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3 years ago

A dealership pays $580 for each electronic unit and sells it for $940. What is the percent markup in the price of each unit

Mathematics

1 answer:

topjm [15]3 years ago

6 0

Answer:

62.06%

Step-by-step explanation:

Given that,

CP of an electronic unit = $580

SP of an electronic unit = $940

We need to find the percent markup in the price of each unit. It can be done as follows :

$\%=\dfrac{940-580}{580}\times 100\\\\=62.06\%$

So, the required answer is 62.06%.

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Two less than four times a number is 6

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Answer: 2 - 4x = 6

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3 years ago

The art museum had a total of 22for visitors on Tuesday. Visitors older than 18 paid $12 for admission. Visitors 18 years or you

Firlakuza [10]

Answer:

42 im pretty sure

Step-by-step explanation:

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3 years ago

If 2/3 of the distance from the y-axis to point A (−30, −45) is equal to 1/4 of the distance from the x-axis to point B(a, a), w

Yuliya22 [10]

The value of a is 80

Step-by-step explanation:

The distance of a point $(x_0,y_0)$ from the y-axis can be written as

$d_y = |x_0|$

because the x-coordinate of the y-axis is zero.

Similarly, the distance of a point $(x_0,y_0)$ from the x-axis can be written as

$d_x=|y_0|$

Since the y-coordinate of the x-axis is zero.

In this problem:

- The distance of the point A (−30, −45) from the y-axis can be written as

$d_A = |-30|=30$

- The distance of point B (a,a) from the x-axis can be written as

$d_B = |a| = a$

Since $a>0$ .

We are told that 2/3 of the distance from the y-axis to point A (−30, −45) is equal to 1/4 of the distance from the x-axis to point B(a, a), which means

$\frac{2}{3}d_A = \frac{1}{4}d_B$

Therefore,

$\frac{2}{3}(30)=\frac{1}{4}a$

And solving for a,

$20=\frac{1}{4}a\\a=80$

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brainly.com/question/3969582

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3 years ago

1+1=hfgdvfhftftfrgrggg

Margarita [4]

Answer:

hfgdvfhftftfrgrggg equals 2

Step-by-step explanation:

Imagine you have one apple on your left hand. Your friend just gave your another apple. Now, you have two apples. So, the hfgdvfhftftfrgrggg would be 2.

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3 years ago

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A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midni

Alex17521 [72]

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, $\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13$

p = 0.20

So, hypothesis:

$H_0:p=\hat{p}\\\\H_a:\hat{p}$

so, test statistic value would be

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604$

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

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3 years ago

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