ASAP PLZZZ HELP

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

Answer:

group one c d f h group two a g e b

Step-by-step explanation:

Hope this helps

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figure a is a scale image of figure b the scale image that maps figure a onto figure b is 1:7 1/4 enter the value of x

skad [1K]

Answer: $x=21.75$

Step-by-step explanation:

You need to convert from mixed number to decimal number.

In order to do this, you can follow the steps shown below:

1. You must divide the numerator of the fraction by the denominator of the fraction.

2. Now you must add the quotient obtained to the Whole number part.

Then, you get:

$7\frac{1}{4}=7+0.25=7.25$

Then, the scale image that maps "Figure A" onto "Figure B" is:

$1:7.25$

Finally, in order to find the value of "x", you need to multiply the given length of the corresponding side of "Figure A" by $7.25$ .

Therefore, the result is:

$x=(3)(7.25)\\\\x=21.75$

3 0

3 years ago

g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib

algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .