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3 years ago

The cost of a whole box of chips are $60. I have $48. What percentage of the total cost do I have?

Mathematics

1 answer:

goblinko [34]3 years ago

4 0

Answer:

Percentage of total I have = 80%

Step-by-step explanation:

Given:

Price of Box = $60

I Have = $48

To Find:

Percentage I have = ?

Solution:

In this question we are just supposed to find out the percentage of the Value we have so

Percentage can be found by the formula

Percentage $=\frac{Given Value}{Total}*100$ %

Now we have the values of all

So putting in the values

Percentage $=\frac{48}{60}*100$ %

$=\frac{4800}{60}$ %

= 80%

So the Percentage of total cost I have is 80%

Percentage of total = 80%

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3 years ago

Find an equation for the line with the given properties. Perpendicular to the line 7x - 3y = 68; containing the point (8, -8)

Sever21 [200]

Answer:

$y=\dfrac{-3}{7}x-\dfrac{32}{7}$

Step-by-step explanation:

Given that,

A line 7x - 3y = 68 and containing the point (8, -8).

The equation can be written as :

$-3y=68-7x\\\\y=\dfrac{68}{-3}+\dfrac{7x}{3}\\\\y=\dfrac{7x}{3}+(\dfrac{-68}{3})$

The slope is :7/3

Line is perpendicular so use m = –3/7

$-8=(-\dfrac{3}{7})\times 8+b\\\\-8+\dfrac{3}{7}\times 8=b\\\\b=\dfrac{-32}{7}$

So, required equation is :

$y=\dfrac{-3}{7}x-\dfrac{32}{7}$

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3 years ago

Every day, Luann walks to the bus stop and the amount of time she will have to wait for the bus is between 0 and 12 minutes, wit

Nana76 [90]

Answer:

a. 341.902.

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.

n instances of a normal variable:

For n instances of a normal variable, the mean is $n\mu$ and the standard deviation is $s = \sigma\sqrt{n}$