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3 years ago

14

The butler family and Clark family each used their sprinklers last summer. The butler family's sprinklers was used for 40 hours.

The Clark family's sprinkler was used for 30 hours. There was a combined total output of 2200L of water. What was the water output rate for each sprinkler if the sum of the two rates was 60l per hour?

1 answer:

sergiy2304 [10]3 years ago

5 0

Answer: the sprinkling rate of the butler family's sprinkler is 40L per hour

the sprinkling rate of the Clark family's sprinkler is 20L per hour

Step-by-step explanation:

Let x represent the sprinkling rate of the butler family's sprinkler.

Let y represent the sprinkling rate of the Clark family's sprinkler.

The butler family's sprinklers was used for 40 hours. The Clark family's sprinkler was used for 30 hours. There was a combined total output of 2200L of water. This means that

40x + 30y = 2200 - - - - - - - - - - -1

if the sum of the two rates was 60l per hour, it means that

x + y = 60

Substituting x = 60 - y into equation 1, it becomes

40(60 - y) + 30y = 2200

2400 - 40y + 30y = 2200

- 40y + 30y = 2200 - 2400

- 10y = - 200

y = - 200/ - 10

y = 20

Substituting y = 20 into x = 60 - y, it becomes

x = 60 - 20 = 40

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(High points) please solve with explanation

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Answer:

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<u>Area = 160 cm^2</u>
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Step-by-step explanation:

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As you can see in the picture, the base is 16 cm and the height is 12 cm, then we replace in the formula:

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Area of a triangle = (base * height) / 2

The height of the triangle is 8 cm, and the base is 8 cm too, because you subtract to the base of the rectangle (16 cm) the measurements in the upper part (16 - 4 - 4 = 8), Now, we replace in the formula:

Area of a triangle = (8 cm * 8 cm) / 2
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We subtract to the found area:

Area of the picture = 192 cm^2 - 32 cm^2
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To find the perimeter, you must add all the sides of the picture, but, as you can see, there is a side that doesn't have the measurent, this is the hypotenuse of the triangle used before, but how we know the other sides, we can use Pythagorean theorem:

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Where:

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So, we replace in the theorem:

$a^{2}+b^{2}=c^{2}$

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The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

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Answer:

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

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Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

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.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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