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Kay [80]
3 years ago
15

Help ............. ​

Mathematics
1 answer:
NARA [144]3 years ago
5 0

What kinda of math is this so i can answer..

:)

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Pentagon DEFGH is translated 4 units to the right to create pentagon UVXYZ. DE is 5 cm long. What is the length of UV?
Wewaii [24]
Without seeing the picture, we don't know how DE and UV are related.
3 0
3 years ago
A solid cylinder has a radius of 4.9 cm and a height of 3.5 cm. Work out the total surface area of the cylinder. Give your answe
Novosadov [1.4K]

Answer:

252 cm²

Step-by-step Explanation:

=>Given:

r = 4.9 cm, and h = 3.3 cm of a solid cylinder, π = 3.14

=> Required:

Total Surface Area (T.S.A) = 2πrh + 2πr²

T.S.A = 2*3.14*4.9*3.3 + 2*3.14*4.9²

= 101.5476 + 150.7828

= 252.3304

= 252 cm² (approximated)

8 0
3 years ago
What is the domain of this function?
frutty [35]

Answer:

Step-by-step explanation:

The closed dot at (0, 9) indicates that that's where the graph begins.  The arrow at the other end indicates that it has no end. Since the domain covers x values only (NOT Y VALUES!), we only need be concerned with the x values. It starts at x = 0 and never ends, so the domain is properly stated as

{x | x ≥ 0}

3 0
3 years ago
6j-k=23<br>3k+6j=11<br>Solve j and k ​using the elimination method on the simultaneous equations
frez [133]

Answer:

Step-by-step explanation:

6j - k = 23

6j + 3k = 11.....multiply by -1

---------------

6j - k = 23

-6j - 3k = -11 (result of multiplying by -1)

---------------

-4k = 12

k = -12/4

k = - 3 <=====

6j - k = 23

6j - (-3) = 23

6j + 3 = 23

6j = 23 - 3

6j = 20

j = 20/6

j = 10/3 <===

solution is : j = 10/3 and k = -3

5 0
3 years ago
A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. Find the te
vredina [299]

Answer:

b. 2.333

Step-by-step explanation:

Test if the mean transaction time exceeds 60 seconds.

At the null hypothesis, we test if the mean transaction time is of 60 seconds, that is:

H_0: \mu = 60

At the alternate hypothesis, we test if it exceeds, that is:

H_1: \mu > 60

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

60 is tested at the null hypothesis:

This means that \mu = 60

A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds.

This means that n = 16, X = 67, s = 12

Value of the test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{67 - 60}{\frac{12}{\sqrt{16}}}

t = \frac{7}{3}

t = 2.333

Thus, the correct answer is given by option b.

5 0
2 years ago
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