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Reika [66]
3 years ago
6

Lamar buys $75 worth of clothes. He pays $81, including sales tax. What is the tax rate?

Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer:

C.) 8%

Step-by-step explanation:

75$ × 8% = 6

75 + 6 = 81

i added a picture from google for those who claim my answer is wrong

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The sum of the three angles of a triangle is 180°. Two angles of a triangle measure 15 and 73. Of x is the third angle, set up a
Agata [3.3K]

Answer:

The measure of the third angle is 92°

Step-by-step explanation:

We first have to make an equation that represents this equality. To do it we simply have to add the values ​​that give us with x and this will result in 180 °

15° + 73° + x = 180°

once we have the equation what we have to do is solve x, so we will pass all the terms that are adding to x by subtracting 180

x = 180° - 73° - 15°

now that we have cleared the x we ​​only have to perform the indicated operation and we will obtain the value of x

x = 92°

The measure of the third angle is 92°

8 0
3 years ago
What is the slope and y-intercept of the function shown in the table? <br> X: 1,4,7 <br> Y:6,12,18
NemiM [27]

Answer:

y=2x*6

Step-by-step explanation:

5 0
3 years ago
Please answer QUICKLY!!!
sergiy2304 [10]

Answer:

the range is {7,5,1}

Step-by-step explanation:

To find the range, just plug in the domain into the function.

f(-1) = -2(-1) + 5 = 7

f (0) = -2(0) + 5 = 5

f(2) = -2(2) + 5 = 1

8 0
2 years ago
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Sergio039 [100]
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7 0
3 years ago
A trucking firm has a large inventory of spare parts that have been in storage for a long time. It knowsthat some proportion of
Debora [2.8K]

Answer:

0.1546\leq \widehat{p}\leq 0.3513

Step-by-step explanation:

The firm tests 75 parts, and finds that 0.25 of them are notusable

n = 75

x = 0.25 \times 75 = 18.75≈19

\widehat{p}=\frac{x}{n}

\widehat{p}=\frac{19}{75}

\widehat{p}=0.253

Confidence level = 95%

So, Z_\alpha at 95% = 1.96

Formula of confidence interval of one sample proportion:

=\widehat{p}-Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}\leq \widehat{p}\leq \widehat{p}+Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}

=0.253-(1.96)\sqrt{\frac{0.253(1-0.253)}{75}}\leq \widehat{p}\leq0.253+(1.96)\sqrt{\frac{0.253(1-0.253}{75}}

Confidence interval =0.1546\leq \widehat{p}\leq 0.3513

7 0
4 years ago
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