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Volgvan
2 years ago
11

Pls help me to answer this​

Mathematics
1 answer:
babunello [35]2 years ago
3 0

I'll do the first two questions of each activity to get you started. The remaining unanswered questions will follow similar steps compared to the other questions in the same activity group.

=======================================================

Activity 5, Problem 1

To find the mean, we first add up the values

4+8+2+8+5+9 = 36

Then we divide over n = 6 because there are 6 values in this data set.

36/n = 36/6 = 6

<h3>Answer: Mean = 6</h3>

=======================================================

Activity 5, Problem 2

Follow the same idea as the previous problem.

Add the values:  9+8+8+13+12 = 50

Divide by the number of values: 50/n = 50/5 = 10

<h3>Answer:  Mean = 10</h3>

=======================================================

Activity 6, Problem 1

The median is the middle-most number. Think of a median of a highway or freeway.

In order to find the median, we first need to sort the data values from smallest to largest.

The original set {7,3,9,9,2,5,8} sorts to {2, 3, 5, 7, 8, 9, 9}

If we erased the first and last items, we get this smaller subset

{3, 5, 7, 8, 9}

repeat again to get

{5,7,8}

At this point its clear the middle most value is 7

But if you wanted, you could cross the first and last values from the list to end up with {7}

<h3>Answer: Median = 7</h3>

=======================================================

Activity 6, Problem 2

We could do the same trick as the previous problem, but I'll use a slightly different route this time.

We'll need to sort {10,8,4,11,9} into {4, 8, 9, 10, 11}

There are n = 5 items in that set. The middle slot would be at slot n/2 = 5/2 = 2.5 = 3

The value 9 is in slot 3 of the sorted items.

<h3>Answer: Median = 9</h3>

=======================================================

Activity 7, Problem 1

The mode is the most frequent value of a set. It repeats the most often.

The set {5,6,7,3,5,4} sorts to {3,4,5,5,6,7}

The value "5" shows up the most. It shows up twice when everything else shows up exactly once only.

<h3>Answer: Mode = 5</h3>

=======================================================

Activity 7, Problem 2

Using the same idea in the previous problem, the set {7,4,11,9,7,3} has the mode 7 because it shows up the most of any other value.

Here's the sorted set: {3,4,7,7,9,11}

Sorting is optional with finding the mode in contrast to being mandatory when it comes to finding the median.

<h3>Answer: Mode = 7</h3>
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Samantha has a shelf that is 95 over 4 inches wide. How many books can Samantha arrange on the shelf if each book is 5 over 4 in
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Answer:

Option 1 - 19, because the number of books is 95 over 4 divided by 5 over 4

Step-by-step explanation:

Given : Samantha has a shelf that is 95 over 4 inches wide.

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Solution :

The thickness of shelf = \frac{95}{4} inches

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\text{Number\ of\ books\ arranged}=\dfrac{\text{Thickness\ of\ shelf}}{\text{Thickness\ of\ one\ book}}

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19, because the number of books is 95 over 4 divided by 5 over 4

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π

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lim(x→0) sin(π cos²x) / x²

Use Pythagorean identity:

lim(x→0) sin(π (1 − sin²x)) / x²

lim(x→0) sin(π − π sin²x) / x²

Use angle difference formula:

lim(x→0) [ sin(π) cos(-π sin²x) − cos(π) sin(-π sin²x) ] / x²

lim(x→0) -sin(-π sin²x) / x²

Use angle reflection formula:

lim(x→0) sin(π sin²x) / x²

Now we multiply by π sin²x / π sin²x.

lim(x→0) [ sin(π sin²x) / x² ] (π sin²x / π sin²x)

lim(x→0) [ sin(π sin²x) / π sin²x] (π sin²x / x²)

lim(x→0) [ sin(π sin²x) / π sin²x] lim(x→0) (π sin²x / x²)

π lim(x→0) [ sin(π sin²x) / π sin²x] [lim(x→0) (sin x / x)]²

Use identity lim(u→0) (sin u / u) = 1.

π (1) (1)²

π

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If we plug in x = 0, the limit evaluates to 0/0.  So using L'Hopital's rule:

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lim(x→0) [ -π cos(π cos²x) sin(2x) ] / 2x

-π/2 lim(x→0) [ cos(π cos²x) sin(2x) ] / x

Again, the limit evaluates to 0/0.  So using L'Hopital's rule one more time:

-π/2 lim(x→0) [ cos(π cos²x) (2 cos(2x)) + (-sin(π cos²x) (-2π cos x sin x)) sin(2x) ] / 1

-π/2 lim(x→0) [ 2 cos(π cos²x) cos(2x) + π sin(π cos²x) sin²(2x) ]

-π/2 (-2)

π

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