Can someone help me solve for X?

The volume of a cone is 18π cubic feet. If the height of the cone is 6 ft, what is the radius of the cone? (V = 1 3 πr2h)

svetoff [14.1K]

Answer:

Volume = 1/3πr^2h

18π = 1/3πr^2 x 6

18 = 2r^2

9 = r^2

r = 3 ft.

Hope this helps you!

8 0

4 years ago

The length of a rectangle is 5 times the width. If the area is 500 square feet, then what is the length of the rectangle, in fee

tester [92]

Answer:

The length of the rectangle is 50 feet, while the width of the rectangle is 10 feet.

Step-by-step explanation:

Since the length of a rectangle is 5 times the width, to determine the length of the rectangle if the area is 500 square feet, the following calculation must be performed:

Area = Base x height

500 = X x 5X

500/5 = X x X

100 = X x X

100 = 10 x 10

X = 10

Thus, the length of the rectangle is 50 feet, while the width of the rectangle is 10 feet.

8 0

3 years ago

Aging workers of the Neotropical termite, Neocapritermes taracua, develop blue crystal containing glands ("backpacks") on their

Kaylis [27]

Answer:

The blue liquid is toxic

Step-by-step explanation:

From the calculations I have done and attached to this problem

Null hypothesis:

H0 : p1 <= p2

Alternate hypothesis:

H1 : p1 > p2

^p1 = 0.925

^p2 = 0.2250

^P = (x1 + x2)/(n1+n2)

= 0.575

After calculating, I got test statistics = 6.33 ( check attachment)

P(z > 6.33) = 0.0000

In conclusion, our pvalue has been found to be less than the level of significance 0.05 so we reject null hypothesis and accept that we have enough sample evidence that the blue liquid is more toxic in comparison to white one.

Please check attachment for calculated test statistics.

7 0

4 years ago

The vertices of an isosceles triangle are A (-10, 1), B (-6, 3) and C (-4, 7).

adelina 88 [10]

check the picture below.

so, those are the points, now, the line of symmetry, the red one, is the line that will cut the vertex is half and will be perpendicular to the opposite side.

now, the slope of the black line we can just get it off the grid, notice the rise and run, rise/run = 6/6 = 1.

now a perpendicular line to that one will have a negative reciprocal slope, thus

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{1\implies \cfrac{1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{1}\implies -1}}$

so the slope of that perpendicular line will be -1, so we're really looking for the equation of a line whose slope is -1 and runs through -6,3.

$\bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad \qquad \qquad slope = m\implies -1 \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=-1[x-(-6)] \\\\\\ y-3=-1(x+6)\implies y-3=-x-6\implies y=-x-3$

5 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