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2 years ago

the train ride at the zoo covers a distance of 2 1/2 miles in 1/3 of and hour.How many miles per hour does the train go?

Mathematics

1 answer:

TEA [102]2 years ago

7 0

Answer:

7 1/2 mph

Step-by-step explanation:

2 1/2=1/3 hour

5 = 2/3 hour

7 1/2 = 1 hour

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2) Puppy B weighs 8 pounds, which is about 75% of its adult weight. What will be the

masha68 [24]

Answer:

Puppy A weighs 8 pounds, which is about 25% of its adult weight. What will be the adult weight of Puppy A? Puppy B weighs 8 pounds, which is about 75% of its adult weight. What will be the adult weight of Puppy B?

Step-by-step explanation:

6 0

3 years ago

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Ex plane how knowing 50 x 4 = 200 helps you find 500 x 400

Alex777 [14]

Answer:

Step-by-step explanation:

if you know 50*4=200, you must also know that 50*10=500 and that 4*100=400 so 500*400= 50*10*4*100

multiplication is comutative so

50*4*10*100=200*1,000= 200, 000

8 0

3 years ago

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The airplane Li’s family will be flying on can seat up to 149 passengers. If 96 passengers are currently on the plane, which ine

Ivenika [448]

For this case, the first thing you should do is define a variable.
We have then:
x: number of passengers remaining who can board the plane.
We have as data:
1) They can board up to 149 passengers
2) There are 96 passengers currently aboard.
Writing inequality we have:
$x + 96 \leq 149
$
Answer:
An inequality that can be used to determine how many more people can board is:
$x + 96 \leq 149$

3 0

3 years ago

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What's 2+2?<br> A) 5<br> B) 2<br> C) 4<br> D) 10

Triss [41]

Answer:

C) 4

Step-by-step explanation:

I hope this helps :)

3 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