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xz_007 [3.2K]
2 years ago
7

If the volume of two similar spheres are 128m^3 and 1458m^3.

Mathematics
1 answer:
Musya8 [376]2 years ago
7 0

Answer:

ZOOM PLZ COME WE R BORED

MEETING ID 798 4170 2552

PASSWORD:XCNeV3

Step-by-step explanation:

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7x15 can be written as
Mrrafil [7]
105 is the product of 15 and 7.
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3 years ago
ANSWER ASAP PLEASE<br> look at the photo
Liula [17]

Step-by-step explanation:

Area of trapezium = 1/2(a+b)h

= 1/2(10+24)12

= 1/2×34×12

= 204 square units

3 0
2 years ago
1.
nirvana33 [79]

Step-by-step explanation:

I think you mean triangle XYZ. and angle XYZ.

so, the angle at Y is 90°.

that makes XZ the Hypotenuse (baseline opposite of the 90° angle) of the right-angled triangle.

so, we can actually use Pythagoras (we don't need the area of the triangle to solve) :

13² = 12² + YZ²

169 = 144 + YZ²

25 = YZ²

YZ = 5 cm

just to check - the area of this triangle is (remember, it is a right-angled triangle, so XY and YZ serve also as baseline and height) :

area = 12 × 5 / 2 = 60/2 = 30 cm²

correct, it fits.

3 0
1 year ago
Who knows how to answer this. It's Dividing complex numbers.​
Elodia [21]

Answer:

answe is 1+i

<em><u>mark this as brainliest!!</u></em>

<em><u>mark this as brainliest!!follow me</u></em>!

4 0
3 years ago
Read 2 more answers
Question 7 of 10
andriy [413]

Answer:

C. n = 90; p = 0.8

Step-by-step explanation:

According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10

Analyzing  the given options, we have;

A. n = 45, p = 0.8

∴ n·p = 45 × 0.8 = 36 > 10

n·(1 - p) = 45 × (1 - 0.8) = 9 < 10

Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

B. n = 90, p = 0.9

∴ n·p = 90 × 0.9 = 81 > 10

n·(1 - p) = 90 × (1 - 0.9) = 9 < 10

Given that for n = 90, p = 0.9, n·(1 - p) = 9  < 10, a normal distribution can not be used to approximate the sampling distribution

C. n = 90, p = 0.8

∴ n·p = 90 × 0.8 = 72 > 10

n·(1 - p) = 90 × (1 - 0.8) = 18 > 10

Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution

D. n = 45, p = 0.9

∴ n·p = 45 × 0.9 = 40.5 > 10

n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10

Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution

A sampling distribution Normal Curve

45 × (1 - 0.8) = 9

90 × (1 - 0.9) = 9

90 × (1 - 0.8) = 18

45 × (1 - 0.9) = 4.5

Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words

Therefore;

A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8

3 0
2 years ago
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