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2 years ago

Round 259.98991 to the nearest hundredth

Mathematics

2 answers:

FrozenT [24]2 years ago

8 0

259.99.................................................................

raketka [301]2 years ago

8 0

Answer:

259.99

Step-by-step explanation:

hope this helps just did a quiz this is the right answer :)

please mark brainliest

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A man bought a cow and a horse for $500. He sold the cow at a profit of 10% and the horse at a loss of 10% and suffered a loss o

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Answer:

Step-by-step explanation:

<u>System of Equations</u>

Let's call:

x = price of the cow

y = price of the horse

The man bought the cow and the horse for $500, thus

x + y = 500 [1]

The cow was sold at a profit of 10%, thus:

Sale price of the cow= 1.1x

The horse was sold at a loss of 10%, thus:

Sale price of the horse= 0.9y

The total operation was a 2% loss, i.e. 0.98*500=490. Thus, we have:

1.1x + 0.9y = 490 [2]

From [1]:

y = 500 - x

Substituting in [2]:

1.1x + 0.9(500 - x) = 490

Operating:

1.1x + 450 - 0.9x = 490

0.2x = 490 - 450 = 40

x = 40/0.2

x = 200

The man paid $200 for the cow

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3 years ago

Pls help. Evaluate the expression 3.14(a2 + ab) when a = 3 and b = 4. (Input decimals only, such as 12.71, as the answer.)

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3 years ago

4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

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Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Rearranging these two equations:

$\\ -0.6*\sigma = 100-\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

$\\ 0.6*\sigma = -100+\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

Summing both equations, we obtain the following equation:

$\\ 3.0*\sigma = 90$

Then

$\\ \sigma = \frac{90}{3.0} = 30$

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

$\\ 2.4*30 = 190-\mu$ (2)

$\\ 2.4*30 - 190 = -\mu$

$\\ -2.4*30 + 190 = \mu$

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Moreover, because this affects to all diabetic patients, they all should be included even if they do not usually eat foods with arficial sweeteners.

Based on this, the best population is all diabetic patients.

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