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babunello [35]
3 years ago
11

Mario can eat 21 hot dogs in 6 minutes. At the same place, how many minutes would it take him to eat 35 hot dogs? ​

Mathematics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

10 minutes

Step-by-step explanation:

Mario can eat e21 hot dog in 6 minutes

Let 35 hot dog  eats in x minutes :

\frac{35}{x} =\frac{21}{6} \\x=\frac{35(6)}{21} \\x=10 min

Therefore, mario would eat 35 dogs in 10 min.

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Julli [10]

The annual salary of Mrs. Fredrick is $ 39397.2

<u>Solution:</u>

Given, Mrs. Frederick is paid semimonthly.  

Her semi-monthly salary is $1.641.55.  

Now, let us find her monthly salary first.

<em>monthly salary = 2 x semi – monthly salary </em>

So, monthly salary = 2 x 1,641.55 = $ 3283.1

Since there 12 months in a year, we obtain the annual salary as follows:

Now, the<em> annual salary = 12 x monthly salary </em>

Annual salary = 12 x 3283.1  = $ 39397.2

Hence, the annual salary of Mrs. Fredrick is $ 39397.2

5 0
3 years ago
How many arrangements are possible using all of the letters in WHISTLER?
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7 0
3 years ago
A snack bar offers chili and/or cheese on ithot dogs for an additional fee.
Sloan [31]

Answer:

Hi Im Pride and i would like to take you to your answer your answer to this question is

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Step-by-step explanation:

please mark brainly

5 0
2 years ago
Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have
boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after  10 days and 20 days, respectively.

We will use half life formula.

A=a\cdot(\frac{1}{2})^{\frac{t}{h}}, where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}

A=1\text{ kg}\cdot(\frac{1}{2})^{1}

A=1\text{ kg}\cdot\frac{1}{2}

A=0.5\text{ kg}

Therefore, amount of the material left after 10 days would be 0.5 kg.

A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}

A=1\text{ kg}\cdot(\frac{1}{2})^{2}

A=1\text{ kg}\cdot\frac{1^2}{2^2}

A=1\text{ kg}\cdot\frac{1}{4}

A=0.25\text{ kg}

Therefore, amount of the material left after 20 days would be 0.25 kg.

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3 years ago
What is an obtuse angle?
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