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2 years ago

A dog is fed 3 1/2 cups of food. A puppy is fed 3/5 of that. How much is the puppy fed?

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

4 0

Answer:

9 puppies can be fed.

Step-by-step explanation:

Given that:

A dog is fed 312 cups of food.
A puppy is fed 35 of that.

To Find:

How much is the puppy fed?

Let us assume:

x puppies can be fed.

Finding the number of puppy:

Puppy : Food = Puppy : Food

⟶ 35 : 1 = 312 : x

⟶ 35/1 = 312/x

Cross multiplication.

⟶ 35x = 312

⟶ x = 312/35

⟶ x = 9 (approx.)

Hence,

9 puppies can be fed.

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3 years ago

Jason is watering his garden. He uses 2.75 gallons of water each minute. How much

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8.25 Gallons

Step-by-step explanation:

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2 years ago

How long is the segment from(-5,2)to(-5,-8)

dusya [7]

Answer:

Step-by-step explanation:

To calculate the distance d use the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 5, 2) and (x₂, y₂ ) = (- 5, - 8)

d = $\sqrt{(-5+5)^2+(-8-2)^2}$

= $\sqrt{0^2+(-10)^2}$

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6 0

2 years ago

Oliver traveled 124.25 miles in 3.5 hours at a constant speed. What is the constant of proportionality in miles per hour?

Elan Coil [88]

Answer:

Simple!

All you have to do is divide 3.5 by 124.25

124.25 / 3.5 = 35.5

Your answer is C.

Hope this helps!

3 0

3 years ago

Read 2 more answers

(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Roun

love history [14]

Answer:

a) $f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

b) $P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

c) $P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857$

d) $P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

$P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5$

Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

$X \sim Unif (a=4, b =11)$

Part a

The probability density function is given by the following expression:

$f(x) = \frac{1}{b-a} , a \leq x \leq b$

$f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11$

Part b

We want this probability:

$P(X \leq 7)$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}$

And replacing we got:

$P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286$

Part c

We want this probability:

$P(5 < X \leq 7)$

And we can use the CDF again and we have:

$P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857$

Part d

We want this conditional probabilty:

$P(X >5 | X \leq 7)$

And we can find this probability with this formula from the Bayes theorem:

$P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5$

7 0

2 years ago