This season, Ramon played 14 soccer games. Sofia played 12 soccer games. Which

Let x=number of cds and y=number of dvds

Elis [28]

X + y = 40.....multiply by -4
4x + 6y = 180
-------------------
-4x - 4y = -160 (result of multiplying by -4)
4x + 6y = 180
------------------add
2y = 20
y = 20/2
y = 10 <===

x = 30

6 0

3 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

$Number = 0.4 * 150$

$Number = 60$

Hence, 60 seats were occupied by concessionary ticket holders.

5 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95