2 years ago

If a student ate 3/4 of their meals away from home, what percent of the total day is spent other than at home?

Mathematics

1 answer:

xxTIMURxx [149]2 years ago

3 0

Its 6%, just divide the eating section by four, then vwolaaa

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There is a harvest festival each year on the planet Bozone. During that time, most of the Bozone residents sit down to rejoice a

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Answer:

The price would be decreased by 18 bozats

Step-by-step explanation:

The following information is given in the question

x = number of kilograms of snig sold

P = Price per kilogram

And, the equation is

p = 300 - 18x

Now if an extra kilogram is sold so it should be x+1

Now the new price is

New price = 300 - 18(x + 1)

= 300 - 18x - 18

Therefore the price would be decreased by 18 bozats

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5 1/2 feet tall because the shadow is 1 foot shorter and so must the person be 1 foot shorter

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A coach gives her players the option of running around their field twice or around their entire stadium once. The following diag

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Get a call to me when I can do it on Saturday

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What is the most precise name for quadrilateral ABCD with vertices A(-2,4), B(5,6), C(12,4) and D(5,2)?

Dmitry_Shevchenko [17]

I think its going to be around B or D

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Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of

sasho [114]

Answer:

a)0,45119

b)1

Step-by-step explanation:

For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)

$P=\frac{1}{365} *\frac{1}{365}=\frac{1}{133225} \\$

Now the poisson approximation is used

λ=nP=80000*1/133225=0,6

Now, let X be the number of couples that birth April 30

P(X ≥ 1) =

1 − P(X = 0) =

$1-\frac{(e^-0.6)*(-0,6)^{0} }{0!}$

P(X ≥ 1) = 0,45119

B) Now want to find the

probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday

$P=52*\frac{1}{365} *\frac{1}{365}=\frac{52}{133225} \\$

Now the poisson approximation is used

λ=nP=80000*52/133225=31.225

Now, let X be the number of couples that birth same day

P(X ≥ 1) =

1 − P(X = 0) =

$1-\frac{(e^-31.225)*(-31.225)^{0} }{0!}$

P(X ≥ 1) = 1

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2 years ago