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3 years ago

A pond at a hotel holds 4,290 gallons of water.

Mathematics

1 answer:

Kisachek [45]3 years ago

5 0

Answer:

55 hours

Step-by-step explanation:

Write the problem as an equation.

4290-78x=0

Where 'x' is each hour the pond is being drained.

Then, solve.

4290=78x Add 78x to each side of the equation

55=x Divide both sides by 78

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Please help with these two math problems<br><br> 1. 7^x ÷ 7^y =<br> 2. z^10x^y ÷ z^5 =

IceJOKER [234]

Answer:

1. 7^(x-y)

2. z^5x^y

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)/(a^c) = a^(b-c)

$\dfrac{7^x}{7^y}=\boxed{7^{x-y}}$

$\dfrac{z^{10}x^y}{z^5}=x^yz^{10-5}=\boxed{z^5x^y}$

8 0

3 years ago

5 / 3 x + 1 / 3x equals 13 + 1 / 3 x + 8 / 3 x

NNADVOKAT [17]

Answer: x = - 12/9 ( the equation is negative)

Step-by-step explanation:

\frac{5}{3}x+\frac{1}{3}=13+\frac{1}{3}x+\frac{8}{3}x

\frac{5}{3}x=3x+\frac{38}{3}

\frac{5}{3}x-3x=3x+\frac{38}{3}-3x

-\frac{4}{3}x=\frac{38}{3}

3\left(-\frac{4}{3}x\right)=\frac{38\cdot \:3}{3}

-4x=38

\frac{-4x}{-4}=\frac{38}{-4}

x=-\frac{19}{2}

8 0

3 years ago

Read 2 more answers

Jake has decided to invest in three business ventures. The total cost of investing in all three ventures is $15,000. The combine

Arte-miy333 [17]

X = first venture, y = second venture, z = third venture

x + y + z = 15,000
x + z = y + 7000
3x + 2y + 2z = 39,000
these are ur equations.....

x + y + z = 15,000
x - y + z = 7000
--------------------add
2x + 2z = 22,000

x + y + z = 15,000....multiply by -2
3x + 2y + 2z = 39,000
-------------------
-2x - 2y - 2z = - 30,000 (result of multiplying by -2)
3x + 2y + 2z = 39,000
------------------add
x = 9,000

2x + 2z = 22,000
2(9000) + 2z = 22000
18,000 + 2z = 22000
2z = 22000 - 18000
2z = 4000
z = 4000/2
z = 2,000

x + y + z = 15,000
9000 + y + 2000 = 15,000
11,000 + y = 15,000
y = 15,000 - 11,000
y = 4,000

first venture (x) = 9,000 <==
second venture (y) = 4,000 <==
third venture (z) = 2,000 <==

7 0

2 years ago

Read 2 more answers

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

A rectangle is 3V5 meters wide and 5V5 meters long.

ad-work [718]

Answer:

i think the answer is b or e but most likely b

Step-by-step explanation:

5 0

2 years ago