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Nezavi [6.7K]
3 years ago
5

What is greater, 500 g or 0.5 kg?

Mathematics
2 answers:
jeyben [28]3 years ago
6 0
5kg is greater than 500g
natima [27]3 years ago
5 0

Answer:

they are equal

0.5kg is 500g

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a family has two cars. the first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of
Kryger [21]
Let the number of gallons consumed be the first car be x, and that of the second car, y. Then:
x + y = 50 . . . . . . . (1)
20x + 15y = 850. . . (2)

(1) x 20 => 20x + 20y = 1000 . . . (3)
(2) - (3) => -5y = -150 => y = -150/-5 = 30

From (1), x + 30 = 50 => x = 50 - 30 = 20

Therefore, first car consumed 20 gallons and the second consumed 30 gallons.
3 0
3 years ago
There are 48 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time
Stella [2.4K]

Answer:

a) 64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

b) The hardness distribution is not given. But you would have to find s when n = 39, then the probability would be 1 subtracted by the pvalue of Z when X = 51.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

For the sum of n trials, the mean is \mu*n and the standard deviation is s = \sigma\sqrt{n}

In this question:

n = 48, \mu = 48*5 = 240, s = 4\sqrt{48} = 27.71

These values are in minutes.

(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?

From 6:50 PM to 11 PM there are 4 hours and 10 minutes, so 4*60 + 10 = 250 minutes. This probability is the pvalue of Z when X = 250. So

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

By the Central Limit Theorem

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

Z = \frac{250 - 240}{27.71}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?

The hardness distribution is not given. But you would have to find s when n = 39(using the standard deviation of the population divided by the square root of 39, since it is not a sum here), then the probability would be 1 subtracted by the pvalue of Z when X = 51.

5 0
3 years ago
Module 07: Project Option 1
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3 years ago
X^3+7x^2+13x+4=0 Show all work
Softa [21]
   
\displaystyle\\
x^3+7x^2+13x+4=0\\\\
x^3+\underbrace{4x^2+3x^2}_{7x^2}+\underbrace{12x+x}_{13x}+4=0\\\\
(x^3+4x^2) + (3x^2 + 12x)+(x+4)=0\\\\
x^2(x+4)+3x(x+4)+(x+4)=0\\\\
(x+4)(x^2 + 3x +1)=0\\\\
x+4 = 0~~~\Longrightarrow~~~\boxed{x_1 = -4}\\\\
x^2 + 3x +1=0\\\\
x_{23}= \frac{-b\pm  \sqrt{b^2-4ac} }{2a}= \frac{-3\pm  \sqrt{3^2-4\cdot 1 \cdot 1} }{2\cdot 1}= \frac{-3\pm  \sqrt{5} }{2}\\\\
\boxed{x_2 = \frac{-3-\sqrt{5} }{2}}\\\\
\boxed{x_3 = \frac{-3+\sqrt{5} }{2}}



4 0
4 years ago
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