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Gelneren [198K]
3 years ago
12

Which measure is always the 50th percentile? a. mean b. median c. upper quartile

Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

The 50th percentile of a data set is always equal to the median. percentile means the order of a point in a set. for example, 97th percentile means your score is 97% higher than the rest. Thus 50th percentile also applies fot the median of the set.
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14. MULTIPLE CHOICE The number of students in the marching band increased from 100 to 125. What is the percent of increase? (Ski
slamgirl [31]

The percentage of the increase will be 25% when the number of students in the marching band increased from 100 to 125.

<h3>What is the percentage?</h3>

The Percentage is defined as representing any number with respect to the 100. It is denoted by the sign %.

It is given in the question that the number of students in the marching band increased from 100 to 125 so we need to find the percentage increase of the students.

The percentage increase of the students will be calculated as:-

% increase=[(125-100)÷(100)]x100

% increase=[(25)÷(100)]x100

% increase= [0.25x100]

% increase= 25%

Hence the percentage of the increase will be 25% when the number of students in the marching band increased from 100 to 125.

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Answer:

The ones selected are correct

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80% of salmon pass through a single dam unharmed. By what percent does the number of salmon decrease when passing through a sing
aalyn [17]

Answer:20%

Step-by-step explanation:

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3 years ago
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Which expressions are equivalent to the one below? Check all that apply.
dem82 [27]

Answer:

Recall from Logarithm law Log a - Log b = Log (a/b)

so, Log 2 - Log 4 = Log ( 2/4)

                            = Log ( 1/2).

Also from the law

Log (a/b) = Log a - Log b

i.e Log (1/2) = Log 1 - Log 2

Recall: Log 1 = 0

which implies : Log 1 - Log 2 = 0 - Log 2

                                               = - Log 2

Step-by-step explanation:

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Calc BC Problem. No random answers plz
stepladder [879]

Answer:

Part A)

f(1)=2, \; f^{-1}(1)=0, \; f^\prime(1)=1.4, \; (f^{-1})^\prime(1)=\frac{10}{7}

Part B)

y=\frac{5}{14}x+\frac{4}{7}

Step-by-step explanation:

Please refer to the table of values.

Part A)

A. 1)

We want to find f(1).

According to the table, when x=1, f(x)=2.

Hence, f(1)=2.

A. 2)

We want to find f⁻¹(1).

Notice that when x=0, f(x)=1.

So, f(0)=1.

Then by definition of inverses, f⁻¹(1)=0.

A. 3)

We want to find f’(1).

According to the table, when x=1, f’(x)=1.4.

Hence, f’(1)=1.4.

A. 4)

We will need to do some calculus.

Let g(x) equal to f⁻¹(x). Then by the definition of inverses:

f(g(x))=x

Take the derivative of both sides with respect to x. On the left, this will require the chain rule. Therefore:

f^\prime(g(x))\cdot g^\prime(x)=1

Solve for g’(x):

g^\prime(x)=\frac{1}{f^\prime(g(x))}

Substituting back f⁻¹(x) for g(x) yields:

(f^{-1})^\prime(x)=\frac{1}{f^\prime({f^{-1}(x)})}

Therefore:

(f^{-1})^\prime(1)=\frac{1}{f^\prime({f^{-1}(1)})}

We already determined previously that f⁻¹(1) is 0. Therefore:

(f^{-1})^\prime(1)=\frac{1}{f^\prime(0)}

According to the table, f’(0) is 0.7. So:

(f^{-1})^\prime(1)=\frac{1}{0.7}=\frac{10}{7}

Hence, (f⁻¹)’(1)=10/7.

Part B)

We want to find the equation of the tangent line of y=f⁻¹(x) at x=4.

First, let’s determine the points. Since f(2)=4, this means that f⁻¹(4)=2.

Hence, our point is (4, 2).

We will now need to find our slope. This will be the derivative at x=4. Therefore:

(f^{-1})^\prime(4)=\frac{1}{f^\prime({f^{-1}(4)})}

We know that f⁻¹(4)=2. So:

(f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}

Evaluate:

(f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}=\frac{1}{2.8}=\frac{10}{28}=\frac{5}{14}

Now, we can use the point slope form. Our point is (4, 2) and our slope at that point is 5/14.

So:

y-2=\frac{5}{14}(x-4)

Solve for y:

y-2=\frac{5}{14}x-\frac{20}{14}

Adding 2 to both sides yields:

y=\frac{5}{14}x-\frac{20}{14}+\frac{28}{14}

Hence, our equation is:

y=\frac{5}{14}x+\frac{4}{7}

7 0
3 years ago
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