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3 years ago

Please help it would be a greatly supported.

Mathematics

1 answer:

USPshnik [31]3 years ago

8 0

Answer:

D. N = w² + 8w + 12

Step-by-step explanation:

The current size is w by w+4, so the area is ...

A = w(w+4)

When each dimension is increased by 2, the new size is (w+2) by (w+4+2). The latter dimension can be simplified to (w+6). Now, the new area is ...

N = (w+2)(w+6) = w² +8w +12 . . . . . . . matches choice D

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a rectangle has an area of 30 square meters and a perimeter of 34 meters. what are the dimensions of the triangle?

gayaneshka [121]

The dimensions for the rectangle are 15 by 2.

15 times 2 is 30, which is the area.

15+15+2+2= 34 which is the perimeter

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3 years ago

Find (18 × 106) + (5 × 104). A) 1.8005 × 106 B) 1.8005 × 107 C) 1.805 × 106 D) 1.805 × 107

Andrews [41]

The answer is D. Because (18 x 10^6) + (5 x 10^4) = 18050000. And the answer to 1.805 x 10^7 also equals 18050000, if you do the math.

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3 years ago

Read 2 more answers

What’s the correct answer

ale4655 [162]

$45x=11\qquad\text{divide both sides by 45}\\\\\boxed{x=\dfrac{11}{45}}=0.2\overline{4}\\\\Answer:\ \boxed{x\approx0.24}$

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3 years ago

Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.

Georgia [21]

Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

$z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ 
z-score =2.326$

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

5 0

3 years ago

Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu

Luda [366]

Answer:

11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

P(z ≤ -6) ≈ 9.86588×10^-10

and

P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

0.50^t = 0.000290372

t·log(0.50) = log(0.00290372) . . . . take logarithms

t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0

3 years ago