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Mars2501 [29]
3 years ago
13

an item is purchased for a wholesale price of $48 and will be sold at a 55 percent markup. which expression should be used to fi

nd the amount of the markup
Mathematics
2 answers:
Xelga [282]3 years ago
8 0

Answer:

$48(0.55)

Step-by-step explanation:

Reil [10]3 years ago
7 0
48 times 1.55 equals 74.4.
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Someone answer quick please for brainliest !
Rina8888 [55]

Answer:

The equation of the line is

y =  -  \frac{1}{2} x +  \frac{5}{2}

Step-by-step explanation:

Equation of a line is

y = mx + c

Where m is the slope

c is the y intercept

y = 2x + 3

Comparing with the above formula

m is 2

Since the lines are perpendicular the slope of the other line is the negative inverse of the original line .

That's

m = - 1/2

Equation of the line using point (1,2) and slope - 1/2 is

y - 2 = -1/2(x - 1)

y - 2 = -1/2x + 1/2

y = -1/2x + 1/2 + 2

The final answer is

y =   - \frac{1}{2} x +  \frac{5}{2}

Hope this helps you.

8 0
3 years ago
Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
Find the GCF for the pair 4, 49. ​
IgorC [24]
I believe the GCF for the park 4 and 49 is 1.
6 0
3 years ago
Does the graph represent a function?
AlekseyPX

Answer:

no it does not you can identify that by doing the vertical line test

Step-by-step explanation:

7 0
3 years ago
SOLVE<br> r(s 2) (t) when r = 2, s = 3, and t = 5 <br><br><br> pls help
Keith_Richards [23]

Answer:

320 or 40 sorry if I´m wrong

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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