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2 years ago

There is a three-Legged

1 answer:

7 0

The 4 legged stool will rock. the 4 ends of the legs form a plane. they cannot sit on an uneven surface. the area they sit on must be planar. the 3 legged stool, however, can sit without rocking, because any 3 points define a plane entirely

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Solve the equation below for x. <br> 500 = 25e ^ (2x)

Zigmanuir [339]

Answer:

x = i π n + log(20)/2 for n element Z

Step-by-step explanation:

Solve for x:

500 = 25 e^(2 x)

500 = 25 e^(2 x) is equivalent to 25 e^(2 x) = 500:

25 e^(2 x) = 500

Divide both sides by 25:

e^(2 x) = 20

Take the natural logarithm of both sides:

2 x = 2 i π n + log(20) for n element Z

Divide both sides by 2:

Answer: x = i π n + log(20)/2 for n element Z

7 0

3 years ago

Lines sand tare perpendicular. If the slope of line sis 5, what is the slope of<br> line t?

Sveta_85 [38]

Answer:

-1/5

Step-by-step explanation:

Perpendicular lines have slopes that are the opposite reciprocal of one another.

The slope of line t is the opposite reciprocal of the slope of line s:

slope of t = -1/(slope of s)

slope of t = -1/5

6 0

3 years ago

I’m on a test so I need help

sladkih [1.3K]

The answer is 2 y2 -9

4 0

3 years ago

Help soon as possible :)

Marianna [84]

Answer:

(3x+5y)(3x-5y)

Step-by-step explanation:

6 0

2 years ago

The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample me

julia-pushkina [17]

Answer:

The sample does not contradicts the manufacturer's claim.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 200 mg

Sample mean, $\bar{x}$ = 211.5 mg

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 18.5 mg

a) First, we design the null and the alternate hypothesis for a two tailed test

$H_{0}: \mu = 200\\H_A: \mu \neq 200$

We use Two-tailed t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799$

c) Now,

$t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930$

Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.

Thus, the sample does not contradicts the manufacturer's claim.

d) P-value = 0.011934

Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.

Yes, both approach the critical value and the p-value approach gave the same results.

4 0

3 years ago