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trapecia [35]
2 years ago
8

How many meters is 7.3 kilometers? 1km=1000m

Mathematics
2 answers:
serg [7]2 years ago
8 0

Answer:

7,300 meters

Step-by-step explanation:

7.3×1000 = 7300

jarptica [38.1K]2 years ago
3 0
It is 7300 bc 7.3x1000=7300
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25 km<br> 7 km<br> 18 km<br> 24 km<br> Find the area!!
Anna35 [415]

Answer:

75600k^4 m^4 Hope that help !

Step-by-step explanation:

1. Take out the constants (25×7×18×24) kkkkmmmm

2. Simplify 25×7=175 (175×18×24)kkkkmmmm

3. Simplify 175×18=3150 (3150×24)kkkkmmmm

4. 3150×24= 75600kkkkmmmm

5. The answer would be 75600k^4 m^4

7 0
3 years ago
Find the measure of each marked angle.<br> (9x-27)<br> (4x)
valkas [14]

Answer:

(9×-27)

= (243×4)

= 972

5 0
3 years ago
What does the value of y have to be so that<br> (3, y) and (-5,6) have a slope of -1 between them?
Natasha_Volkova [10]

Answer:

y= -2

Step-by-step explanation:

Please see attached picture for full solution.

(From the 4th to 5th line, I multiplied both sides by 8)

3 0
3 years ago
Factor completely: 10ax + 5bx - 8ay - 4by
kiruha [24]
The answer is (5x - 4y)(2a + b)

10ax + 5bx - 8ay - 4by = 10ax + 5bx - (8ay + 4by) =<span>
                                      = 5x * 2a + 5x * b - (2a * 4y + b * 4y) =
                                      = 5x(2a + b) - 4y(2a + b) = 
                                      = (5x - 4y)(2a + b)</span>
4 0
3 years ago
How does the sample size affect the validity of an empirical​ argument? A. The larger the sample size the better. B. The smaller
astra-53 [7]

Answer:

A. The larger the sample size the better.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

We have to look at the standard error, which is:

s = \frac{\sigma}{\sqrt{n}}

This means that an increase in the sample size reduces the standard error, and thus, the larger the sample size the better, and the correct answer is given by option a.

7 0
3 years ago
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