What is the Evaluation

Can someone please help me with this question? Thanks.

katrin2010 [14]

Area = length x width

Area = 3x + 5 Cm x 8 cm

Multiply each term by each term.

Area = 24x + 40 square cm

3 0

3 years ago

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Ppppplllssss heeellppppppppppppllllpllklgfgtfchgrrdgvcfyrf

zubka84 [21]

Answer:

Equation: 3x - 11 = 85

Solution: x = 32

Step-by-step explanation:

3x - 11 = 85

You will have to do the opposite to the equation shown on the screen.

3x = 96 (85 + 11)

x = 32 (96 / 3)

3 0

11 months ago

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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:

$\mu = 591, \sigma = 42$

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{42}$

$Z = 0.21$

$Z = 0.21$ has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{50}} = 5.94$

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{5.94}$

$Z = 1.515$

$Z = 1.515$ has a pvalue of 0.9351

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{100}} = 4.2$

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

$Z = \frac{600 - 591}{4.2}$

$Z = 2.14$

$Z = 2.14$ has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

8 0

2 years ago

Which of the following is the area of the special trapezoid if AB = 10, CD = 10, and the height is 8.

Lelu [443]

Area of a trapezoid is A=height over 2 B1 +B2, so 10+10=20, then 20(8)=160, 160/2=80. Answer is b.

3 0

2 years ago

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The annual energy consumption of the town where Camilla lives in creases at a rate that is onal at any time to the energy consum

boyakko [2]

Answer:

6.575 trillion BTUs

Step-by-step explanation:

Let represent the annual energy consumption of the town as E

The rate of annual energy consumption * energy consumption at time past

dE/dt * E

dE/dt =K

k = the proportionality constant

c= the integration constant

(dE/dt=) kdt

lnE = kt + c

E(t) = e^kt+c ⇒ e^c e^kt e^c is a constant, and e^c = E₀

E(t) = E₀ e^kt

The initial consumption of energy is E(0)=4.4TBTU

set t = 0 then

4.4 = E₀ e⇒ E₀ (1)

E₀ = 4.4

E (t) = 4.4e^kt

The consumption after 5 years is t = 5, e(5) = 5.5TBTU

so,

E(5) = 5.5 = 4.4e^k(5)

e^5k = 5/4

We now take the log 5kln = ln(5/4)

5k(1) = ln(5/4)

k = 1/5 ln(5/4) = 0.04463

We find the town's annual energy consumption, after 9 years

we set t=9

E(9) = 4.4e^0.04463(9)

= 4.4(1.494301) = 6.5749TBTUs

Therefore the annual energy consumption of the town after 9 years is

= 6.575 trillion BTUs

3 0

3 years ago