Find −a2 − 3b3 + c2 + 2b3 − c2 if a = 3, b = 2, and c = −3.

Two answers are possible.
One is with the repetition of digits.
The other is without repetition of digits.
ANSWER 1: With Repetition
Numbers: 1,2,4,5,7,8
There are 4 digits __ __ __ __
1st digit can be filled by any of the 6 numbers.
Similarly 2nd, 3rd and 4th digits.
Hence each digit will have 6 possibilities.
Therefore no.of 4 digit numbers that can be formed are 6 x 6 x 6 x 6 = 1296
ANSWER 2: Without Repetition
4 digits __ __ __ __
Now the first digit can be filled by any of the six numbers. Therefore there are 6 possibilities
The second digit will have only 5 possibilities as one of the numbers gets used up by the 1st digit.
Similarly the 3rd and 4th digits will have 4 and 3 possibilities respectively.
Therefore no.of 4 digit numbers that can be formed are 6 x 5 x 4 x 3 = 360

8 0

3 years ago

True / False: ANOVA, Part I. Determine if the following statements are true or false in ANOVA, and explain your reasoning for st

Juliette [100K]

The answer is B hope that helps

7 0

3 years ago

A professional basketball team made 37.9% of its three-point field goals in one season. If 80 three-point field goal attempts ar

kompoz [17]

Answer:

The probability that more than 35 were made is 0.14686.

Step-by-step explanation:

We are given that a professional basketball team made 37.9% of its three-point field goals in one season.

80 three-point field goal attempts are randomly selected from the season.

Let $\hat p$ = <em>sample proportion of three-point field goals made in one season.</em>

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

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

where, p = population proprotion of three-point field goals = 37.9% = 0.379

n = sample of three-point field goal attempts = 80

$\hat p$ = sample proportion of three-point field goals = $\frac{35}{80}$ = 0.4375

Now, if 80 three-point field goal attempts are randomly selected from the season, the probability that more than 35 were made is given by = P( $\hat p$ > 0.4375)

P( $\hat p$ > 0.4375) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ > $\frac{0.4375-0.379}{\sqrt{\frac{0.4375(1-0.4375)}{80}} }$ ) = P(Z > 1.05) = 1 - P(Z $\leq$ 1.05)

= 1 - 0.85314 = <u>0.14686</u>

The above probability is calculated by looking at the value of x = 1.05 in the z table which has an area of 0.85314.

4 0

4 years ago