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1 year ago

Which expression represents the volume of this prism?

Mathematics

2 answers:

Mila [183]1 year ago

8 0

We know for any prism the volume formula is

$\\ \tt\hookrightarrow Area\:of\:base\times Height$

Area of base=1×2=2units

Volume:-

$\\ \tt\hookrightarrow 2(8)units^3$

Ronch [10]1 year ago

4 0

Answer:

Option A

Step-by-step explanation:

We know that:

Volume of prism: L x B x H
W = 8 units
L = 2 units
H = 1 unit

Solution:

L x W x H
=> 2 x 8 x 1
=> 2 x 8
=> 16 units³

Looking at the options, we can see that our result matches with Option A.

Hence, Option A is correct.

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Anvisha [2.4K]

Answer:

Step-by-step explanation:

using the rule of exponents

• $(a/b)^{n}$ = $a^{n}$ / $b^{n}$ , hence

$(7/4)^{11}$ = $7^{11}$ / $4^{11}$ → D

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3 years ago

Complete the following exercises by applying polynomial identities to complex numbers.

erik [133]

1.)

=(x-8i)(x+8i)

x^2+8ix-8ix-64i^2

x^2-64i^2

x^2-64(-1)

x^2+64

2.)

=(4x-7i)(4x+7i)

16x^2+28ix-28ix-49i^2

16x^2-49i^2

16x^2-49(-1)

16x^2+49

3.)

=(x+9i)(x+9i)

x^2+9ix+9ix+81i^2

x^2+18ix+81(-1)

x^2+18ix-81

4.)

=(x-2i)(x-2i)

x^2-2ix-2ix+4i^2

x^2-4ix+4(-1)

x^2-4ix-4

5.)

=[x+(3+5i)]^2

(x+5i+3)^2

(x+5i+3)(x+5i+3)

x^2+5ix+3x+5ix+25i^2+15i+3x+15i+9

x^2+6x+10ix+30i+25i^2+9

x^2+6x+10ix+30i+25(-1)+9

x^2+6x+10ix+30i-25+9

x^2+6x+10ix+30i-16

Hope this helps :)

6 0

2 years ago

Interval of increase x^(1/3)(x+4)

Agata [3.3K]

Your function f(x)=(x^(1/3))(x+4)

What your going to want to do first is calculate your first derivative.
You will want to use both the power rule and the product rule.

f '(x)=(x+4)/x^(2/3)+x^(1/3)

f '(x)=4/3(x^(-2/3))(x+1)

Now find the critical numbers

4/3(x^(-2/3))=0 x+1=0
x=0, x=-1

Now plug numbers on the intervals (-infinity, -1),(-1,0) and (0,infinity) into the derivative of your function. The numbers I would choose are -2, -1/2 and 1.

Since f '(-2) is negative, f '(-1/2) is positive and f '(1) is positive, (-1,-3) and (1,5) are relative extrema. This means that the function f(x)=(x^(1/3))(x+4) is increasing on the interval (-1,0) U (0,infinity).

7 0

3 years ago

The capacity of an elevator is 10 people or 1730 pounds. The capacity will be exceeded if 10 people have weights with a mean gre

Nadusha1986 [10]

Answer:

a. 48.80%

b. 46.02%

c. 57.93% probability of the sample mean weight being above the weight limit, which is a high probability, meaning that the elevator does not appear to have the correct weight limit

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 175, \sigma = 31$

a. find the probability that if a person is randomly selected, his weight will be greater than 176 pounds.

This is 1 subtracted by the pvalue of Z when X = 176. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{176 - 175}{31}$

$Z = 0.03$

$Z = 0.03$ has a pvalue of 0.5120

1 - 0.5120 = 0.4880

48.80% probability that if a person is randomly selected, his weight will be greater than 176 pounds.

b. Find the probability that 10 randomly selected people will have a neam that is greater than 176 pounds.

Now $n = 10, s = \frac{31}{\sqrt{10}} = 9.8$

This is 1 subtracted by the pvalue of Z when X = 176. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Thorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{176 - 175}{9.8}$

$Z = 0.1$

$Z = 0.1$ has a pvalue of 0.5398

1 - 0.5398 = 0.4602

46.02% probability that 10 randomly selected people will have a neam that is greater than 176 pounds.

c. Does the elevator appear to have the correct weight limit?

The weight limit is 173 pounds in a sample of 10.

The probability that the mean weight is larger than this is 1 subtracted by the pvalue of Z when X = 173. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{173 - 175}{9.8}$

$Z = -0.2$

$Z = -0.2$ has a pvalue of 0.4207

1 - 0.4207 = 0.5793

57.93% probability of the sample mean weight being above the weight limit, which is a high probability, meaning that the elevator does not appear to have the correct weight limit

8 0

2 years ago

"Quantos números inteiros há entre -6 e 4?"

LekaFEV [45]

Think: -5, -4, -3, -2, -1, 0, 1, 2 3. How many integers is that?

Piensa: -5, -4, -3, -2, -1, 0, 1, 2 3. ¿Cuántos enteros es eso?

6 0

2 years ago