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2 years ago

Francis is 9 years older than Vincent. Francis is 17 years old. Let v represent how old Vincent is.

Mathematics

2 answers:

Marysya12 [62]2 years ago

7 0

Its b because since vincent is 9 years younger so just subtract 9 from the age and voila

baherus [9]2 years ago

5 0

Wait should we be adding cause if u add idk what it will get that is high school stuff

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for the given kite find the length of side LM write your answer in simplest radical form. show all work.

Gnesinka [82]

Answer:

13sqrt(2) units

Step-by-step explanation:

LM² = 7² + 17²

LM² = 338

LM = sqrt(338)

LM = 13sqrt(2)

sqrt: square root

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The answer to this problem?

LenaWriter [7]

The simplified expression of $\sqrt{\frac{162x^9}{2x^{27}}}$ is $\frac{9}{x^9}$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\sqrt{\frac{162x^9}{2x^{27}}}$

Divide 162 and 2 by 2

$\sqrt{\frac{81x^9}{x^{27}}}$

Take the square root of 81

$9\sqrt{\frac{x^9}{x^{27}}}$

Apply the quotient rule of indices

$9\sqrt{\frac{1}{x^{27-9}}}$

Evaluate the difference

$9\sqrt{\frac{1}{x^{18}}}$

Take the square root of x^18

$\frac{9}{x^9}$

Hence, the simplified expression of $\sqrt{\frac{162x^9}{2x^{27}}}$ is $\frac{9}{x^9}$

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1 year ago

What is the Value of X please help!

Brums [2.3K]

Answer:

x = 7cm

Step-by-step explanation:

$\frac{36}{6} = \frac{42}{x}\\\\6 = \frac{42}{x}\\\\x = \frac{42}{6} = 7$

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2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago