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3 years ago

Help me find the surface area plz and tell me how you got this

Mathematics

2 answers:

podryga [215]3 years ago

8 0

All you have to do is find the area of all the shapes and add them so length times height besides the triangle which is 1/2(l*h)

Angelina_Jolie [31]3 years ago

8 0

24 x 10 = 240

24 x 10 = 240

24 x 12 = 288

(8 x 12) / 2 = 48

(8 x 12) / 2 = 48

add them

240 + 240 +288 + 48 + 48 = 864

answer 864 cm^2

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HELPP!!

s344n2d4d5 [400]

Answer:

2 km

Step-by-step explanation:

basically do the inverse:

3×4(instead of divide)=12

12-8=4

it says jose ran 2× as many km as karen so you divide 4 by 2 giving you 2

sorry if its incorrect

6 0

2 years ago

Solve for p.<br><br> 13 − 2p = 5<br><br> p =

Naya [18.7K]

Answer:

P = 4

Step-by-step explanation:

2 x 4 is 8

so 13 - 8 = 5

7 0

3 years ago

In reciprocal is the denominator one unit

docker41 [41]

Not always if you're saying it will always be something like 1/2 becomes 2/1 with a unit of one. It's just flipped, so 2/6 will be 6/2, 4/5 becomes 5/4, 22/16 becomes 16/22 etc. That is all there is to it. Make sure you simplify your answers.

5 0

3 years ago

a weight lifter can lift 550 pounds. his goal is to improve by 20% during the next year. how many pounds does he want to be able

sattari [20]

The answer is 660 lbs.

Hope this helps!

7 0

3 years ago

A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21

sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{340119}{10} = 34011.9$

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

$S.D = \sqrt{\frac{2711418821}{9}} = 17357.09$

Confidence interval:

$\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621$

$34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)$

7 0

3 years ago