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3 years ago

A drawer contains 12 brown socks, 8 white socks, and 2 blue socks. One sock is selected at random.

Mathematics

2 answers:

kaheart [24]3 years ago

5 0

Answer:

The likelihood is 1/5

Step-by-step explanation:

In order to find this we take the number of blue socks (2) and divide by the sample size. We can find the sample size by adding all of the socks that it could be. Since it must not be brown, we have to add blue to white.

2 + 8 = 10

So we divide blue by that number

2/10 = 1/5

Katyanochek1 [597]3 years ago

3 0

Answer:

1/11

Step-by-step explanation:

12 + 8 + 2

22 / 2

1/11

There is always a pair of socks... A pair is 2. Add everything and divide it.

Hope this helps....

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What is the area of the following shape

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Answer:

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Over a period of 12 hours, the temperature dropped 60 degrees. What was the average change in temperature in one hour?

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Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by y=√x and y=x^2. Find V either by sl

Ede4ka [16]

Answer:

The volume of the solid is $3\pi/10$

Step-by-step explanation:

In this case, the washer method seems to be easier and thus, it is the one I will use.

Since the rotation is around the y-axis we need to change de dependency of our variables to have $f(x)\rightarrow f(y)$ . Thus, our functions with $y$ as independent variable are:

$x=\sqrt{y}\\ x=y^2$

For the washer method, we need to find the area function, which is given by:

$A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]$

By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function $x=\sqrt{y}$ and the inner one by $x=y^2$ . Thus, the area function is:

$A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)$

Now we just need to integrate. The integration limits are easy to find by just solving the equation $\sqrt(y)=y^2$ , which has two solutions $y=0$ and $y=1$ . These are then, our integration limits.

$V=\pi\int_{0}^1 (y-y^4)dy\\ V=\pi (\int_{0}^1 ydy - \int_{0}^1 y^4dy)\\ V=\pi/2-\pi/5\\\boxed{V=3\pi/10}$

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3 years ago

The American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day. Legacy

inessss [21]

Answer:

We conclude that the null hypothesis is rejected which means residents of Legacy Ranch use less water on average.

Step-by-step explanation:

We are given that the American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day.

Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day.

Let $\mu$ = average water usage by residents of Legacy Ranch.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 67 gallons {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}

Alternate Hypothesis, $H_A$ : $\mu$ < 67 gallons {means that the residents of Legacy Ranch use less water on average}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean water use per day = 63 gallons

s = sample standard deviation = 8.1 gallons

n = sample of responded = 23

So, test statistics = $\frac{63-67}{\frac{8.1}{\sqrt{23} } }$ ~ $t_2_2$

= -2.368

The value of the test statistics is -2.368.

Now at 0.025 significance level, the t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test. Since our test statistics is less than the critical values of t as -2.368 < -2.074, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the residents of Legacy Ranch use less water on average.

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