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aleksley [76]
2 years ago
11

Find the surface area of the prism.

Mathematics
2 answers:
Mrac [35]2 years ago
8 0

Answer: 136

Step-by-step explanation: 5x7=35 5x7=35 6x4/2=12 6x4/2=12 7x6=42 35+35+12+12+42=136

V125BC [204]2 years ago
3 0
The surface area is 136 m
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It's not clear what polynomial you posted, but let's say you had the polynomial

5x^3+7x^2-9x+10

This is a cubic polynomial as the largest exponent is 3. The degree is also 3. The degree of the polynomial is equal to the largest exponent.

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Find the area enclosed by the curve y^2=x^2-x^4
yulyashka [42]

Answer: 4/3

Step-by-step explanation:

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3 years ago
Which phrase best describes the translation from the graph y = (x – 5)2 + 7 to the graph of y = (x + 1)2 – 2?
Andreas93 [3]

The parent. function shifted up by 6 units to produce x + 1 and to the left by 9 units

<h3>What is translation?</h3>

This is a way of changing the position of an object on an xy-plane.

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8 0
2 years ago
If you borrow $100 at 50% interest, how much interest do you think you will have to pay?
Elena L [17]

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the answer is 50 dollars

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6 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
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