Which of the following are equivalent to sin 300°

n200080 [17]

The equation is 4(x-3) -(5x+1) = 3

The next step is solving the equation is to apply the distributive property and the equation becomes 4x - 12 - 5x - 5 = 3.

So the correct answer is the third option.

6 0

4 years ago

Read 2 more answers

the length of a rectangle is 8 less than twice the width. the perimeter is 38 feet. Find the length and width. someone please he

Aleksandr-060686 [28]

Answer:L=4 W=15

Step-by-step explanation:

P=2L+2W

38-8=30

If length is 8 less than twice the width and the perimeter is 38, then subtract 38-8 which is 30, then divide 30 by 2 to get 15.

P=2(4) +2(15)

P=38

7 0

3 years ago

If h(x)=6-x,which is the value of (h*h)(10)

Andrews [41]

Answer:

16

* usually means multiplication

If you have an open circle there instead, please inform me.

Step-by-step explanation:

* usually means multiplication

If you had an open circle there, please let me know.

Find h(10) and then multiply that answer to itself.

h(10)=6-10=-4

So the answer is -4*-4=16

7 0

3 years ago

Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string ha

emmasim [6.3K]

Answer:

45/1024
1/4
15/128
193/512
9/512

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

p(2 1-bits) = 45/1024

__

b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

p(b0=0 & b9=0) = 1/4

__

c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

p(7 1-bits) = 120/1024 = 15/128

__

d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

p(more 0 bits) = 386/1024 = 193/512

__

e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

p(2 1-bits | first is a 1-bit) = 9/512

6 0

3 years ago

Which interval for the graphed function contams the local<br> maximum?

Natasha2012 [34]

Answer:

$[0,2]$