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2 years ago

Log10^x - log10^3 - 1

Mathematics

1 answer:

Pavel [41]2 years ago

8 0

Answer:

1^x-2

Step-by-step explanation:

First you put log(10)^x into the calculator, the product is 1^x. Then you subtract that by log(10)^3 which you you get 1^x-1 then you subtract that answer by 1 to get 1^x-2.

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Use the formula for the area of a circle to find the area of the bullseye and the next ring together ?!?!?!

Anon25 [30]

I got D for the answer pi x 12mmsquare cause you're only measuring the area of the bullseye and i got 452.39

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3 years ago

Read 2 more answers

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach

nlexa [21]

Answer:

A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions can be 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that $\mu = 130, \sigma = 30$

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 130}{30}$

$X - 130 = 0.675*30$

$X = 150.25$

A score of 150.25 is necessary to reach the 75th percentile.

7 0

3 years ago

Find the area of a trapezoid<br> Explain please

djverab [1.8K]

Answer:

Step-by-step explanation:

Area = a+b/2 * h

6+10/2= 8

8*4 = 32

7 0

2 years ago

The graph below shows the solution to which system of inequalities?

nlexa [21]

Answer:

Option D.

Step-by-step explanation:

Consider option D. $y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3$

Take point (0,0)

On putting this point in inequation $y\leq \frac{3x}{4}+10$ , we get

$0\leq 10$ which is true . So, solution is region towards the origin i,e region below the line $y= \frac{3x}{4}+10$ including the line itself .

On putting (0,0) in inequation $y\leq \frac{-x}{2}-3$ , we get $0\leq -3$ which is false , so solution is region away from the origin i.e region below line $y= \frac{-x}{2}-3$ including the line itself .