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2 years ago

One-third of a number yields twelve. Which equation could be used to find the number?

Mathematics

2 answers:

NeX [460]2 years ago

7 0

1/3n=12

this is the answer

LenaWriter [7]2 years ago

4 0

Answer:

1/3n=12

Step-by-step explanation:

The reason why it is 1/3n=12, is because you know that the number in front of n is 1/3. By figuring that out, you have already ruled out 3n=12 and 3n=-12. Finally, you know that 12 is a positive number leaving you with 1/3n=12.

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PLEASE HELP. ITS A MINI-QUIZ / HOMEWORK AND ITS TRIANGLES.

polet [3.4K]

Answer:

76 degrees

180-38-38 is 104

180-104 is 76 since it's a straight line which equals to 180 degrees

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2 years ago

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In this diagram, bac~edf. if the area of bac= 6 in.², what is the area of edf? PLZ HELP PLZ PLZ PLZ

NemiM [27]

Answer:

2.7 in²

Step-by-step explanation:

Area of ∆BAC : ∆Area of EDF = BC² : EF² (based on the area of similar triangles theorem)

Thus:

$6 in^2 : x in^2 = (3 in)^2 : (2 in)^2$

$\frac{6}{x} = \frac{3^2}{2^2}$

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2 years ago

Evaluate x(-y + z) for x = 5, y = 9, and z = 1.

NemiM [27]

Answer:

-20

Step-by-step explanation:

x(-y+z) [Switch in the values]

=5(-5+1) [Use distributive property and multiply]

=(-25)+5 [do the basic math]

=(-20) [final answer]

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2 years ago

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Today, both the soccer team and the basket ball team had games (starting on Monday). The soccer team plays every 3 days and the

yarga [219]

I think eight because if you add those days together they will play again in eight days so that they play on the same day.

Sorry if I got it wrong I tried and that's what counts, right?

3 0

2 years ago

Assume that 33.4% of people have sleepwalked. Assume that in a random sample of 1459 adults, 526 have sleepwalked.

lutik1710 [3]

Answer:

Step-by-step explanation:

Given that 33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p = $\frac{526}{1459} \\=0.361$

Sample proportion p is normal for large samples with mean = 0.334 and

std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.334(1-0.334)}{1459} } \\=0.0123$

a) P(526 or more of the 1459 adults have sleepwalked.)

$=P(p\geq 0.361)\\=P(Z\geq \frac{0.361-0.334}{0.0123} \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139$

b) Yes, because hardly 1.4% is the probability

c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.

7 0

2 years ago