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Rama09 [41]
2 years ago
15

Find the sum of ∑4/k=1 (-4k)

Mathematics
1 answer:
Annette [7]2 years ago
4 0

Answer:

Hello,

Answer C: -40

Step-by-step explanation:

\displaystyle \sum_{k=1}^4\ (-4k)\\\\=-4*\sum_{k=1}^4\ k\\\\=-4*4*\dfrac{1+4}{2} \\\\=-4*5*2\\\\=-40\\

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What’s 35 divide by 4?
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35/4 = 8.75 or 8 3/4

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2 years ago
Suppose you can somehow choose two people at random who took the SAT in 2014. A reminder that scores were Normally distributed w
Sindrei [870]

Answer:

22.29% probability that both of them scored above a 1520

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 = 1497, \sigma = 322

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

So

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

Z = \frac{1520 - 1497}{322}

Z = 0.07

Z = 0.07 has a pvalue of 0.5279

1 - 0.5279 = 0.4721

Each students has a 0.4721 probability of scoring above 1520.

What is the probability that both of them scored above a 1520?

Each students has a 0.4721 probability of scoring above 1520. So

P = 0.4721*0.4721 = 0.2229

22.29% probability that both of them scored above a 1520

8 0
3 years ago
Suppose the polynomial f(x) has the following end behavior: as x approaches infinity, f(x) approaches infinity, and as x approac
galina1969 [7]

Answer:

C. x³+10x²−5x+5

E. 7x⁵+4x²

F. x+8

Step-by-step explanation:

The graph they described should look something like the one i drew below.

So all you have to do is look at the number with the highest exponent.

If the exponent is an odd number and the x is positive then that function will have an end behavior like the picture i posted.

x³ is good

-x³ is bad because of the negative sign

x² is bad because exponent is even

x⁷ is good

123x¹ is good

6 0
3 years ago
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