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3 years ago

Round 78482 to the nearest thousand

Mathematics

1 answer:

shtirl [24]3 years ago

6 0

Answer:

78,000

Explanation:

8,000 is in the thousands place, the 4 in the hundreds place is too low to round it up to 9,000, so it gets rounded down to 78,000.

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What is the value of x? The figure is not drawn to scale. Show work

Alecsey [184]

Answer:

x = 21.

Step-by-step explanation:

AS there are parallel lines:

4/12 = 7/x

4x = 84

x = 84/4 = 21.

8 0

2 years ago

A band leader randomly selects 2 students out of 80 band members. If you and your brother are both band members, what is the pro

drek231 [11]

The first time the band leader selects one member, the probability that either you or your brother gets selected would be 2 out of 80 or in fraction 2/80. Now, once one of you gets chosen, for the next selection, there are only 79 members left and the probability that the remaining one (between you and your brother) gets selected after is 1/79. So, if you are both to be selected, the combined probability for that to happen will be (2/80)(1/79) = 1/3160. Therefore, the answer is 1/3160<span>.</span>

6 0

3 years ago

A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

lesantik [10]

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

$P(failure)=\frac{1}{10000}$

⇒

$P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999$

For components D,E,G and H we have :

$P(failure)=\frac{3}{10000}$

⇒

$P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997$

Finally, for components I and J :

$P(failure)=\frac{5}{10000}$

⇒

$P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995$

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = $(0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402$

0.997402 = 99.7402 %

4 0

3 years ago

HELP PLEASE I DONT UNDERSTAND AND NEED TO PASS

Nina [5.8K]

Answer:

20 for de first one and answer (B) for the second.

Step-by-step explanation:

20 for de first one and answer (B) for the second.

hope this helped;3

7 0

2 years ago

Please help with this question

natita [175]

Answer:

$12\sin{(2\theta)}\cos{(\theta)}$

Step-by-step explanation:

The appropriate trig identity is ...

$\sin{a}+\sin{b}=2\sin{\left(\dfrac{a+b}{2}\right)}\cos{\left(\dfrac{a-b}{2}\right)}$

Here, you have a scale factor of 6 and a=3θ, b=θ. Filling in these values gives ...

$6\sin{3\theta}+6\sin{\theta}=12\sin{\left(\dfrac{3\theta+\theta}{2}\right)}\cos{\left(\dfrac{3\theta-\theta}{2}\right)}\\\\=12\sin{(2\theta)}\cos{(\theta)}$

8 0

2 years ago