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Anastaziya [24]
3 years ago
5

Find the domain of the function. (Enter your answers using interval notation.) f(s, t) = square root of s^2+t^2.

Mathematics
1 answer:
stealth61 [152]3 years ago
7 0

Answer:

S=\{\forall \,s,t\in \mathbb{R}|s\in(-\infty,+\infty), t\in(-\infty,+\infty)\}

Step-by-step explanation:

Let be f(s,t) = \sqrt{s^{2}+t^{2}}, the range of a square root compehends the subset of positive real numbers plus zero. Likewise, the range of the square of any number has the same subset with all real set as domain. As square root encloses a sum of squares, the domain of the function is:

S=\{\forall \,s,t\in \mathbb{R}|s\in(-\infty,+\infty), t\in(-\infty,+\infty)\}

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In January 1995, each student in a random sample of 150 physics majors at a large university was asked in how many lab classes h
Mnenie [13.5K]

Answer:

The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.

Step-by-step explanation:

A <em>t</em>-test for difference between means can be used to determine whether the mean number of lab classes taken by physics majors in January 1995 is different from the mean number of lab classes taken in 2015.

The hypothesis is:

<em>H₀</em>: There is no difference between the two population means, i.e. <em>μ</em>₁ = <em>μ</em>₂.

<em>Hₐ</em>: There is a significant difference between the two population means, i.e. <em>μ</em>₁ ≠ <em>μ</em>₂.

The survey was conducted in a similar pattern. Assume that the two population standard deviations are same.

The information provided is:

\bar x_{1}=1.74\\s_{1}=1.49\\\bar x_{2}=1.87\\s_{1}=1.57\\n_{1}=n_{2}=150\\\alpha =0.10

The test statistic is:

t=\frac{\bar x_{1}-\bar x_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}

Here <em>s</em>_{p} is the pooled standard deviation.

Compute the value of pooled standard deviation as follows:

s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(150-1)1.49^{2}+(150-1)1.57^{2}}{150+150-2}}=1.531

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{1.74-1.87}{1.531\times\sqrt{\frac{1}{150}+\frac{1}{150}}}=-0.735

The test statistic value is -0.735.

The decision rule is:

If the <em>p </em>- value of the test is less than the significance level, <em>α</em> = 0.10 then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value as follows:

p-value=2P(t_{298}

The <em>p</em>-value = 0.463 > <em>α</em> = 0.10.

As the <em>p</em>-value is more than the significance level the null hypothesis will not be rejected at 10% level of significance.

Conclusion:

The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.

8 0
3 years ago
In de pas monch. Goran rencad 6 video games and 3 DVDS. The rental price for each video game was $290. The rental price for each
Nimfa-mama [501]

Answer:

$27.30

Step-by-step explanation:

just do 290x6+330x3=2730

but just do this $27.30

so x is $27.30

7 0
3 years ago
Pls help......................
denis23 [38]

Answer:

Reduce the expression, if possible, by cancelling the common factors.

answer is 8/9

Step-by-step explanation:

6 0
3 years ago
Please i need help!
inysia [295]
I think it’s the 3rd one
3 0
2 years ago
Find the distance, c, between (–3, –4) and (1, 2) on the coordinate plane. Round to the nearest tenth.
Anon25 [30]
Answer
7.21
Step-by-step explanation:
Data :
(-3,-4) , (1,2)
Where
x₁ = -3 , x₂ = 1
y₁ = -4 , y₂ = 2
Formula for distance
D = √{(x₂-x₁)² + (y₂-y₁)²}
= √{(l--3)² + (2--4)²} = √(16+36) =√52 =7.21
5 0
2 years ago
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