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3 years ago

Help ASAP giving BRAINLIEST! Am i correct?

Mathematics

2 answers:

drek231 [11]3 years ago

6 0

Answer:

Yes

Step-by-step explanation:

We have the inequality:

x-7 > 27

To solve this, we have to get by itself. 7 is being subtracted from x. To undo this, add 7 to both sides, since addition is the opposite of subtraction.

x-7+7>27+7

x>34

So, your answer is correct

murzikaleks [220]3 years ago

3 0

Answer:

The answer is yes

Step-by-step explanation:

x-7 > 27

x> 27 +7

x > 34

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How many 3-letter combinations can be made from this set if no letter can be repeated?

Wittaler [7]

Answer:

B 120 different combinations

Step-by-step explanation:

There are 6 letters to choose from

Picking a letter for the first spot 6

Then there are 5 letters left

Picking a letter for the second spot 5

Then there are 4 letters left

Picking a letter for the second spot 4

6*5*4

= 120

7 0

3 years ago

Solve: 46, 56, 67, 78, 34, 46, 57 mean: Range: Median: Mode:

Serhud [2]

Answer:

mean:54.857142857143

Range:44

median:56

mode:46

5 0

3 years ago

Read 2 more answers

Simply the expression: <br> Pls need help fast!!!!!!

sukhopar [10]

Answer:

4^6

Step-by-step explanation:

6 0

3 years ago

X + 1 / 2 - 1 = 1 / 4

Nimfa-mama [501]

First, make the mixed fraction into improper fraction,

x−12=54

Add 1/2 on both sides

x−12+12=54+12

Find LCM,

LCM is 4

x=54+1⋅22⋅2

x=54+24

Add

x=74

Or,

x=134

Or,

x=1.75

X=1.75 IS YOUR ANSWER!

Hope this helps!

Have a great day: )

- Hailey:)

(Nothing copied or pasted!!!!!)

7 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago