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Oduvanchick [21]
3 years ago
10

A store is having a sell in which all items are reduced by 20% including tax

Mathematics
1 answer:
Andrei [34K]3 years ago
6 0

I’m not sure what the question is, so if you could tell me what the actual question is I might be able to answer it!

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What is the answer for 604.8 36 with work included?​
goblinko [34]

Answer:

16.8

Step-by-step explanation:

You have to do 360 divided my 6048 because you have to do things on both numbers like moving the decimal.

8 0
2 years ago
Into the boxes:<br> 2x + 8 = 1x + 9<br> Solve the equation.
JulsSmile [24]

Answer:

x = 1

Step-by-step explanation:

Rearrange equation by making x's on one side and numbers on the other

2x - x = 9 - 8

x = 1

6 0
3 years ago
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Please help me with this problem!!!
Lilit [14]

3  {}^{4x  - 5}  = ( \frac{1}{27} ) {}^{2x + 10}  \\ note  \: \: that =  \\ ( \frac{1}{27} ) = ( \frac{1}{3 {}^{3} } ) = 3 {}^{ - 3}  \\  \\ 3 {}^{4x - 5}  = (3 {}^{ - 3} )  {}^{2x + 10}  \\ 3 {}^{4x + 5}  = 3 {}^{ - 3(2x + 10)}  \\ equal \:  \: bases =  > equal \: exponents

4x - 5 =  - 3(2x + 10) \\ 4x - 5 =  - 6x - 30 \\ 4x  + 6x =  - 30 + 5 \\ 10x =  - 25 \\ x =   - 2.5

7 0
2 years ago
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The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

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

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error 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 = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0
3 years ago
Please help!!!!!! ASAP!!!
Kitty [74]

Answer:

B. po ang answer sa palagay ko lang po

6 0
2 years ago
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