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4 years ago

14

A warehouse has boxes that are 18 inches tall and other boxes that are 30 inches tall. If 18 inch boxes is stacked next to 30 in

ch boxes, what is the lowest height at which the two boxes will be the same height?

1 answer:

Aleksandr-060686 [28]4 years ago

4 0

90 inches 18 times 5 is 90 and 30 times 3 is 90 so the lowest point which they will be at the height is 90 inches

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4. Let f(x) = 2x +1 and g(x) = x². Find (gºf)(a).

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Answer:

Step-by-step explanation:

hello :

f(x) = 2x +1 and g(x) = x².

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4 years ago

Please help me sir/ma'am i really am confused and dysforic

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2 years ago

(2y+3)(3y^2+4y+5) use the chart to multiply the binomial by the trinomial

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(2y+3)(3y^2+4y+5)

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3 years ago

Sue initially has 5 hours of pop music and 4 hours of classical music in her collection. Every month onwards, the hours of pop m

Snezhnost [94]

<h2>Answer</h2>

f(x) = 5(1.25)x + 4

<h2>Explanation</h2>

To solve this, we are going to use the standard exponential equation:

$f(x)=a(1+b)^x$

where

$a$ is the initial amount

$b$ is the growth rate in decimal form

$x$ is the time (in months for our case)

Since the hours of classic music remain constant, we just need to add them at the end. We know form our problem that Sue initially has 5 hours of pop, so $a=5$ ; we also know that every month onward, the hours of pop music in her collection is 25% more than what she had the previous month, so $b=\frac{25}{100} =0.25$ . Now let's replace the values in our function:

$f(x)=a(1+b)^x$

$f(x)=5(1+0.25)^x$

$f(x)=5(1.25)^x$

Now we can add the hours of classical music to complete our function:

$f(x)=5(1.25)^x+5$

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3 years ago

Read 2 more answers

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

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

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0

2 years ago