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jarptica [38.1K]
2 years ago
6

I need help please Write an equation in slope intercept form for the following graph.

Mathematics
1 answer:
olganol [36]2 years ago
3 0

Answer:

  • y = - 0.5x + 3

Step-by-step explanation:

<u>Take two points on the line:</u>

  • (0, 3) and (2, 2)

<u>The slope-intercept form:</u>

  • y = mx + b, where m- slope, b- y-intercept

<u>Find the slope:</u>

  • m = (2 - 3) / (2 - 0) = - 1 / 2 = - 0.5

The y- intercept is b = 3 according to the first point.

<u>The line is:</u>

  • y = - 0.5x + 3
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6. Let E, F, G be three events. Find expressions for the events that of E, F, G a. only E occurs; b. both E and G but not F occu
marta [7]

Answer:

a.) only E occurs :

E((FUG)')

b.) both E and G but not F

EG(F')

C.) Atleast one of the events :

(E U G U F)

d.) Atleast two of the events

(EG U EF U GF)

E.) all three occurs

(EGF)

F.) none of the events occur :

(E U G U F)'

G.) At most one of the events occur

(EG U EF U GF)'

H.) at most two of them occur

(E' U G' U F')

I.) exactly two of them occur;

(EG U EF U GF) * (EGF)'

j. at most three of them occur.

EFG U (EFG)'

Step-by-step explanation:

Events = {E, F, G}

a.) only E occurs :

E((FUG)') ; E and not F or G

b.) both E and G but not F

EG(F') ; both E and G and not F

C.) Atleast one of the events :

(E U G U F) ; Either E or F or G

d.) Atleast two of the events

(EG U EF U GF) ; Either EG or EF or GF occurs

E.) all three occurs

(EGF) ; All three events, E, G and F occurs

F.) none of the events occur :

(E U G U F)' ; Neither E, G nor F occurs

G.) At most one of the events occur

(EG U EF U GF)' ; complement of atleast two) at most one

H.) at most two of them occur

(E' U G' U F') ; Either E , F or G does not occur

I.) exactly two of them occur;

(EG U EF U GF) * (EGF)' ; Exactly two events occurs and all three events Do not occur

j. at most three of them occur.

EFG U (EFG)' ; All three events occur union all three events do not occur

6 0
3 years ago
A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 93 m
Anna007 [38]
All other components will work indefinitly
5 0
4 years ago
Which equation represents the statement"one-fourth more than a number is equal to two-thirds?
Taya2010 [7]

Answer:

x + 1/4 = 2/3

Step-by-step explanation:

one-fourth more than a number

x + 1/4

4 0
3 years ago
Read 2 more answers
You are going to make “words” using the letters in the word WHISKAS. a) How many seven-letter words can you make? b) How many se
nalin [4]

Hey there! I'm happy to help!

PART A

When we say words, we don't really mean words. We just mean how many different ways we can arrange the letters. We could make a word like Hiskasw and that would work.

We have six different letters: w, h, i, s, k, and a. We have two s's and this will be very important when finding these permutations.

The first thing we do is take the number of letters and find its factorial. In this case, it is seven, so we have 7×6×5×4×3×2×1=5040.

But, this is not how many combinations there are, because we have two s's, and since they are the same many of our combinations are actually identical, but the s's are just switched. So, we take the number of s's (2) and we factorial it, which just still equals two, and then we divide our first factorial by that.

5040/2=2520

There are 2520 ways you can arrange the word WHISKAS.

PART B

Let's think of the seven letter slots in the word. It doesn't matter where you place one of the S's, but a slot next to the first S you place only has one choice of letter, which is another S to make it adjacent. This means that one  specific slot is required to have an S. If we start off filling in our required one with an S, we have six letters left to fill in our other slots, which will give us a result of 6!, which is 720 seven-letter words.

PART C

Now, we have to have A be the first term and H be the last. If we think of the seven letter slots, we have only one choice on the first one, which is A, and only one choice on the last one, which is H. This leaves us with 5! or 120 possibilities, but we also have to divide by two because we have two S's, so there are 60 seven-letter words you can make if the words must begin with A and end with H.

Have a wonderful day!

4 0
3 years ago
Help!!! Please help me I need help on my homework!​
LiRa [457]

Answer:

Using the law of indices

5 X 5^-3

5^1-3

5^-2

8 0
3 years ago
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