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3 years ago

What is the equivalent of 3/4=?/8?

Mathematics

2 answers:

Phoenix [80]3 years ago

8 0

The equivalent to 3/4 is 6/8.

ozzi3 years ago

3 0

3/4= 6/8 that's your answer.

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Anybody wanna help?Asap?

artcher [175]

Answer:

Step-by-step explanation:

its simple

Hope this help you

Red

4 0

3 years ago

Write an expression for angle rst.

levacccp [35]

Answer:

∠RST = 6x - 18

Step-by-step explanation:

Since SQ bisects ∠TSR then ∠RSQ = ∠QST = 3x - 9

and ∠RST = ∠RSQ + ∠QST = 3x - 9 + 3x - 9 = 6x - 18

4 0

4 years ago

In a study to compare two different corrosion inhibitors, specimens of stainless steel were immersed for four hours in a solutio

djverab [1.8K]

Answer:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_A = 242$ sample mean for inhibitor A

$s_A = 20$ sample standard deviation for inhibitor A

$n_A = 47$ sample size for A

$\bar X_B = 220$ sample mean for inhibitor B

$s_B = 31$ sample standard deviation for inhibitor B

$n_B = 42$ sample size for A

Solution to the problem

For this case the confidence interval for the difference of means is given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 47+42-2= 87$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that $t_{\alpha/2}=1.988$

And replacing we got:

$(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862$

$(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138$

And the 95% confidence interval for the difference in the means is given by: $10.862 \leq \mu_A -\mu_B \leq 33.138$

4 0

3 years ago

Help appreciated please and thank you

meriva

Answer:

D. $\frac{4 \sqrt{33} }{33}$

Step-by-step explanation:

$\tan \alpha = \frac{4}{ \sqrt{33} } \\ = \frac{4}{ \sqrt{33} } \times \frac{ \sqrt{33} }{ \sqrt{33} } \\ \\ = \frac{4 \sqrt{33} }{ {( \sqrt{33)} }^{2} } \\ \\ \tan \alpha = \frac{4 \sqrt{33} }{33}$

8 0

3 years ago

Read 2 more answers

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

valentinak56 [21]

Answer:

tan(θ) = 0, 0.577, -0.577

Step-by-step explanation:

3tan³(θ) - tan(θ) = 0

tan(θ)(3tan²(θ) - 1) = 0

tan(θ) = 0

tan²(θ) = ⅓ tan(θ) = +/- sqrt(⅓)

tan(θ) = 0, sqrt(⅓), -sqrt(⅓)

tan(θ) = 0, 0.577, -0.577

To find θ values, domain is required

6 0

4 years ago