The letters of the alphabet are written on 26 cards. Two cards are chosen at random without replacement. What is the probability that at least one of them is a consonant

The probability of selecting at least one consonant when selecting 2 cards is 0.0954.

Step-by-step explanation:

There are 26 letters in the English alphabet series.

Of these 26 letters there are 5 vowels and 21 consonants.

The total number of ways to select 2 letters is:

The number of ways to select at least one consonant is:

n (At least 1 consonant) = N - n (Two vowels)

Compute the probability of selecting at least one consonant as follows:

Thus, the probability of selecting at least one consonant when selecting 2 cards is 0.0954.

your answer is the last one

step-by-step explanation:

divide the length of a'c' by the length of ac to get a scale factor of two

& then reflect vertically (over the y-axis)

so your answer is dilation by a scale factor of 2 followed by reflection over the y-axis

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12