Domain: Range: Increasing: Decreasing:

What does 8 3/4 equal to as a improper fraction

Paul [167]

Answer:

35/4

Step-by-step explanation:

6 0

2 years ago

The average weight of a basketball is 21 1/8 ounces.the average weight of a baseball is 5 2/8 ounces. How many more ounces does

s344n2d4d5 [400]

Answer:

26 3/8

Step-by-step explanation:

5 0

3 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

2 years ago

What is the area of this figure? Enter your answer in the box.

alisha [4.7K]

The answer is 28 I think

7 0

2 years ago

Read 2 more answers

A professor receives, on average, 24.7 emails from students the day before the midterm exam. To compute the probability of recei

MaRussiya [10]

Answer:

Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is $X \sim Poisson(\lambda=24.7)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

The best answer for this case would be:

C. Poisson distribution

Step-by-step explanation:

Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is $X \sim Poisson(\lambda=24.7)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$