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3 years ago

Create an array that models 5×2

Mathematics

1 answer:

Alexxx [7]3 years ago

3 0

Answer:its easy

Step-by-step just do 2 rows of 5

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How many solutions exist for the equation: 8(4-2x)=4(3-5x)+4x

madam [21]

Answer:

Step-by-step explanation:

We have the equation:

$8(4-2x)=4(3-5x)+4x$

Let's expand both sides. Distribute on the left:

$32-16x=4(3-5x)+4x$

Distribute on the right:

$32-16x=12-20x+4x$

Combine like terms on the right:

$32-16x=12-16x$

Add 16x to both sides:

$32\neq12$

This is not true.

So, our equation has no solutions.

This would mean that our two linear equations are parallel to each other.

Our answer is C

4 0

3 years ago

Read 2 more answers

The period T in seconds of a simple pendulum of length L in feet is given by T=2π√L/32. Determine the length in feet of a simple

ASHA 777 [7]

In this situation, you have to use the equation
T = 2π√L / 32
We need to find L so we have:
T / 2π = √L / 32
(T / 2π) ^ 2 = L / 32
L = 32 * (T / 2π) ^ 2

Substitute the values:
L = 32 * (8.72 / (2 * (3.14))) ^ 2
L = 61.69694511 feet
L = 62 feet
So the Answer is
L = 62 feet

5 0

3 years ago

Read 2 more answers

A Halloween trick-or-treat group consists of 3 Trolls, 4 Nazguls, 4 Ents, and 5 Pokemon. A committee of 4 is to be picked to rep

balu736 [363]

Answer:

The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357

Step-by-step explanation:

No. of Trolls = 3

No. of Nazguls =4

No. of Ents = 4

No. of Pokemon = 5

Total Monsters = 16

We are given that A committee of 4 is to be picked to represent the group at the Monster Bash.

(a) Find the probability that the committee will consist of 1 from each type of monster.

So, the probability that the committee will consist of 1 from each type of monster:

= $\frac{^3C_1 \times ^4C_1 \times ^4C_1 \times ^5C_1}{^{16}C_4}$

= $\frac{\frac{3!}{1!(3-1)!} \times \frac{4!}{1!(4-1)!} \times \frac{4!}{1!(4-1)!}\times \frac{5!}{1!(5-1)!}}{\frac{16!}{4!(16-4)!}}$

= $\frac{\frac{3!}{1!(2)!} \times \frac{4!}{1!(3)!} \times \frac{4!}{1!(3)!}\times \frac{5!}{1!(4)!}}{\frac{16!}{4!(12)!}}$

= $0.1318$

b)Suppose that the Pokemon refuse to be on the same committee as a the Trolls.

So, the probability that this type of committee is formed

= $\frac{(^3C_3 \times ^8C_1)+(^3C_2 \times ^8C_2)+(^3C_1 \times ^8C_3)+(^3C_0 \times ^8C_4)+(^5C_4 \times ^8C_0)+(^5C_3 \times ^8C_1)+(^5C_2 \times ^8C_2)+(^5C_1 \times ^8C_3)}{^{16}C_4}$

= $\frac{(\frac{3!}{3!(3-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{3!}{2!(3-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{3!}{1!(3-1)!} \times\frac{8!}{3!(8-3)!})+(\frac{3!}{0!(3-0)!} \times \frac{8!}{4!(8-4)!})+(\frac{5!}{4!(5-4)!} \times \frac{8!}{0!(8-0)!})+(\frac{5!}{3!(5-3)!} \times \frac{8!}{1!(8-1)!})+(\frac{5!}{2!(5-2)!} \times\frac{8!}{2!(8-2)!})+(\frac{5!}{1!(5-1)!} \times \frac{8!}{3!(8-3)!})}{\frac{16!}{4!(16-4)!}}$

= $0.5357$

Hence The probability that the committee will consist of 1 from each type of monster is 0.1318 and Suppose that the Pokemon refuse to be on the same committee as a the Trolls.So,the probability that this type of committee is formed is 0.5357

4 0

3 years ago

Using the Classic Rock data in the Math 108 group on StatCrunch, create a histogram with the

liberstina [14]

The product of the frequency and PlayCount gives the number of songs

played from the information in the histogram.

Correct responses (rounded to the nearest whole integer)

1. 2%

2. 27 PlayCount per song

3. 68%

<h3>How does one analyze an histogram?</h3>

1. The given data from the histogram is presented in a tabular form as follows;

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}f&899&224&126&96&71&52&39&41&44&23&14&12&3&4&3\\x&10&20&30&40&50&60&70&80&90&100&110&120&130&140&150\end{array}\right]$