All categories

3 years ago

A store marks up a laptop 50% from its original price of $299. What is the new price for the computer?

Mathematics

1 answer:

ira [324]3 years ago

6 0

Answer:

$448.50

Step-by-step explanation:

"Marking up" an item by x% means to increase its price by x% of the original price.

Here, when marking up by 50%, you would multiply the original price by 1.5:

1.5*299 = $448.50

You wouldn't just multiply by .5 because that would only give you 50% of $299. You could multiply .5*299 and get $149.5, but then you'd have to add it to $299, which would get you $448.50 again!

The quicker way is to just multiply the original price by 1+x%.

Hope this helps!

You might be interested in

Help! It's geometry. :) 35 points.

Masja [62]

Answer:

g = 21

h = 8

Step-by-step explanation:

3g + 5 + 6g - 14 = 180 (since they are adjacent angles and will equal 180)

Simplify to get g = 21.

Substitute 21 for g in 3g + 5 and set that equal to 9h - 4 and solve for h to get 8 (since they are opposite angles and will equal each other)

8 0

3 years ago

Read 2 more answers

What does |-9.23-1.79| - |7.34-2.57| equal?

liubo4ka [24]

Answer:

11.02-9.91= 1.11

Step-by-step explanation:

|-11.02| - |-9.91| (First solve in absolute)

11.02 - 9.91 (any values outside absolute becomes positive)

1.11

Step-by-step explanation:

4 0

3 years ago

What is the range of y =5/2(x-2)² - 4

blagie [28]

Answer:

vertex:(2,-4)

Focus:(2,- $\frac{39}{10}$ )

Axis of symmetry :x=2

Directrixt: y=- $\frac{41}{10}$

Step-by-step explanation:

7 0

4 years ago

Let g(x)=Intragal from 0 to x f(t) dt, where r is the function whos graph is shown.

leonid [27]

$\displaystyle g(x) = \int_0^x f(t) \, dt$

then g(x) gives the signed area under f(x) over a given interval starting at 0.

In particular,

$\displaystyle g(0) = \int_0^0 f(t) \, dt = 0$

since the integral of any function over a single point is zero;

$\displaystyle g(4) = \int_0^4 f(t) \, dt = 8$

since the area under f(x) over the interval [0, 4] is a right triangle with length and height 4, hence area 1/2 • 4 • 4 = 8;

$\displaystyle g(8) = \int_0^8 f(t) \, dt = 0$

since the area over [4, 8] is the same as the area over [0, 4], but on the opposite side of the t-axis;

$\displaystyle g(12) = \int_0^{12} f(t) \, dt = -8$

since the area over [8, 12] is the same as over [4, 8], but doesn't get canceled;

$\displaystyle g(16) = \int_0^{16} f(t) \, dt = 0$

since the area over [12, 16] is the same as over [0, 4], and all together these four triangle areas cancel to zero;

$\displaystyle g(20) = \int_0^{20} f(t) \, dt = 24$

since the area over [16, 20] is a trapezoid with "bases" 4 and 8, and "height" 4, hence area (4 + 8)/2 • 4 = 24;

$\displaystyle g(24) = \int_0^{24} f(t) \, dt = 64$

since the area over [20, 24] is yet another trapezoid, but with bases 8 and 12, and height 4, hence area (8 + 12)/2 • 4 = 40, which we add to the previous area.

5 0

3 years ago

I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an

Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

$\mu = 0 - 5 = -5$

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

$\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}$

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1 - (-5)}{\sqrt{13}}$

$Z = \frac{6}{\sqrt{13}}$

$Z = 1.66$

$Z = 1.66$ has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

5 0

3 years ago