Is r(l+2h) a formula for a length area or volume

Peyton has collected 120 aluminum cans for recycling. If 20 cans will fit in each blue plastic bag, how many bags will she need

gogolik [260]

Hi Renee!

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We Know:

120 aluminum cans.

20 cans fit in each blue bag.

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Solution:

120 / 20 = 6

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Answer:

She will need 6 bags to carry all the aluminum cans.

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Hope This Helps :)

6 0

3 years ago

The length of a rectangle is 5959 inches greatergreater than twice the width. if the diagonal is 2 inches more than the length,

lorasvet [3.4K]

Length: 2w + 59

width: w

diagonal: (2w + 59) + 2 = 2w + 61

Length² + width² = diagonal²

(2w + 59)² + (w)² = (2w + 61)²

(4w² + 118w + 3481) + w² = 4w² + 122w + 3721

5w² + 118w + 3481 = 4w² + 122w + 3721

w² + 118w + 3481 = 122w + 3721

w² - 4w + 3481 = 3721

w² - 4w - 240 = 0

a = 1, b = -4, c = -240

w = $[-(b) +/- \sqrt{(b)^{2} - 4(a)(c) }]/2(a)$

= $[-(-4) +/- \sqrt{(-4)^{2} - 4(1)(-240) }]/2(1)$

= $[4 +/- \sqrt{(16 + 960) }]/2$

= $[4 +/- \sqrt{(976) }]/2$

= $[2 +/- 4\sqrt{(61) }]/2$

= $1 +/- 2\sqrt{(61) }$

since width cannot be negative, disregard 1 - 2√61

w = 1 + 2√61 ≈ 16.62

Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24

Answer: width = 16.62 in, length = 92.24 in

7 0

3 years ago

Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut

defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .