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3 years ago

12

The given equation is either linear or equivalent to a linear equation. solve the equation. (if there is no solution, enter no s

olution. if all real numbers are solutions, enter reals.) 3 x + 12 = x + 3 3

1 answer:

murzikaleks [220]3 years ago

6 0

2x+12=33
2x=21
x=10.5
The answer is 10.5

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3 There are 5 times as many boys as girls in a school.

Dafna1 [17]

Answer:

Number of boys = 675

Step-by-step explanation:

Given that:

There are 5 times as many boys as girls in a school.

Number of pupils in school = 810

Let,

x be the number of girls

y be the number of boys

According to given statement;

x+y=810 Eqn 1

y = 5x Eqn 2

Putting y=5x in the Eqn 1

x+5x=810

6x=810

Dividing both sides by 6

$\frac{6x}{6}=\frac{810}{6}\\x=135$

Putting x=135 in Eqn 2

y = 5(135)

y = 675

Hence,

Number of boys = 675

8 0

3 years ago

The value of y varies directly with x, and y = 16 when x = 8. What is the value of x

Natali [406]

Answer:

A) 16

Step-by-step explanation:

y = kx

16 = k*8

16/8 = k

k = 2

y = 2x

32 = 2x

32/2 = x

16 = x

5 0

3 years ago

A probability experiment is conducted in which the sample space of the experiment is S 1,2,3,4, 5,6,7,8, 9, 10, 11, 12]. Let eve

Alenkinab [10]

Answer:

$E\ n\ F = \{1, 5, 6,7\}$

$F\ n\ G =\{9\}$

$F\ U\ G =\{1,5,6, 7, 8,9,10, 11, 12\}$

$P(F\ U\ G) = \frac{3}{4}$

Step-by-step explanation:

Given

$S = \{1,2,3,4, 5,6,7,8, 9, 10, 11, 12\}$

$E = \{1,2, 3, 4, 5, 6,7\}$

$F =\{1,5,6, 7, 8,9\}$

$G = \{9,10, 11, 12\}$

$H = \{2, 3, 4\}$

Solving (5): E and F

Sets of E and F = E n F

$E\ n\ F = \{1,2, 3, 4, 5, 6,7\}\ n\ \{1,5,6, 7, 8,9\}$

List out common elements

$E\ n\ F = \{1, 5, 6,7\}$

<em>They are not mutually exclusive because </em> $n(E\ n\ F) \neq 0$ <em />

Solving (6): F and G

Sets of F and G = F n G

$F\ n\ G =\{1,5,6, 7, 8,9\}\ n\ \{9,10, 11, 12\}$

List out common elements

$F\ n\ G =\{9\}$

<em>They are not mutually exclusive because </em> $n(F\ n\ G) \neq 0$ <em />

Solving (7): F or G and P(F or G)

Sets of F or G = F U G

$F\ U\ G =\{1,5,6, 7, 8,9\}\ U\ \{9,10, 11, 12\}$

List all elements without repetition

$F\ U\ G =\{1,5,6, 7, 8,9,10, 11, 12\}$

Solving P(F U G)

$P(F\ U\ G) = \frac{n(F\ U\ G)}{n(S)}$

$n(F\ U\ G) = 9$

$n(S) = 12$

Hence;

$P(F\ U\ G) = \frac{9}{12}$

Divide by 3

$P(F\ U\ G) = \frac{3}{4}$

6 0

3 years ago

I'm tired just did over 10 works help please thank you

Dafna11 [192]

Part A. For finding the quartiles, box plot would be the best to represent the range of scores of the students.

Part B. The first step is to order the data: 2 3 3 3 3 4 5 6 7 8 14. Next, you will find the median. The median is 4. So, second the quartile = 4 The median splits the remaining data into two sets. The first set is 2 3 3 3 3 and the median of this set is 3. So, first quartile = 3. The second set is 5 6 7 8 14 and the median of this set is 7. And the third quartile = 7

I did this test today i know

3 0

3 years ago

Javier bought a painting for $150. Each year the painting's value increases by a factor of 1.15. What's the expression that give

lubasha [3.4K]

Multiply 1.15 times 7 to find the painting value

8 0

3 years ago

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