What is the 12th term of the sequence? 3, −9, 27, −81, 243, ...

We are told that the two players, taken together, scored a total of 49 points. What is the probability that player A scored firs

mario62 [17]

Answer:

1/2 or 50% probability

Step-by-step explanation:

There are just two players, player A and player B. Since it is between this two players, for player A to score first we would have:

AB where player A is the first player to score, we wouldn't have BA since player would not be the first here

Probability = number of favorable outcomes/total number of outcomes

Therefore probability of A being the first player = number of A/total number of players

=1/2 = 0.50

There is therefore a 50% chance of A being the first player

7 0

1 year ago

Cameron buys 2.45 pounds of apples and 1.65 pounds of pears. Apples and pears each cost c dollars per pound. If the total cost a

Leto [7]

Answer:

2.45c + 1.65c = 4.12 + 0.75

Step-by-step explanation:

To write an equation to find the value for c, we need to declare what c is first.

c = price of fruit

2.45c + 1.65c = 4.12 + 0.75

Now we multiplied c to 2.45 and 1.65 and added them together, because whatever the value of c is will give us the equivalence of the sum of 4.12 + 0.75.

Now to check if the equation is right, let's solve for c.

2.45c + 1.65c = 4.12 + 0.75

4.1c = 4.87

Now to get the value of c, we divide both sides of the equation by 4.1.

$\dfrac{4.1c}{4.1}=\dfrac{4.87}{4.1}$

c = 1.19

Now let's substitute the value of c in the equation to see if we got it right.

2.45(1.19) + 1.65(1.19) = 4.12 + 0.75

2.92 + 1.96 = 4.87

4.87 = 4.87

Therefore concluding that the value of c is 1.19.

7 0

2 years ago

The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its preci

Anni [7]

Answer:

The probability that a defective ball bearing was manufactured on a Friday = 0.375

Step-by-step explanation:

Let the event of making a mistake = M

The event of making a precision ball bearing production on Monday = Mo

The event of making a precision ball bearing production on Tuesday = T

The event of making a precision ball bearing production on Wednesday = W

The event of making a precision ball bearing production on Thursday = Th

The event of making a precision ball bearing production on Friday = F

the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, P(M|T) = 4% = 0.04

4% on Wednesday, P(M|W) = 0.04

4% on Thursday, P(M|Th) = 0.04

8% on Monday, P(M|Mo) = 0.08

and 12% on Friday = P(M|F) = 0.12

The Company manufactures an equal amount of ball bearings (20 %) on each weekday, Hence, the probability that a random precision ball bearing was made on a particular day of the week, is mostly the same for all the five working days.

P(Mo) = 0.20

P(T) = 0.20

P(W) = 0.20

P(Th) = 0.20

P(F) 0.20

The probability that a defective ball bearing was manufactured on a Friday = P(F|M)

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F)

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

We can obtain each of these probabilities by using the expression for conditional probability.

P(Mo n M) = P(M|Mo) × P(Mo) = 0.08 × 0.20 = 0.016

P(T n M) = P(M|T) × P(T) = 0.04 × 0.20 = 0.008

P(W n M) = P(M|W) × P(W) = 0.04 × 0.20 = 0.008

P(Th n M) = P(M|Th) × P(Th) = 0.04 × 0.20 = 0.008

P(F n M) = P(M|F) × P(F) = 0.12 × 0.20 = 0.024

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

P(M) = 0.016 + 0.008 + 0.008 + 0 008 + 0.024 = 0.064

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F) = 0.024

P(M) = 0.064

P(F|M) = P(F n M) ÷ P(M) = (0.024/0.064) = 0.375

Hope this Helps!

4 0

2 years ago

For each exercise draw circle O with radius 12. Then draw radii OA and OB to form an angle with the measure named. Find the leng

guajiro [1.7K]

You are asked to draw circle O with radius 12. Then draw radii OA and OB to form an angle with the measure named. You are asked to find the length of AB. The answers for each of the following measures are:
1)Measure of AOB=90, AB = 16.97 units
2)measure of AOB=180, AB = 24 units
3) measure of AOB=60, AB = 10.39 units
3) measure of AOB=120, AB = 10.39 units

5 0

2 years ago

The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou

lions [1.4K]

Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = the amount spent on daily transportation

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = average annual amount American households spend on daily transportation = $6,312

$\sigma$ = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

P(X < $1,000) = 0.05

P( $\frac{X-\mu}{\sigma}$ < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

P(Z < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

$\frac{\$1000-\$6312}{\sigma}=-1.645$

$\sigma=\frac{-\$5312}{-1.645}$ = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

P($4000 < X < $6000) = P(X < $6000) - P(X $\leq$ $4000)

P(X < $6000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\$6000-\$6312}{\$3229.18}$ ) = P(Z < -0.09) = 1 - P(Z $\leq$ 0.09)

= 1 - 0.5359 = 0.4641

P(X $\leq$ $4000) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{\$4000-\$6312}{\$3229.18}$ ) = P(Z $\leq$ -0.72) = 1 - P(Z < 0.72)

= 1 - 0.7642 = 0.2358

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

P(X > x) = 0.03 {where x is the required range}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-\$6312}{3229.18}$ ) = 0.03

P(Z > $\frac{x-\$6312}{3229.18}$ ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

$\frac{x-\$6312}{3229.18}=1.88$

${x-\$6312}=1.88\times 3229.18$

x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

8 0

2 years ago