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olga2289 [7]
3 years ago
15

Seven more than twice a number is equal to 21. Find all the numbers that make this sentence is true.

Mathematics
1 answer:
julsineya [31]3 years ago
6 0

x = a number

7 + 2x = 21

Subtract 7 from both sides

2x = 14

Divide both sides by 2

x = 7

7 is the only number that makes this sentence true.

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almond37 [142]

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If the perimeter is 102, how long is the longest side?
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25.5
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Selecting an Instructor At a convention there are 7 mathematics instructors, 5 computer science instructors, 3 statistics instru
LiRa [457]

Answer:

11/19

Step-by-step explanation:

Firstly, we need to know the total number of instructors. This is equal to 7+5+4+3 = 19 instructors.

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The probability of selecting a math instructor is 7/19

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4 0
3 years ago
You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? Yes No (b) Find P(1 on green die an
navik [9.2K]

Answer:

1) yes ; 2) 1/36 ; 3) 1/36 ; 4) 1/18

Step-by-step explanation:

Given the following :

Two fair dice : 1 green ; 1 red

A) Are the outcomes on the dice independent:

Yes, becomes the outcome of the green dice does not have any effect on the outcome of the red dice.

B) Find P(1 on green die and 5 on red die).

Probability = (number of required outcome) / (total possible outcomes)

Total outcomes of a dice = 6

P(1 on green) = 1 / 6

P(5 on red) = 1/6

P(1 on green die and 5 on red die) :

(1/ 6) × (1/6) = 1/36

C) Find P(5 on green die and 1 on red die)

P(5 on green) = 1/6

P(1 on red) = 1/6

Find P(5 on green die and 1 on red die):

1/6 × 1/6 = 1/36

D) Find P((1 on green die and 5 on red die) or (5 on green die and 1 on red die))

P(5 on green die and 1 on red die) = 1/36

P(1 on green die and 5 on red die) = 1/36

P((1 on green die and 5 on red die) or (5 on green die and 1 on red die)) =

P(5 on green die and 1 on red die) + P(1 on green die and 5 on red die)

= (1/36 + 1/36) = 2 /36 = 1/18

8 0
3 years ago
Let ​f(x)equals=33xminus−​1, ​h(x)equals=startfraction 7 over x plus 5 endfraction 7 x+5 . find ​(hcircle◦​f)(66​).
Ronch [10]
\bf \begin{cases}
f(x)=3x-1\\
h(x)=\cfrac{7}{x+5}\\\\
(h\circ f)(x)=h(~~f(x)~~)
\end{cases}
\\\\\\
f(66)=3(66)-1\implies \boxed{f(66)=197}
\\\\\\
\stackrel{h(~~f(x)~~)}{h(~~f(66)~~)}\implies \stackrel{h(~~f(66)~~)}{h\left(~~\boxed{197}~~ \right)}=\cfrac{7}{\boxed{197}+5}\implies h\left(~~\boxed{197}~~ \right)=\cfrac{7}{202}
7 0
3 years ago
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