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3 years ago

Can someone help to on this question?

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

Answer:the answer is x=5,-3

Step-by-step explanation:

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Graph the function. For the function whose graph is shown below, which is the correct formula for the function? y = –x – 1 y = x

rusak2 [61]

You forgot to mention the graph.
So, I make the graph of both given functions and you will compare your graph with them.

First, y= -x-1
put y=0 you get point (-1,0) called x-intercept
put x=0 you get point (0,-1) called y-intercept
join both the points, you will get the graph
Graph is attached in the picture.

Second, y= x+1
put y=0 you get point (-1,0) called x-intercept
put x=0 you get point (0,1) called y-intercept
join both the points, you will get the graph
Graph is attached in the picture.

4 0

3 years ago

Convert 0.16 into a fraction

VARVARA [1.3K]

Step-by-step explanation:

$0.16 = \frac{16}{100} = \frac{4}{25} \\$

4 0

4 years ago

What is the probability that e) A fair coin lands Heads 6 times in a row? f) A fair coin lands Heads 4 times out of 5 flips? g)

cricket20 [7]

Answer:

e) 1.56%

f) 15.62%

h) 0.879%

g) 11.72%

Step-by-step explanation:

What we will do is solve point by point.

e) A fair coin lands Heads 6 times in a row?

We have the following:

Total number of possible outcomes = 2 ^ 6 = 64

Number of favorable outcomes = 1

Required probability = 1/64 = 1.56%

f) A fair coin lands Heads 4 times out of 5 flips

We have the following:

Total number of possible outcomes = 2 ^ 5 = 32

Number of favorable outcomes = 5C4

nCr = n! / (r! * (n-r)!)

5C4 = 5! / (4! * (5-4)!) = 5

Required probability = 5/32 = 15.62%

g) he bit string has exactly two 1s, given that the string begins with a 1 if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

Number of ways in which a position excluding the start of the string can be chosen is 9C1

Total number of favorable outcomes = 9C1

9C1 = 9! / (1! * (9-1)!) = 9

Required probability = 9/1024 = 0.879%

h)The bit string has the sum of its digits equal to seven if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

For the sum of the digits to be 7 there has to be 7 ones.

Number of ways in which 7 position can be chosen is 10C7.

Total number of favorable outcomes = 10C7

10C7 = 10! / (7! * (10-7)!) = 120

Required probability = 120/1024 = 11.72%

5 0

4 years ago

Someone please help

Strike441 [17]

The answer is A,D and E because they all have the same shape and size therefore they’re congruent.

4 0

4 years ago

Tina's Treats charges a $4.35 fee for each delivery. Which table best represents the relationship between f, the amount made fro

vovikov84 [41]

Answer:

Is there any answer choices? If not I would think its 5$

Step-by-step explanation:

5 0

3 years ago