All categories

3 years ago

Please help explanation if possible

Mathematics

1 answer:

Eva8 [605]3 years ago

5 0

Answer:

612 adults

361 students

Step-by-step explanation:

To solve this question, set two equations:

Let x be number of adults and y be number of students.

As there are in total 937 people, the equation would be the sum of both adults and children:

$x+y=937$

$x=937-y$ ...... ( 1 )

As the total sale amount is $1109, the equation would be to add up the ticket fee:

$2x+0.75y=1,109$ ...... ( 2 )

Put ( 1 ) into ( 2 ):

$2(937-y)+0.75y=1,109$

$1874-2y+0.75y=1,109$

$-1.25y=-765$

$y=763/1.25$

$y=612$

Put y into ( 1 ):

$x=973-612$

$x=361$

Therefore there are 612 adults and 361 students.

You might be interested in

Which is more, 13 feet or 4 yards?

NikAS [45]

Answer:

it is 13 feet

Step-by-step explanation:

because it is 4.33333 yards

3 0

3 years ago

Please, Answer is needed.

Yuki888 [10]

Answer:

540 mins | 9 hours | 540, 9

420 mins | 7 hours | 420, 7

300 mins | 5 hours | 300, 5

180 mins | 3 hours | 180, 3

Step-by-step explanation:

1 hour = 60 mins

To convert hours into minutes, multiply the number of hours by 60 minutes:

9 hours × 60 mins = 540 mins

7 hours × 60 mins = 420 mins

5 hours × 60 mins = 300 mins

3 hours × 60 mins = 180 mins

5 0

3 years ago

Read 2 more answers

The ratio of boys to girls in a group is 7:3. If there are 40 more boys than girls, work out how many people there are in total.

LenaWriter [7]

Answer:

Step-by-step explanation:

7 - 3 = 4

7 + 3 = 10

4/10 = 40

10/10 = ?

10/10 x 40 x 10/4

= 100 people

7 0

3 years ago

Helppp meee!!!!!!!!!!

swat32

C because 180-230 would be negative b as well as three quarters of seven

4 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago