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Tems11 [23]
3 years ago
13

A pet store sells puppies and kittens in the ratio 5 : 4 . If their sales of puppies and kittens combined came to 27, how many p

uppies did they sell?
Mathematics
2 answers:
Anika [276]3 years ago
4 0

Answer:

15 puppies

Step-by-step explanation:

A pet store sells puppies and kittens in the ratio 5 : 4

As per the ratio, total puppies and kittens is 5+4= 9

If their sales of puppies and kittens combined came to 27

The ratio of puppies to the total is 5:9

LEt x be the puppies sold

RAtio of puppies sold to the total is x: 27

Now make a proportion using the ratio and solve for x

\frac{5}{9} =\frac{x}{27}

cross multiply it

135=9x

Divide both sides by 9

x=15

S, 15 puppies

dimulka [17.4K]3 years ago
3 0
So puppies will be 5x and Kittens 4 x so they combined will be 5x+4x=27
9x=27
X=3
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Cone, pyramid and triangular cube

Step-by-step explanation:

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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

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Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

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3 years ago
In ΔXYZ, ∠Y=90° and ∠X=73°. ∠ZWY=80° and XW=80. Find the length of ZY to the nearest 100th.
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Answer:

Solution given:

Y=90°

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wy=\frac{yz}{5.67}

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E<u>q</u><u>u</u><u>a</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>equation</u><u> </u><u>1</u><u>&</u><u>2</u>

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