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abruzzese [7]
3 years ago
14

PLEASE??can someone give me help ?

Mathematics
2 answers:
rjkz [21]3 years ago
7 0
The correct one is B. 
It's false because you can live anywhere in the US. But the converse is true because if you live in Illinois then you do live in the US. 
fiasKO [112]3 years ago
5 0
It is D that makes scene

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A long distance runner ran 4x4x4x4x4x4 miles last year. How many miles is this?
Archy [21]

Answer:

4096

Step-by-step explanation:

4x4x4x4x4x4=4^6

Group them to multiply to make it easier:

4x4=16

4x4=16

4x4=16

16x16x16=4096

honestly just plug it into a calculator

3 0
3 years ago
Read 2 more answers
What is the estimate of 11,942 ​
ki77a [65]

Answer:

10000 (1sf)

12000(2sf)

11900(3sf)

11940(4sf)

Step-by-step explanation:

I would go for 12000 (2sf) if it the degree of accuracy isn't specified in the question

6 0
3 years ago
Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
3 years ago
1.) An American tourist changed US $925 into Trinidad and Tobago currency at the exchange rate US $1.00= TT $ 4.25
storchak [24]

Step-by-step explanation:

1)a) 925×4.25=3931,25

b) 1806.25:4.25=425

I don't know second one.

3)a 55×1.12=61.6

b)

8 0
3 years ago
Which choice is equivalent to the expression below when y2 0?<br> √y^2 + √16y^3 – 4y√y
DanielleElmas [232]

Answer:

Option C.

Step-by-step explanation:

We start with the expression:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}

And:

y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}

If we apply these two to our initial expression, we can rewrite it as:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

\sqrt{y^3}  + 4*\sqrt{y^3} - 4\sqrt{y^3} = \sqrt{y^3}

Here we can use the second simplification again, to rewrite:

\sqrt{y^3} = y*\sqrt{y}

So, concluding, we have:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y} = y*\sqrt{y}

Then the correct option is C.

8 0
3 years ago
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