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1 year ago

A printer takes 7 minutes to print a document with 63 pages. How many pages

Mathematics

1 answer:

Artemon [7]1 year ago

5 0

Create a proportion.

Since it takes 7 minutes to print 63 pages, and half an hour is 30 minutes, you create two ratios(unit per unit) and line up the units(minutes to minutes, pages to pages).

$\frac{7 min}{63 pages} = \frac{30 min}{Xpages}$

Now you cross multiply, 63 * 30 = 1890 and 7 * x = 7x.

So we have 1890 = 7x.

Then we divide both sides by seven.

$\frac{1890}{7} = \frac{7x}{7}$

270 = x, therefore you can print 270 pages in half an hour :)

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Which of the following sets of ordered pairs represents a function

Feliz [49]

Answer: A

Step-by-step explanation:

Each value of x maps onto only one value of y.

8 0

2 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago

Several paint mixtures are made by mixing blue paint and white paint. Paint Colors Blue Paint White Paint Mixture A 5 cups 12 cu

Nesterboy [21]

Answer:

The mixture C is the correct option

Step-by-step explanation:

According to the given scenario, the calculation is as follows:

For Mixture A

Blue Paint - 5 cups

White Paint - 12 cups

The ratio between them is 5:12

For Mixture B

Blue Paint - 6 cups

White Paint - 6 cups

The ratio between them is 6:6 = 12:12

It came by multiply the numerator and denominator by 12

For Mixture C

Blue Paint - 4 cups

White Paint - 12 cups

The ratio between them is 4:12

For Mixture D

Blue Paint - 5 cups

White Paint - 6 cups

The ratio between them is 5:6 = 10:12

It came by multiply the numerator and denominator by 12

As it can be seen that in all four mixtures the denominator is the same so for calculating the lowest ratio we have to see the small value in the numerator

As it can be seen that there is a small value of 4

hence, the mixture C is the correct option

5 0

3 years ago

Read 2 more answers

Problem 5.A miner is trapped inside a mine and has access to 3 doors. The first door leads to a tunneland can take him to safety

Likurg_2 [28]

Answer:

Expected time is 15 hours for him to get to safety.

Step-by-step explanation:

We define X as the time that this miner would get to safety.

We define Y as the door he chooses initially.

P(Y= 1) = P(Y=2)=P(Y=3) = 1/3

We have E[X|Y=1] = 3

E[X|Y] = 5 hours + E[X}

E[X|Y] = 7 hours + E[X]

Then we have

E[X] = 1/3(3 + 5 + E[X] + 7 + E[X])

We cross multiply

3*E[X] = (15 + 2E[x])

3E[X] - 2E[X] = 15

E[X] = 15

So the time it would take to get him to safety is 15 hours

5 0

3 years ago

Use two 2-Column proofs to show that either diagonal of a rectangle will create two congruent triangles?

tangare [24]

Given: Rectangle ABCD
Prove: ∆ABD≅∆CBD
Solution:

 Statement Reason

ABCD is a parallelogram Rectangles are parallelograms since the definition of a parallelogram is a quadrilateral with two pairs of parallel sides.

Segment AD = Segment BC The opposite sides of a parallelogram are Segment AB = Segment CD congruent. This is a theorem about the parallelograms.

∆ABD≅∆CBD SSS postulate: three sides of ΔABD is equal to the three sides of ∆CBD

Given: Rectangle ABCD
Prove: ∆ABC≅∆ADC
Solution:

 Statement Reason

Angle A and Angle C Definition of a rectangle: A quadrilateral
are right angles with four right angles.

Angle A = Angle C Since both are right angles, they are congruent

Segment AB = Segment DC The opposite sides of a parallelogram are Segment AD = Segment BC congruent. This is a theorem about the parallelograms.

∆ABC≅∆ADC SAS postulate: two sides and included angle of ΔABC is congruent to the two sides and included angle of ∆CBD

6 0

3 years ago