The exponent indicates how many times the ( blank) is used as a factor.

I am thinking of three positive numbers. The mode is 5 The median is 5 The range is 9 Work out the three numbers.

Answer: either 5, 5, 14

First, we know the median is 5. Thus, the middle value of the data is 5, so the set can now be read x, 5, y. Then, because the mode is 5 and the set is not trimodal, either x or y must be 5. Thus, the set could either be 5, 5, 14, or

-4, 5, 5. However, because it must contain only positive number, the answer is 5, 5, 14

Hope it helps <3

8 0

3 years ago

DJ Giovanni is making a playlist for a friend; he is trying to decide what 12 songs to play and in what order they should be pla

Elena L [17]

Answer:

5.78×10¹⁵

Step-by-step explanation:

The playlist has 12 songs, and he wants the same number of songs for each genre, so he must pick 4 songs per genre.

The number of ways he can choose 4 country songs (ignoring the order) is ₂₂C₄ = 7315.

The number of ways he can choose 4 reggae songs (ignoring the order) is ₁₁C₄ = 330.

The number of ways he can choose 4 pop songs (ignoring the order) is ₅C₄ = 5.

The total number of combinations is 7315 × 330 × 5 = 1.21×10⁷.

Once he has his 12 songs selected, the number of ways he can arrange them is ₁₂P₁₂ = 12! = 4.79×10⁸.

So the total number of possible playlists is:

(1.21×10⁷) × (4.79×10⁸) = 5.78×10¹⁵

4 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

Help <br><br> simply. this math problem

saveliy_v [14]

- 4 1/5 = -21/5 = -42/10

-13 1/10 = -131/10

-42/10 - -131/10=

-42/10 + 131/10 = 89/10 = 8 9/10

answer is A

4 0

3 years ago

Read 2 more answers

What does a slope of 1/4 look like

Reil [10]

Slope is also known as rise over run -
$\frac{rise}{run}$
Here, we have the slope $\frac{1}{4}$ . This means that the rise is $1$ and the run is $4$ . Since both values are positive, we can say that we go up one and we go right four. When we plot this on a graph (the graph is attached), we get the same exact formula we found. **In the graph, I plotted the line $y= \frac{1}{4} x$ **. Hope this helped!

3 0

2 years ago