I need help with my assignment but dont know how to do it/ what should i do?

Mathematics

1 answer:

ohaa [14]2 years ago

5 0

Answer:

first explain what you need help with specifically then maybe people can help, to bland of a question

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If four pounds of a certain candy cost $10 how manu pounds of that candy can be purchased for $12

dangina [55]

4.8 pounds, set it up as 4 over ten times x over 12, 4 times 12 divided by ten equals 4.8

8 0

2 years ago

A bike store sells scooters at a 54% markup. If the store bought each scooter for $29.25, what is the resale price to the neares

7nadin3 [17]

The purchasing price of the store = $29.25.

They want to resale it at 54% markup of the purchasing price.

54% of 29.25 = 0.54 × 29.25.

On multiplying 0.54 and 29.25, we get 15.795.

Resale price = The purchasing price + Markup

= 29.25 + 15.795.

On adding 29.25 and 15.795, we get 45.045.

We can round it to 45 to the nearest dollar.

<h3>Therefore, the resale price to the nearest dollar is $45.</h3>

5 0

3 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.