At a back to school sale, backpacks are marked down 15%. If the original price was $34, what is the sale price?

1 answer:

3 0

There are two ways to do this problem the first: multiply the original price by the percentage marked down ($34x.15=5.10) then subtract 5.10 from 34 and that gives you $28.90. The second way: subtract 15% from 100% (85%) and multiply $34 by .85 which will give you $28.90. :) whichever is easiest :)

You might be interested in

test worth 70 points; test score of 90% write a proportion to find how many points a student needs to earn on the test to get th

GaryK [48]

70/70 = 100%
35/70 =50%
63/70=90%

6 0

3 years ago

ashton is saving money to buy a new bike. he needs $120 but has only saved 60% so far. How much more money does Ashton need to b

Tasya [4]

Ashton has saved: 60% of $120=(60/100)*$120=(60*$120)/100=$72
Ashton need to buy the bike=$120-$72=$48

He needs to buy the bike=$48

5 0

3 years ago

The Ohio Department of Agriculture tested 203 fuel samples across the state

Rus_ich [418]

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: