{Distance covered by Jingua on return trip}=\frac{2\text{ Miles}}{\text{ Hour}}\times 2\frac{1}{2}\text{ Hours}$

$\text{Distance covered by Jingua on return trip}=2\text{ Miles}\times \frac{5}{2}$

$\text{Distance covered by Jingua on return trip}=5\text{ Miles}$

Therefore, Jingua hiked 5 miles on her return trip.

8 0

2 years ago

A has the coordinates (–4, 3) and B has the coordinates (4, 4). If DO,1/2(x, y) is a dilation of △ABC, what is true about the im

fomenos

A dilation is a transformation $D_{o,k}(x,y)$ , with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'. The center O is a fixed point, P' is the image of P, points O, P and P' are on the same line.

In a dilation of $D_{o, \frac{1}{2} }(x,y)$ , the scale factor, $\frac{1}{2}$ is mapping the original figure to the image in such a way that the distances from O to the vertices of the image are half the distances from O to the original figure. Also the size of the image is half the size of the original figure.

Therefore, <span>If $D_{o, \frac{1}{2} }(x,y)$ is a dilation of △ABC, the truth about the image △A'B'C'</span> are:

<span>AB is parallel to A'B'.

$D_{o, \frac{1}{2} }(x,y)=( \frac{1}{2}x, \frac{1}{2}y)$

The distance from A' to the origin is half the distance from A to the origin.</span>

3 0

2 years ago

Read 2 more answers