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3 years ago

Edward entered into a 5-kilometer race. How many meters will he need to run?

Mathematics

1 answer:

EleoNora [17]3 years ago

3 0

Answer:

5000 meters

Step-by-step explanation:

I kilometer=1000 meters

5 kilometers=5000 meters

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What is the value of y in the equation 3(3y – 12) = 0? (5 points)

natta225 [31]

Set equal to 0

3=0, nope

3y-12=0
add 12
3y=12
divide 3
y=4

5p-4p-8=-2+3
p-8=1
p=9
1 solution

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4 years ago

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If I got $5 a week how much money would I have after 3 weeks

tensa zangetsu [6.8K]

You would get $15 in total because 5x3=15.

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3 years ago

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if the computer screen shows this image as calinda is downloading a file, how big is the file she’s loading ?

just olya [345]

Answer:

the file is 20 kb

Step-by-step explanation:

16kb/80%

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1600=80%

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3 years ago

If there are 17 green marbles in a jar of 65 marbles, what is the probability of drawing 2 green marbles at random without repla

erastovalidia [21]

Answer:

Option b) $\frac{17}{260}$

Step-by-step explanation:

We are given the following information in the question:

Total number of marbles = 65

Number of green marbles = 17

We have to find the probability of drawing 2 green marbles at random without replacement.

Formula:

$Probability = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$\text{P(Drawing 2 green marbles at random without replacement)} = \\\text{Probability of drawing }1^{st} \text{ green marble}\times \text{Probability of drawing }2^{nd} \text{ green marble}$

$\text{Probability of drawing }1^{st} \text{ green marble} = \displaystyle\frac{17}{65}\\\\\text{Probability of drawing }2^{nd} \text{ green marble} = \displaystyle\frac{16}{64}$

$\text{P(Drawing 2 green marbles at random without replacement)} =\displaystyle\frac{17}{65}\times \frac{16}{64}=\frac{17}{260}$

Option b) $\frac{17}{260}$

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3 years ago

What is 1 1/2 minus 1/4

vfiekz [6]

1 1/2 - 1/4
You have to have a common denominator which is 4 in this istance.
That makes it 1 2/4 - 1/4. That equals 1 3/4.

Hope that this helped! :)

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3 years ago

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