Ask question

Ask question

All categories

3 years ago

7

Coach Hutchison is the basketball coach at school . she asks players to record their heights. The histogram shows the data she h

as collected
What unit of measure is used in the Study?
__________________________________

O A. Inches

O B. Feet

O C. Number of playerd

O D. Number of observations

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

The unit of measure used in the study is "inches"

You might be interested in

Can someone please help me with this question?? (no links!!!)

Tpy6a [65]

Answer:

180ft

Step-by-step explanation:

By using the formula, we do Pi (3) x the radius which is 6² then x by the height, so, we do 3 x 6² x 5. 6² is 36. 3 x 36 = 108. Next, we do 5 x 108. This gives us 540. Now, we have to divide by 3. This makes our answer 180ft.

7 0

3 years ago

What is the sum of 4.9 × 102 and 6.8 × 103?

dusya [7]

Answer:

The sum is

$7.92\times 10^3$

Step-by-step explanation:

We want to add

$4.9\times 10^2 and 6.8\times 10^3$

The sum of these two numbers is:

$4.9\times 10^2 +6.8\times 10^3$

This is the same as:

$4.9\times 100 +6.8\times 1000$

$490 +6800=7290$

We rewrite 7290 in standard form to get;

$7.92\times 10^3$

6 0

3 years ago

Transversal relations

Shalnov [3]

Answer:

The answer is 75

Step-by-step explanation:

By way of parallelism

$x + 15 = 90 \\ x = 75$

7 0

3 years ago

Read 2 more answers

Stacy sells art prints for $12 each. Her expenses are $2.50 per print, plus $38 for equipment. How many prints must she sell for

charle [14.2K]

4 prints
12x = 2.5x + 38
at 4, these expressions are equal

7 0

3 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

3 years ago