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2 years ago

runben sees 14 wheels on a total of 6 bicycles and tricycles how many bicycles and tricycles are there

Mathematics

1 answer:

olganol [36]2 years ago

8 0

Answer:

B = 4

T = 2

Step-by-step explanation:

First, figure out the equation.

We know that bicycles have 2 wheels, that there are 3 on a tricycle, and there are a total of 14 wheels and a total of 6 bicycles and tricycles.

Let b stand for bicycles and t stand for tricycles:

14 = 2b + 3t

6 = b + t

We can figure out the amount of either by rearranging the second equation to isolate one variable. I will solve it in two ways

In the first way, I will solve for b

(-t) 6 = b + t (-t)

6 - t = b

Plug this into the first equation and solve for remaining variable

14 = 2(6 - t) +3t

14 = 12 - 2t + 3t

14 = 12 +t

-12 -12

2 = t

6 - 2 = b

b = 4

The second way was to solve for t first

(-b) 6 = b + t (-b)

6 - b = t

14 = 2b + 3(6 - b)

14 = 2b + 18 - 3b

(-18) 14 = 18 -b (-18)

-4/-1 = -b/-1

b = 4

6 - 4 = t

t = 2

It doesn't matter which way you go, they both give you the exact same answer.

Sooo, recap!

1) write equations

2) switch the easier of the two to isolate one variable

3) substitute to find other variable (x2)

4) Find answers! =D

Hope this helps!

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K is the midpoint of PQ, P has

Pachacha [2.7K]

Answer:

Coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

Step-by-step explanation:

We are given:

K is the midpoint of PQ

Coordinates of P = (-9,-4)

Coordinates of K = (-1,6)

We need to find coordinates of Q $(x_2,y_2)$

We will use the formula of midpoint: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We are given midpoint K and $x_1,y_1$ the coordinates of P we need to find $x_2,y_2$ the coordinates of Q.

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(-1,6)=(\frac{-9+x_2}{2},\frac{-4+y_2}{2})\\$

Now, we can write

$-1=\frac{-9+x_2}{2}, 6=\frac{-4+y_2}{2}\\Simplifying:\\-2=-9+x_2\:,\: 12=-4+y_2\\-2+9=x_2\:,\: 12+4=+y_2\\x_2=7\:,\:y_2=16$

So, we get coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

6 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

Zigmanuir [339]

Answer:

Ai. Arithmetic sequence

Aii. Tn = 5 + 7n

Bi. Geometric

Bii. Tn = 8 × 2ⁿ¯¹

Step-by-step explanation:

To successfully answer the questions given above, note the following:

1. If the sequence is Arithmetic, then:

2nd – 1st = 3rd – 2nd = common difference (d)

2. If the sequence is geometric, then,

2nd / 1st = 3rd / 2nd = common ratio (r)

3. A sequence can not be arithmetic geometric at the same time.

4. The nth term of arithmetic sequence is:

Tn = a + (n – 1)d

5. The nth term of geometric sequence is:

Tn = arⁿ¯¹

A. Sequence => 12, 19, 26

i. Determination of the type of sequence.

We'll begin by calculating the common difference

1st term = 12

2nd term = 19

3rd term = 26

Common difference (d) = 2nd – 1st

d = 19 – 12 = 7

d = 3rd – 2nd

d = 26 – 19 = 7

Since a common difference exist in the sequence, the sequence is arithmetic sequence.

ii. Determination of the nth term.

Common difference (d) = 7

1st term (a) = 12

nth term (Tn) =?

Tn = a + (n – 1)d

Tn = 12 + (n – 1)7

Tn = 12 + 7n – 7

Tn = 5 + 7n

B. Sequence => 8, 16, 32

Bi. Determination of the type of sequence.

Let us begin by calculating the common ratio.

1st term = 8

2nd term = 16

3rd term = 32

Common ratio (r) = 2nd / 1st

r = 16 / 8

r = 2

r = 3rd / 2nd

r = 32 / 16

r = 2

Since a common ratio exist in the sequence, the sequence is geometric.

Bii. Determination of the nth term.

Common ratio(r) = 2

1st term (a) = 8

nth term =?

Tn = arⁿ¯¹

Tn = 8 × 2ⁿ¯¹

8 0

3 years ago

What is the percent of change if 20 is increased to 25?

natima [27]

That would be 25%

that is because 5 is 25% of 20

Hope this helps!

6 0

3 years ago

Read 2 more answers

What is an estimation strategy<br> This is for 6th grade math please help

Lunna [17]

Answer:

Estimation is a mental process of coming up with an answer that is relatively close, to allow decisions to be made. The types of estimation are quantity, computation and measurement.

An estimation strategy is simply coming up with a good estimate based on evidence and already known facts. Coming to a conclusion as close to the real answer as you can. It can also be called an educated guess.

8 0

3 years ago