The original price of the ring is <span>£</span>1400
Answer:
a. 9.5x + 6.5(x+c) < 8 when c>0
b. Must be one child more than the no. of adults.
Step-by-step explanation:
For Cinema 1:
for adult = $9.50
for child = $6.50
For Cinema 2:
Per person regardless of age = $8.00
First of all, we will find out the condition when per person rates in both cinema are equal.
Assume x = no. of adults
y = no. of children
Rate per person in Cinema I = Rate per person in Cinema II
(9.5x + 6.5y)/(x+y) = 8
9.5x + 6.5y = 8(x+y)
9.5x + 6.5y = 8x + 8y
9.5x-8x = 8y-6.5y
=> x = y
So rates are equal when no. of adults equals no. of children
For Cinema I to have better rates, no. of children should be atleast 1 more than the no. of adult. In this way the rate per person of Cinema I will be less than 8
Hence we form an inequality when y = x+c and c > 0
9.5x + 6.5(x+c) < 8 when c>0
Hence there must be 1 more children than the no. of adults attending Cinema I for it to be a better deal.
Assuming the coin is not weighted and is a fair and standard coin - the chance of flipping head is 1/2. You can either flip head or tails, there are no other possible outcomes.
Step-by-step explanation:
Using Binomial Expansion,
(x + y)³
= 3C0 * x³ + 3C1 * x²y + 3C2 * xy² + 3C3 * y³.
Therefore the coefficient of xy² is 3C2 = 3.
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have 
, 
, 
and 
, 
, 
. The pooled estimate is given by
a. We want to test 
vs 
(two-tailed alternative).
The test statistic is 
and the observed value is 
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value 
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by 
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where 
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)
