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3 years ago

11

Four water tanks can hold a total of 432.4 gallons of water. On average, how much water can one tank hold? How much water is in

a tank that is half-full?

2 answers:

Sauron [17]3 years ago

4 0

Answer:

108.1 gallons of water, 54.05 gallons of water.

Step-by-step explanation:

Four water tanks can hold water = 432.4 gallons

one tank can hold water = 432.4 ÷ 4 = 108.1 gallon

One tank can hold 108.1 gallons of water.

If a tank is half-full then how much water is in a tank

One tank, fully filled can hold water = 108.1 gallon

A tank that is half-full of water = 108.1 × $\frac{1}{2}$ = 54.05 gallons

There is 54.05 gallons water in a tank if it is half-full.

andreyandreev [35.5K]3 years ago

3 0

On average one tank holds 108 gallons because 432.4 divided by 4 is 108.1. One half full tank would hold about 54 gallons (54.05 to be exact) because half of 108.1 gallons (a full tank) is 54.05 gallons.

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The scheduled arrival time for a daily flight from Boston to New York is 9:35 am. Historical data show that the arrival time fol

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Answer:

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean is:

$M = \frac{a+b}{2}$

The standard deviation is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Arrival time of 9:18 am and a late arrival time of 9:41 am.

9:41 is 23 minutes from 9:18. So the time is uniformily distributed between 0 and 23 minutes, so a = 0, b = 23.

Mean:

$M = \frac{a+b}{2} = \frac{0+23}{2} = 11.5$

Standard deviation:

$S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(23 - 0)^{2}}{12}} = 6.64$

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes

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2 years ago

A study was conducted to investigate the relationship between the resale price, y (in hundreds of dollars), and the age, x (in y

GaryK [48]

Answer:

a. $121.07

b. $60.9

C. $20.03

Step-by-step explanation:

From the equation given

Y=181.7-20.21x

Where y is in dollars and X is in years

a. To find the resale price after 3years we have, we substitute x=3 into the given equation.

We have

y=181.7-20.21(3)

y=181.7-60.63

y=121.07

The resale price after 3years is $121.07

b. To find the resale price after 6years we have, we substitute x=6 into the given equation.

We have

y=181.7-20.21(6)

y=181.7-120.72

y=60.98

The resale price after 3years is $60.98

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[(x=3)-(x=6)]/3

=(121.07-60.98)/3

$20.03

Hence the average annual decrease is $20.03

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3 years ago

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A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

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