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2 years ago

Which ratio of the number of games won to the number of games played could also be included in this table?

Mathematics

2 answers:

juin [17]2 years ago

8 0

Answer:

3:5 is the ratio

Step-by-step explanation:

to get the ratio you got to pick any one of them (like 6 and 10 or maybe 24 and 40) and simplify it till you can't simplify it anymore, okay hope it helped <3

Zielflug [23.3K]2 years ago

6 0

I need points for a test to answer please don’t report

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If the angle NME = x+97, the angle EML = x+66, and the angle NML = 141, find the angle EML

defon

Answer:

∠ EML = 94°

Step-by-step explanation:

See the attached diagram.

The three angles ∠ NME, ∠ EML and ∠ NML together make 360°.

Now, ∠ NME = x + 97, ∠ EML = x + 66 and ∠ NML = 141°

Hence, (x + 97) + ( x + 66) + 141 = 360

⇒ 2x + 304 = 360

⇒ 2x = 56

⇒ x = 28

So, ∠ EML = x + 66 = 28 + 66 = 94° (Answer)

6 0

3 years ago

Which expression is the factored equivalent of x2 - 5x + 4?

ozzi

Answer:

Step-by-step explanation:

5 0

3 years ago

1. f(x) = x2 - 21<br> f(5)

matrenka [14]

Answer:

x=13

Step-by-step explanation:

if you will substitute the value of f(x) with 5, you will get the following equation:

5=2x-21

2x=26/÷2

x=13

(I didn't know what you ment when you wrote x2, is it x square or 2X)

I really hope it was the second one, cause that's how I solved it.

8 0

3 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago

the tables show the typical number of minutes spent exercising each week for a group of fourth-grade students and a group of sev

Ann [662]

What table? please provide a picture and I'll try and answer.

7 0

3 years ago