If 4 pounds of potatoes is 3 dollars, how many dollars is 1 pound of potatoes?

Do you need to ask a question, but don't have points.....

horsena [70]

Yes, felt that. thanks and have a lovely day!:D

8 0

2 years ago

A cinema can hold 270 people at one performance 5/9 of the seats were occupied of the occupied seats 40% we occupied by concessi

Elza [17]

Question:

A cinema can hold 270 people at one performance 5/9 of the seats were occupied of the occupied seats 40% we occupied by concessionary ticket holders.

What is the number of seats occupied by concessionary ticket holders?.

Answer:

60 seats

Step-by-step explanation:

Given

Number of seats = 270

Occupied Seats = 5/9

Concessionary ticket holders = 40% of occupied Seats

Required

The number of seats occupied by concessionary ticket holders

First the number of occupied seat has to be calculated.

$Occupied\ Seats = \frac{5}{9} * 270$

$Occupied\ Seats = \frac{1350}{9}$

$Occupied\ Seats = 150$

Next is to determine the number of seats occupied by concessionary ticket holders.

$Number = 40\%\ of\ occupied\ seats$

$Number = 40\%\ of\ 150$

Convert percentage to decimal

$Number = 0.4 * 150$

$Number = 60$

<em>Hence, 60 seats were occupied by concessionary ticket holders.</em>

5 0

2 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.