2% Of 144.4 is what is what

Jen Butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 79.

xxMikexx [17]

<u>Answer:</u>

Price of the adult's ticket = $13

Price of a child's ticket = $ 10

<u>Step-by-step explanation:</u>

Given:

Cost of train fares for Three adults and four children = $79

Cost of train fares Two adults and three children = $ 56

To Find:

Price of the adult's ticket and the price of a child's ticket =?

Solution:

Let the cost of one adult's ticket be x and

Let the cost of one child's ticket be y

Then

Three adults and four children must pay $ 79 be

3x + 4y = 79-----------------------------------------------(1)

Two adults and three children must pay $56

2x + 3y = 56----------------------------------------------(2)

multiply eq(1) by 2, we get

6x + 8y = 158----------------------(3)

multiply eq(2) by 3, we get

6x + 9y = 168----------------------(4)

Subracting (3) from (4)

6x + 9y = 168

6x + 8y = 158

(-) (-) (-)

------------------------------

0x + y = 10

------------------------------

y=10

Now substitute the value of y in eq (1) to get the value of x

3x + 4(10)= 79

3x + 40 = 79

3x = 79 - 40

3x = 39

x = $\frac{39}{3}$

x= 13

3 0

4 years ago

ethan earned 5% of the maximum number of points possible . It ethan earned 40 points , what is the maximum number of points poss

Morgarella [4.7K]

Answer:

3840

Step-by-step explanation:

4 0

3 years ago

3. Find the measure of Part I: An inscribed angle is an angle formed by two chords of a circle which share an endpoint. Find the

nevsk [136]

Answer:

ther answer equation 7x10^2 equals to the left rectangle

Step-by-step explanation: take the left diciple and minus it by the left exponent leaving you with a hokie pokie into obamas back yard

5 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

What is x in 3x-7=5x+13

Elodia [21]

3 x - 7 = 5 x + 13

3 x - 5x = 13 + 7

-2x = 20

x = 20 / -2

x = - 10

7 0

3 years ago

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