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2 years ago

Find the volume of the composite solid below round your answer to the nearest tenth

Mathematics

1 answer:

gladu [14]2 years ago

4 0

First find the volume of the cylinder.

The formula is: V=πr²h

Plug the numbers in the figure into the formula.

V=πr²h

V=π×3²×10

V=π×9×10

V=π90

V=282.743

Therefore, the volume of the cylinder is approximately 282.7in³

Find the volume of the rectangluar prism.

V=lwh

V=12×7×4

V=84×4

V=336

Therefore, the volume of the rectangular prism is 336in³

Add the volumes:

282.7+336=618.7

The volume of the composite solid is approximately 618.7in^3

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Find the midpoint of NP⎯⎯⎯⎯⎯ given N(2a, 2b) and P(2a, 0).

geniusboy [140]

Answer:

(2a, b )

Step-by-step explanation:

Given the endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

[ $\frac{1}{2}$ (x₁ + x₂ ), $\frac{1}{2}$ (y₁ + y₂ ) ]

Here (x₁, y₁ ) = N(2a, 2b) and (x₂, y₂ ) = P(2a, 0), thus

midpoint = [ $\frac{1}{2}$ (2a + 2a), $\frac{1}{2}$ (2b + 0 ) ] = (2a, b )

4 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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3 years ago

Evaluate g(n- 7) if g(x) = x^2-6/7x

noname [10]

Step-by-step explanation:

g(n-7) = (n-7) ^2-6/7 (n-7) = n^2 -14n+49-6/7 n +6= n^2-104/7 n +55

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3 years ago

Geometry it's about relations within triangles. need help solving for variables

elena-14-01-66 [18.8K]

Answer:

`18. x=9