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DanielleElmas [232]
3 years ago
15

Write 3 to the 9th power as an expression

Mathematics
1 answer:
8_murik_8 [283]3 years ago
3 0
The expression that best represents the statement <span>3 to the 9th power would be written as:

3^9.

The value of 9 here is the exponent in the expression and 3 is the base. </span><span>This process of using </span>exponents<span> is called "raising to a power", where the </span>exponent<span> is the "power".</span>
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Uma pessoa deseja restaurar seu veículo antigo e para isso deverá contratar serviços mecânico, de funilaria e de autoelétrica. P
Leni [432]

Answer:

CI=P*(1 + R/100)^18

A=(CI + P) = P(1+R/100)^18

13500/P=1(100+R/100)^18

A/P=(100+R/100)^18

A/P=(100+R/100)^18

A=13500$        as (750 * 18)

(13500)/P=(1 +1.15/100)18

(13500)/P=(1+1.15/100)18

13500=((1.0115)^18

P=R$10989.02

Step-by-step explanation:

CI=Compound Interest

A=Amount

P=Principal.

3 0
3 years ago
The ratio to hearts to moons is | The ratio of moons to all shapes is. I need help due in 10 minutes
Alinara [238K]

Hearts to moon= 6:4

                         = 3:2 (simplified)

Moon to all shapes= 4:10

                               = 2:5 (simplified)

4 0
3 years ago
Read 2 more answers
Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando
DENIUS [597]

Answer:

n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

6 0
3 years ago
Which quantity is use to generate the error bars on a graph?
rodikova [14]
The error bars are used to indicate the variability of the data presented in a graph.

There are several quantities that can be used to generate error bars in the graph. These are:
standard deviation
standard error
confidence interval

Usually, one standard deviation above and below the mean is used although it is advised to indicate which variability data is used to generate the error bars in the graph since the 3 quantity given are not equal.
7 0
3 years ago
732,178 + 167 - 542,137
klio [65]

Answer:

190,208

Step-by-step explanation:

732,178 + 167 = 732,345

732,345 - 542,137 = 190,208

6 0
3 years ago
Read 2 more answers
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