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3 years ago

Measure 4 gallons of water using a 3 gallon jar and a 5 gallon jar

Mathematics

2 answers:

vlada-n [284]3 years ago

8 0

Answer

Fill the 5 gallon jug.

Pour the 5 gallon jug into the 3 gallon jug.

Empty the 3 gallon jug and pour the remaining 2 gallons in the 3 gallon jug.

Fill the 5 gallon jug and pour 1 gallon into the 3 gallon jug (remember it will only take one gallon)

Viola the remaining 5 gallon jug will have exactly 4 gallons remaining.

V125BC [204]3 years ago

3 0

Answer:

use 4/5 of the 5 gallon jar or 1 and 1/3 of the 3 gallon jar

Step-by-step explanation:

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Given sin(u)= -7/25 and cos(v) = -4/5, what is the exact value of cos(u-v) if both angles are in quadrant 3

solmaris [256]

Given:

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

To find:

The exact value of cos(u-v) if both angles are in quadrant 3.

Solution:

In 3rd quadrant, cos and sin both trigonometric ratios are negative.

We have,

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

Now,

$\cos (u)=-\sqrt{1-\sin^2 (u)}$

$\cos (u)=-\sqrt{1-(-\dfrac{7}{25})^2}$

$\cos (u)=-\sqrt{1-\dfrac{49}{625}}$

$\cos (u)=-\sqrt{\dfrac{625-49}{625}}$

On further simplification, we get

$\cos (u)=-\sqrt{\dfrac{576}{625}}$

$\cos (u)=-\dfrac{24}{25}$

Similarly,

$\sin (v)=-\sqrt{1-\cos^2 (v)}$

$\sin (v)=-\sqrt{1-(-\dfrac{4}{5})^2}$

$\sin (v)=-\sqrt{1-\dfrac{16}{25}}$

$\sin (v)=-\sqrt{\dfrac{25-16}{25}}$

$\sin (v)=-\sqrt{\dfrac{9}{25}}$

$\sin (v)=-\dfrac{3}{5}$

Now,

$\cos (u-v)=\cos u\cos v+\sin u\sin v$

$\cos (u-v)=\left(-\dfrac{24}{25}\right)\left(-\dfrac{4}{5}\right)+\left(-\dfrac{7}{25}\right)\left(-\dfrac{3}{25}\right)$

$\cos (u-v)=\dfrac{96}{625}+\dfrac{21}{625}$

$\cos (u-v)=\dfrac{1 17}{625}$

Therefore, the value of cos (u-v) is 0.1872.

6 0

3 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