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3 years ago

6 fluid ounces per day. Express this rate of flow in cubic meters per year.

Mathematics

1 answer:

stealth61 [152]3 years ago

3 0

Answer:

0.06 cubic meters per year

Step-by-step explanation:

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Using the digits 0 to 9

Gre4nikov [31]

Supplementary Angles:

$\boxed{100°} and \boxed{80°}$

Arithmetically speaking, the closest 2 supplementary angles can get (in 3 and 2 digits respectively) is the upwritten.

Complementary Angles:

$\boxed{45°} and \boxed{45°}$

Simply, in this case, for angles to be numerically as close as possible - make both the angles 45°.

8 0

2 years ago

X+2 = 7/8 mulltiplied by -15x

Marysya12 [62]

Answer:

x=16/113

Step-by-step explanation:

look at them in order

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2 years ago

Traveling at 65 mph how many feet can you travel in 22 minutes?

baherus [9]

If you are travelling at 65 mph, you have to think of it as if you are moving at a constant speed. By dividing 65 by 60, you are calculating how many miles you are moving each minute. In doing so, you can calculate that you are moving at 1.083333 miles each minute. That's approximately 5720 feet per minute. (This is calculated by multiplying 1.083333 by 5280 (the number of feet in a mile)). 5720 x 22 minutes, you will have traveled 125,840 feet.

Hope this helps!

6 0

3 years ago

Read 2 more answers

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

3 years ago

A dance studio charges $12 per

klemol [59]

Answer:

you will need to subtract 15% from $12 and get 10.20 now times that by 5 and you will get $51.

5 0

3 years ago