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2 years ago

Which Exponentail expression represents The numerical expression 1/12 . 1/12 . 1/12 . 1/12?

Mathematics

1 answer:

melomori [17]2 years ago

6 0

Answer:

1/12^4 the for is small and raised

Step-by-step explanation:

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Which graph represents the sequence: 80, -40, 20

ANEK [815]

80/-40=-40/20=-2,
the sequence: 80, -40, 20 is a geometric sequence
its general formula is Vn+1 = q Vn, where q= -2,
if we put Vn+1 = f(x)
 Vn = x

so we have f(x)= -2x so the graph that represents the sequence is graph of linear equation

8 0

3 years ago

The table shows results of an experiment that was replicated.

KiRa [710]

The option that best describes the experiment is accurate and reproducible.

<h3>What option describes the data?</h3>

All the values from the experiment are close in value to the accepted value. This indicates that the experiment is accurate. Two experiments yield the same values. This indicates that the experiment is reproducible.

Here is the table used in answering the question:

Accepted Value: 130

Experiment 1 129

Experiment 2 131

Experiment 3 129

Experiment 4 132

To learn more about experiments, please check: brainly.com/question/14019529

#SPJ1

4 0

2 years ago

If y(x) = 4x, what is x when y(x) = 4 A. 1 B. 4 C. 2 D. 0

Eva8 [605]

Answer:

Step-by-step explanation:

cause if :

y=4

4(x)=4

4(1)=4

4=4

hope this helps

4 0

3 years ago

I'm not understanding this can you help me

hoa [83]

Answer:

gawin mo pag isipan mo wag puro brainly

3 0

3 years ago

Read 2 more answers

A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers

Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

$\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57$

Using normal distribution table, $z_{0.005$ is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

$e= z_{0.005}\frac{\sigma}{\sqrt{n} } = 2.57*\frac{3900}{\sqrt{100} } =1002.3$

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0

3 years ago