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weqwewe [10]
3 years ago
14

Your friend earns $10.50 per hour. This is 125% of her hourly wage last year. How much did your friend earn per hour last year?

Mathematics
1 answer:
qwelly [4]3 years ago
8 0
Well 
10.50 / 125%=8.4
your friend earned $8.40 per hour last year
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A thief steals a number of rare plants from a nursery. On the way​ out, the thief meets three security​ guards, one after anothe
DedPeter [7]

The thief originally stole 64 plants.

<h3>What is arithmetic?</h3>

In mathematics, it deals with numbers of operations according to the statements.

Here,
Let the number of plants originally stolen by the thief be x.
On the way​ out, the thief meets three security​ guards, one after another. To each security​ guard, the thief is forced to give one-half of the plants that he still​ had, plus 2 more.​

Here,
For gourd 1 the planet is given = 1/2 x
For gourd 2 the planet is given = 1/2 * 1/2 x = 1/4 x
For gourd 3 the planet is given = 1/2*1/4 x = 1/8 x
Total plant given to the gourds = 1/2x +1/4x +1/8x + 2 = 7/x/8x +2
Finally, the thief leaves the nursery with 10 plants
x - 7x/8 + 2 = 10
x/8 = 8
x = 64

Thus, the thief stole 64 plants.

Learn more about arithmetic here:

brainly.com/question/14753192

<h3>#SPJ1</h3>


7 0
2 years ago
What is the length of the diagonal of a poster board with dimensions 22 inches by 28 inches? Round to the nearest tenth.
devlian [24]

Answer:

The length of the diagonal of a poster board is 35.6\ in

Step-by-step explanation:

Let

x----> the length of the diagonal of a poster board

we know that

Applying the Pythagoras Theorem

x^{2}=22^{2}+28^{2} \\ \\x^{2}=1,268\\ \\x=35.6\ in

4 0
3 years ago
Read 2 more answers
A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.
son4ous [18]

Answer:

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.01 = 0.99, so z = 2.325

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

3 0
3 years ago
Can someone help me with these 3
photoshop1234 [79]
24. 244 cm

25. 65 mm

26. 37/80 or 0.4625

8 0
4 years ago
The average of 1/2006 and 1/2007 is equal to half of their?
erma4kov [3.2K]
Add it. average= sum of 2 counts/2
5 0
3 years ago
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