What is 9/16 more than 11/12

Sveta_85 [38]

The answer to 20+1

=21

6 0

3 years ago

You are at a restaurant and the check comes to a total of $20. If you want to leave a 20% tip, how much total money should you p

grandymaker [24]

Answer:

the total tip should be $24.00

7 0

3 years ago

An airline promotion to business travelers is based on the assumption that two-thirds of business travelers use a laptop compute

skad [1K]

Answer:

a

The null hypothesis is $H_o : p = 0.67$

The alternative hypothesis is $H_a : p \ne 0.67$

b

The sample proportion is $\^ p = 0.65$

Step-by-step explanation:

From the question we are told that

The sample size is n = 546

The number of people that use a laptop overnight is $k = 355$

The population proportion is $p = \frac{2}{3} = 0.67$

Generally

The null hypothesis is $H_o : p = 0.67$

The alternative hypothesis is $H_a : p \ne 0.67$

Gnerally the sample proportion is mathematically represented as

$\^ p = \frac{ k}{n}$

=> $\^ p = \frac{ 355}{546}$

=> $\^ p = 0.65$

4 0

3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

Inessa05 [86]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

5 0

3 years ago

A study of all the students at a small college showed a mean age of 20.9 and a standard deviation of 2.9 years. These numbers ar

Ganezh [65]

Answer:

These numbers are - parameters.

Step-by-step explanation:

A study of all the students at a small college showed a mean age of 20.9 and a standard deviation of 2.9 years.

These numbers are - parameters.

These numbers are parameters as they are for all the students, and merely not a sample.

7 0

3 years ago