A motorist drove from town P to town Q, a distance of 80 km, in 30 minutes . What is his average speed?

An experiment consists of first rolling a die and then tossing a coin: A. How many elements are there in the sample space? B. Le

stepladder [879]

Answer:

A) There are 12 elements in the sample space.

B) The required probability is 1/4 or 0.25.

C) Yes, the events are mutually exclusive.

Step-by-step explanation:

Consider the provided information.

Part(A)

An experiment consists of first rolling a die and then tossing a coin.

Thus the sample space is:

(1,H), (2,H) (3,H) (4,H) (5,H), (6,H)

(1,T), (2,T) (3,T) (4,T) (5,T), (6,T)

Hence, there are 12 elements in the sample space.

Part(B) Let A be the event that either a 2, 3 or 4 is rolled first, followed by landing a head on the coin toss.

Now consider the above sample space. There are 3 possible case in which 2, 3 or 4 is rolled first, followed by landing a head on the coin toss.

i.e (2,H) (3,H) (4,H)

The required probability is: P(A)= $\frac{3}{12}=\frac{1}{4}=0.25$

Hence, the required probability is 1/4 or 0.25.

Part(C)

Mutually exclusive, means that they cannot occur at the same time.

From part (B) Event A = (2,H) (3,H) (4,H)

Let B be the event that a number less than 2 is rolled, followed by landing a head on the coin toss.

Thus, the Event B = (1,H).

Therefore, Event A and B can't be occur at the same time as P(A and B)=0.

Hence, yes, the events are mutually exclusive.

7 0

3 years ago

(15-26)/3 + (-8+19)/2

Gre4nikov [31]

Answer:

11/6

Step-by-step explanation:

(15-26)/3 + (-8+19)/2

Do parentheses:

-11/3 + 11/2

Find the common denominator:

-11(2)/3(2) + 11(3)/2(3)

Multiply:

-22/6 + 33/6

Add the fractions:

11/6

7 0

3 years ago

Read 2 more answers

The sum of two binomials is 7x 2 - 15. If one of the binomials is 4x 2-8. what is the other binomial?

SashulF [63]

Answer:

c

Step-by-step explanation:

6 0

3 years ago

If sin θ = 0.55 what is the value of cos θ? How do I solve this?

givi [52]

0.55^2 + cos^2 = 1
cos^2 = 1 - 0.55^2
cos^2 = 1 - 0.3025
cos^2 = 0.6975
cos = 0.8351

4 0

3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, a

Masja [62]

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

= [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50 {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

So, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

8 0

3 years ago

A motorist drove from town P to town Q, a distance of 80 km, in 30 minutes . What is his average speed?​

A motorist drove from town P to town Q, a distance of 80 km, in 30 minutes . What is his average speed?