Bob had 18 crackers Sam eat 1/6 of the crackers and sue ate 1/3 of the crackers. how many crackers did Bob have left

2 answers:

6 0

We can add 1/6 & 1/3 together to find out the total amount of crackers that were eaten:

1/6 + 1/3 (Find the Common Denominator. In this case it is 6.)
= 1/6 + 2/6
= 3/6 (Now find the Lowest Terms. Both 6 & 3 divide into 3)
= 1/2 (In Lowest Terms)

This means that 1/2 of the Crackers have been eaten.

1/2 of 18 = 9

Therefore, Bob has 9 Crackers left.

Basile [38]3 years ago

3 0

The first thing to do is make both fractions over 6:
1/3 = 2/6
Sue at 2/6 of the crackers.

Now we can add the fractions:
2/6 + 1/6 = 3/6

The next thing to do is cancel down the fraction.
3/3 = 1
6/3 = 2
3/6 = 1/2

Now that we know that half of the crackers have been eaten, we halve the total amount:
18 X 1/2 = 18 / 2 = 9

Bob had 9 crackers left.

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Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

aleksandr82 [10.1K]

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$ , then $f(\pi)=e^{\sin\pi}=e^0=1$ . Recall the definition of the derivative of a function $f(x)$ at a point $x=c$ :

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$

4 0

3 years ago

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deff fn [24]

Since the sum of the probabilities of all possible outcomes must be 100%, we can deduce the following:

Cooking in under 20 minutes: 10%
Cooking between 20 and 30 minutes: 85%
Cooking in more than 30 minutes: 5%

In fact, the probabilities of cooking in less than 20 or more than 30 sum up to 15%, which means that the remaining outcome (i.e. cooking time between 20 and 30) must complete this probability to 15, and in fact 15+85=100.

That being said, all three answers are simply a combination of these three scenarios: let C be the cooking time, for aesthetic reasons:

$P(C\geq 20) = P(20 \leq C \leq 30)+P(C\geq30) = 85\%+5\%=90\%$