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Bingel [31]
2 years ago
8

Members of a junior high basketball team want to brag about how tall they are. First, they measured each player’s height in inch

es. Now they have to decide whether the mean or the median is better. Which measure of center should the team use? Basketball team’s height (in inches) 64, 67, 83, 65, 66, 62, 69 The team’s mean height is. The team’s median height is. If the team members wants to brag about how tall they are, they should use their height.
Mathematics
2 answers:
nirvana33 [79]2 years ago
5 0

Answer:

The team’s mean height is

✔ 68

The team’s median height is

✔ 66

If the team members wants to brag about how tall they are, they should use their

✔ mean

height.

Step-by-step explanation:

stealth61 [152]2 years ago
3 0

Answer: mean is 66 and the median height is 68.

Step-by-step explanation: uh I dont know what to put here....?

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n chemistry, the pH of a solution is defined by: pH space equals space minus log space left square bracket straight H plus right
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Answer:

The correct option is;

pH = 4.1

Step-by-step explanation:

The given information are;

The hydrogen ion [H⁺] concentration in tomato juice = 10 ^{-4.1}

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4 0
3 years ago
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6 0
2 years ago
Read 2 more answers
A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the
Naya [18.7K]

Answer:

The area is growing at a rate of \frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

Step-by-step explanation:

<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 \frac{cm}{min} and we identify such as the following differential rate:

\frac{dr}{dt} = 3\,\frac{cm}{min}

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find  \frac{dA}{dt}.

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

A=\pi\,r^2

We now apply the derivative operator with respect to time (\frac{d}{dt}) to this equation, and use chain rule as we find the quadratic form of the radius:

\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}

Now we replace the known values of the rate at which the radius is growing ( \frac{dr}{dt} = 3\,\frac{cm}{min}), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\

which we can round to one decimal place as:

\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

4 0
3 years ago
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