Two distinct solid fuel propellants, type a and type b, are being considered for a space program activity. burning rates of the propellant are crucial. random samples of 12 specimens of propellant a are taken with sample means of 84 cm/sec with a standard deviation of 4. and, random samples of 18 specimens of propellant b are taken with sample means of 77 cm/sec with a standard deviation of 6. find a 99% confidence interval for the difference between the average burning rates for the two propellants. assume the populations to be approximately normally distributed with equal variances.

C.I. =  (2.297, 11.703)

Step-by-step explanation:

The t-statistic for difference of mean is given by,

Here, = 84

= 7

s₁ = 4

n₁ = 12

s₂ = 6

n₂ = 18

Substituting all value in formula,

We get, t = -3.541 at 28 degree of freedom.

Using this formula, we get, t = 1.5342

Therefore, based on the data provided, the 99% confidence interval for the difference between the population means is: 2.297 < < 11.703

which indicates that we are 99% confident that the true difference between population means is contained by the interval (2.297, 11.703)

your answer would be 10x - 1

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5