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2 years ago

B) There are at most 12 students

Mathematics

1 answer:

Vaselesa [24]2 years ago

7 0

ehh??

your question is incomplete I guess

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The Smith Family made an $800 down payment and pays $75 per month for new furniture. At the same time the Cooper family made a $

spin [16.1K]

75m + 800 = 95m + 500
Solve that equation
Which should get you m=15
15 is your answer

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3 years ago

Evan plant grass in a rectangular space behind the clubhouse

Nastasia [14]

A= l x b
70 = 14b
divide both sides by 14
b= 5

6 0

2 years ago

The table below shows field goal statistics for three different kickers

FromTheMoon [43]

Answer:

<h2>From least to greatest: C, B and A</h2>

Step-by-step explanation:

To order each Kicker correctly, we have to find the success rate of each one, which is obtained by dividing and then multiplying by 100 to obtained the percentage:

$R=\frac{made}{attempted}$

<h3>Success rate of A:</h3>

$R_{A}=\frac{12}{14}.100=86\%$

<h3>Success rate of B:</h3>

$R_{B}=\frac{15}{18}.100=83\%$

<h3>Success rate of C:</h3>

$R_{C}=\frac{19}{24}.100=79\%$

So, comparing all three, from least to greatest we have:

C with 79%, then B with 83%, and A with 86%.

3 0

3 years ago

Read 2 more answers

Which of the following Is a like radical to first one is question rest are answers

madam [21]

Y√5 is a like radical to 3x√5

8 0

3 years ago

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Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

6 0

3 years ago