All categories

3 years ago

3. If the lateral surface area of a

Mathematics

2 answers:

azamat3 years ago

6 0

Step-by-step explanation:

natulia [17]3 years ago

3 0

Answer:

I) R = 3 cm;

II) V = 14.3 cm³.

Step-by-step explanation:

I) C = 94.2 ÷ 5 = 18.84

A cylinder has a base with a circumference of 18.84cm.

$R=\dfrac{C}{2\pi } =\dfrac{18.84}{2 \cdot 3.14} =\dfrac{18.84}{6.28} =3 \: (cm)$

II) $V=\pi R^{2}H=3.14 \cdot 3^{2} \cdot 5 = 141.3 \: (cm^{3})$

You might be interested in

This purse is roughly a trapezoid. The bottom base is 16 inches long. The top base is 10. It stands 13 inches tall. The latch is

balandron [24]

To find the amount of fabric for one side of the purse you will find the area of the trapezoidal space that is created. The latch is on the top, so the 2 in is not necessary information.

A = 1/2 h(b1 + b2)
1/2 x 13 x (10 + 16)
A = 169 square inches of fabric will be needed.

3 0

3 years ago

Recall that Rn denotes the right-endpoint approximation using n rectangles, Ln denotes the left-endpoint approximation using n r

pogonyaev

Answer:

$R_5=1.12$

Step-by-step explanation:

We want to calculate the right-endpoint approximation (the right Riemann sum) for the function:

$f(x)=x^2+x$

On the interval [-1, 1] using five equal rectangles.

Find the width of each rectangle:

$\displaystyle \Delta x=\frac{1-(-1)}{5}=\frac{2}{5}$

List the x-coordinates starting with -1 and ending with 1 with increments of 2/5:

-1, -3/5, -1/5, 1/5, 3/5, 1.

Since we are find the right-hand approximation, we use the five coordinates on the right.

Evaluate the function for each value. This is shown in the table below.

Each area of each rectangle is its area (the y-value) times its width, which is a constant 2/5. Hence, the approximation for the area under the curve of the function f(x) over the interval [-1, 1] using five equal rectangles is:

$\displaystyle R_5=\frac{2}{5}\left(-0.24+-0.16+0.24+0.96+2)= 1.12$

3 0

3 years ago

The weights (in pounds) of 30 preschool children are listed below. which weights are below the 25th percentile? 25 25 26 26.5 27

Leto [7]

from the first 35 through 25

6 0

3 years ago

If you could put wings on any species of animal, what animal would you choose?

denis-greek [22]

Answer:

Humans

Explanation:

So I can have wings :)

5 0

2 years ago

Read 2 more answers

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago