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Effectus [21]
4 years ago
9

Suma a doua numere este 500 iar diferenta lor este 100. aflati numerele

Mathematics
1 answer:
antoniya [11.8K]4 years ago
5 0
Notăm cu a,b  numerele
a+b=500
a-b=100⇒a=100+b
100+b+b=500
100+2b=500
2b=400
b=200
a=200+100⇒a=300
Numerele sunt 200 şi 300
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An individual can win a prize by drawing the highest numbered card out of five cards. After each draw, the card is replaced and
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Step-by-step explanation:

1/5 for one drawing

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Then multiply them together to get 1/25.

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3 years ago
Which inequality describes the range of f(x)
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Answer:

y ≥ -3

Step-by-step explanation:

When we talk of the range of a function, we are referring to the y-values which are possible for the function to take

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3 years ago
Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f
schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{130}{20} = 6.5

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

S.D = \sqrt{\frac{233.5}{19}} = 3.50

b) Standard Error

= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826

c) Point estimate for the mean time required for all college is given by the sample mean.

\bar{x} = 6.5

d) 90% Confidence interval:  

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729  

6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

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Answer:

4m

Step-by-step explanation:

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3 years ago
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