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3 years ago

5

Norma bought a plastic bag that contains 100 ballons the lable says that there are 20 each of blue,yellow, red, ,green and orang

e ballons what is the probability that the ballon will be either red or blue?

1 answer:

Volgvan3 years ago

7 0

Answer:

It is 120

Step-by-step explanation:

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Jamie rode her bike home for 5 blocks before realizing she forgot her math book at school and would need it for her homework. Sh

Alik [6]

B shows her getting closer to home, immediately turning round, gives time for finding book then riding all the way home at a constant speed

5 0

3 years ago

Read 2 more answers

Searches related to Anita needs 5 pounds of bananas to make banana bread for a bake sale. Each pond of bananas cost 0.50. How ca

m_a_m_a [10]

Answer:

$2.5

Step-by-step explanation:

Number of pounds of banana needed = 5 pounds

Cost per pound of banana = 0.50

To solve using place value blocks :

Ignoring the decimal in the cost per pound:

We can first take the product of 5 (Number of pounds of banana) and take cost per pound as 5. Since it is easier to make direct multiplication without decimal.

5 * 5 = $25

Now we can add the one decimal point (in the cost per pound) to our result

Then we have $2.5

Hence, total cost = $2.5

4 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

<u>Differentiation</u> (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

<u>Differential</u> (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

2 years ago

A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze, and the second brand is 90% pure anti

Murrr4er [49]

With amounts measured in gallons, let

x = amount of 65% antifreeze

y = amount of 90% antifreeze

1 gal of the 65% brand contains 0.65 gal of pure antifreeze; x gal would contain 0.65x gal. Similarly, y gal of the 90% brand contains 0.90y gal of pure antifreeze.

To obtain 120 gal of 80% antifreeze solution (which contains 0.80•120 = 96 gal of pure antifreeze), we must have

x + y = 120 … … … … … [total volume of antifreeze solution]

0.65x + 0.90y = 96 … [total volume of pure antifreeze]

Solve the first equation for y :

y = 120 - x

Substitute this into the second equation and solve for x :

0.65x + 0.90 (120 - x) = 96

0.65x + 108 - 0.90x = 96

0.25x = 12

x = 48

Solve for y :

y = 120 - 48

y = 72

6 0

2 years ago

Find the probability on #4<br> Please

julia-pushkina [17]

The probability of drawing a 5 from card numbered 1 - 10 is 1/10 because only one card out of ten is numbered 5.
The probability of rolling a 2 on the cube is 1/6 because there are 6 faces of the cube with 6 different numbers. The probability of picking a 5 and rolling a 2 together is 1/60 because
1/10 x 1/6 = 1/60. You multiply them together because there will be 60 combinations: 10 card possibilities and 6 cube possibilities.
Hope this helps!

3 0

3 years ago