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DanielleElmas [232]
3 years ago
12

Two large numbers of the Fibonacci sequence are F49= 7,778,742,049 and F50= 12,586,269,025. If these two numbers are added toget

her, what number results?
Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0
It is 20,365,011,074
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this week maddie worked 2 1/2 hours on monday, 2 2/3 hours on tuesday and 3 1/4 hours on wednesday. how many more hours will mad
ddd [48]
For this you will have to know how to add fractions. First of all i like to add the whole numbers together first, so 2+2+3= 7. so now what you have to do is add the fractions together. so    2/3 + 1/4 + 1/2 = ?   you will have to find a common denominator for 3, 4, and 2; which is 12.  Multiply each fraction by the needed number to get each denominator to 12. so 2/3 will be multiplied by 4, 1/4 will be multiplied by 3, and 1/2 will be multiplied by 6. The new equation and answer will be:   8/12 + 3/12 + 6/12 = 17/12. 17/12 as a mixed fraction is 1 5/12. now we add 1 5/12 and 7 to equal 8 5/12.  and then you will have to subtract this from 10, using common denominators and such to get your final answer of  2 1/12.   (sorry for the long spiel on how to do this question)
7 0
3 years ago
Pleas I really need help with this
Anna71 [15]

Answer:

4r ^2 +9r + 12

Step-by-step explanation:

Hope this helps! :)

8 0
2 years ago
The matrices shows the number of dance students enrolled in each type of class for two different years.
Scorpion4ik [409]

Answer:

n54

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Does anybody know thisss?????!!!!!
nataly862011 [7]
There are 12 months in most years.
.. 23%/12 ≈ 1.91667%

_____
The answer shown in your picture is correct to 2 decimal places.
3 0
3 years ago
. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person
bulgar [2K]

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or greater

Step-by-step explanation:

Problems of normally distributed samples are solved using 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.

In this problem, we have that:

\mu = 100, \sigma = 5

What is the probability that a person selected at random will have an IQ of 110 or greater?

This is 1 subtracted by the pvalue of Z when X = 110. So

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

Z = \frac{110 - 100}{5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that a person selected at random will have an IQ of 110 or greater

5 0
3 years ago
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