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Sholpan [36]
3 years ago
11

Suppose we want to choose 6 letters, without replacement, from 15 distinct letters. (A) how many ways can this be done, if the o

rder of choices is not taken into consideration? (B) How many ways can this be done, if the order of choices is taken into consideration?
Mathematics
1 answer:
Olenka [21]3 years ago
7 0

Answer:

Below in bold.

Step-by-step explanation:

A.This is the number of combinations of 6 from 15

= 15C6

=  15! / (15-6)! 6!

= 5,005 ways.

B.  This is the number of permutaions of 6 from 15:

= 15! / (15-6)!

= 3,603,600 ways.

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2)<br> 110°<br> 18x + 2<br> A) 8<br> C) 6<br> B) 12<br> D) -10
VARVARA [1.3K]

Answer:

C) 6

Step-by-step explanation:

Alternate interior angles are congruent meaning that they are equal to each other sp 18x+2=110 is simplified to 18x=108. Then divide by 18 to get x equal to 6. Hope this helped and post more :)

3 0
3 years ago
Help me please o(* ̄▽ ̄*)ブ
Vesna [10]

Answer:

Hey there!

Part a) Distance from Wrexham to Oswetry: 15 miles.

Part b) Ruthin to Oswestry: 33 miles.

Total distance travelled for Part b= 33(5), or 165 miles.

6 0
3 years ago
(08.05 MC)
oee [108]

Answer:

x=67

Step-by-step explanation:

5 0
3 years ago
What is the general equation of a sine function with an amplitude of 6, a period of pi/4, and a horizontal shift of pi/2?
agasfer [191]

The general equation of the sine function y=f(x) is defined as

Y=AsinB(x-C)+D

where A is the Amplitude

B represents the frequency of the function with period equals 2\pi/B

C represents the Horizontal shift, For Phase shift= -C/B

D represents the Vertical shift.

The data given that the amplitude of the function A=6

B=\frac{2\pi}{Current\,  period}=\frac{2\pi}{\pi/4}=8

Vertical shift D= 0

Horizontal Shift C= \pi/2

Now plug in \neq A=6, B=8\,  and \, C=\pi/2 in the general equation of sine function, we get

y=6\sin8(x-\pi/2)+0\\y=6\sin 8(x-\pi/2))

4 0
3 years ago
Please help! Identify the minimum value of the function y=3x^2-12x+10.
Dimas [21]
The lowest point is at the vertex, or U-turn.

\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\&#10;\begin{array}{llccclll}&#10;y = &{{ 3}}x^2&{{ -12}}x&{{ +10}}\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)&#10;\\\\\\&#10;\left( -\cfrac{-12}{2(3)}~~,~~10-\cfrac{(-12)^2}{4(3)} \right)\qquad \textit{so the lowest value is at }10-\cfrac{(-12)^2}{4(3)}
6 0
3 years ago
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