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2 years ago

lenny left home wih 2 hundred dollar bill he purchased 3 t-shirts each costing 39.50. how much change did he have left

Mathematics

2 answers:

vekshin12 years ago

5 0

Answer:

Change left = $81.50

Step-by-step explanation:

Total cost of 3 Tshirts = 3×39.50

= $118.50

To find the change

$200.00

- $118.50

= $81.50

Nezavi [6.7K]2 years ago

3 0

Answer:

He gets $81.50 back as change

Step-by-step explanation:

39.50 times 3 is 118.5

you then do 200-118.5 and you get $81.50

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Help on these questions. Please show work it would help a lot. Thanks :)

Murrr4er [49]

10: an isosceles base angles are both equal. A triangle has 3 angles. All three angles added equals 180 in every triangle. since one of the base angle is 72 degrees then the other base angle is 72 degrees. ( a triangle has two base angles). so 72+72= 144

   180-144= 36
The vertex angle = 36 degrees.

11: angles ratio is 2:4:6. Remember that all angles equal 180. Add the ratio amounts, 2+4+6= 12. Then divide 180 by 12 (180/12 = 15).
the smallest angle ratio is 2, so multiply 15 by 2 ( 15 * 2= 30 )
the smallest angle is 30 degrees.

12: all three angles of a triangle adds up to 180. thus, 64 + 74 + x = 180
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x is the third angle. so the third angle is 42 degrees.

6 0

3 years ago

A marketing research company needs to estimate the average total compensation of CEOs in the service industry. Data were randoml

mezya [45]

SOLUTION:

Step 1 :

In this question, we are told that a marketing research company needs to estimate the average total compensation of CEOs in the service industry.

We also have that: Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180).

Then, we are asked to find the margin error for the confidence interval.

Step 2:

We need to recall that:

$\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}$ $\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}$

It means that:

$\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}$ $\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}$

where,

$CI_H\text{ = }$$5,836,180$\text{ }$$$$ $CI_{L\text{ }}=\text{ }2,181,260$

putting the values into the equation for the margin of error, we have that:

$\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{ - }2,181,260\text{ }}{2}$ $\begin{gathered} =\text{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}$

CONCLUSION:

The margin error for the confidence interval is 1, 827, 460

5 0

1 year ago

NEED HELP ASAP!<br> Subtract.<br> (4x^2+8x−2)−(2x^2−4x+3)<br>Answer in standard form

AlekseyPX

$(4x^2+8x-2)-(2x^2-4x+3)\\\\=4x^2+8x-2-2x^2+4x-3\\\\=(4x^2-2x^2)+(8x+4x)+(-2-3)\\\\=\boxed{2x^2+12x-5}$