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3 years ago

The elementary school students are putting on a play as a fundraiser. The student tickets cost $2.00 each and

Mathematics

2 answers:

Lena [83]3 years ago

6 0

Answer:

850 for student and 850 for students

Step-by-step explanation:

Alex73 [517]3 years ago

6 0

<h3>Answer with Step-by-step explanation:</h3>

Let student tickets sold be=x

and adult tickets sold=y

They sold 5 times as many student tickets as adult tickets.

i.e. x=5y

The student tickets cost $2.00 each and

the adult tickets cost $4 each.

They made $140.

i.e. 2x + 4y = 140

Dividing both sides by 6, we get

x + 2y = 70

or 5y + 2y= 70

i.e. 7y=70

i.e. y=10

x= 5y

⇒ x = 50

Hence, Student tickets sold= 50

and Adult tickets sold= 10

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How can sample data be used to from an estimate about the population?

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Answer: A proportion can be solved to relate the sample to the population and solved to form an estimate.

Step-by-step explanation:

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3 years ago

Which multiplication expression is eqeul to 3/5 ÷ 1/2

eduard

Answer:

Lets solve = 3/5 ÷ 1/2

3/5 ÷ 1/2

= 3/5 × 2/1

= 3 × 2/5 × 1

= 6/5

In mixed fraction its

$1 \frac{1}{5}$

____________________

Now lets solve "a" option

3/5 × 2

= 3 × 2/5 × 1

= 6/5

In mixed fraction

$1 \frac{1}{5}$

So option a = 3/5 x 2/1 is ur answer

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2 years ago

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You recently purchased a new snowboard for the upcoming winter. The board, with bindings, came out to be $315.32. You were told

gizmo_the_mogwai [7]

First the original price is reduced by 15%
assume the original price is x
100%-15%=85%..u r left with 85% of the price
thus,85/100 multiply x
the reduced price is reduced again by 5%
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find x
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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.

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3 years ago

Elizabeth is making lemonade. She uses a box of mix to make of a gallon. How much mix will she need to make one gallon?

velikii [3]

The answer is she needs one more

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3 years ago

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