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2 years ago

a box is 10 inches long, 3 inches wide, and 4 inches high. what is the surface area in square inches of the closed box

Mathematics

1 answer:

Ilya [14]2 years ago

3 0

Answer:

Step-by-step explanation:

10+3=13 adding 4 more =17 there you go hope that help

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Solve 2x+3y=2<br> 5x-y=-29<br> By elimination

Andreyy89

Answer:

$\huge\boxed{x=-5,\ y=4\to(-5,\ 4)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}2x+3y=2\\5x-y=-29&\text{multiply both sides by 3}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}2x+3y=2\\15x-3y=-87\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad17x=-85\qquad\text{divide both sides by 17}\\.\qquad\dfrac{17x}{17}=\dfrac{-85}{17}\\.\qquad\boxed{x=-5}\\\\\text{Substitute it to the second equation:}\\\\5(-5)-y=-29\\\\-25-y=-29\qquad\text{add 25 to both sides}\\\\-25+25-y=-29+25\\\\-y=-4\qquad\text{change the signs}\\\\\boxed{y=4}$

5 0

4 years ago

Please help fast....

Doss [256]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find thw product of (x+3)(x-3)(x+7)

Lesechka [4]

Answer:

x^3+7x^2−9x−63

Step-by-step explanation:

can i please have brainliest

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

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

Helppppppppppppp meee

Darina [25.2K]

It will be 14 times 6.5

5 0

3 years ago