Find the value of x.

Luke has a calculator that will only display number less than or equal to 999,999,999. Which of the following products will his

liq [111]

Answer:

≥ 999,999,999

Step-by-step explanation:

that is the less than or equal to sign

hope this helps

8 0

2 years ago

Find the exact value of T Image below

Elanso [62]

Answer:

a) 1/800 or 0.00125

b) i) 0.0013

ii) 0.001

c) 60%

Step-by-step explanation:

T = [tan(2×30)+1][2cos(30)-1] ÷ (y²-x²)

T = (tan60 + 1)(2cos30 - 1) ÷ (41² - 9²)

T = (sqrt(3) + 1)(sqrt(3) - 1) ÷ 1600

T = (3-1)/1600

T = 2/1600

T = 1/800

T = 0.00125

Error: 0.002 - 0.00125

0.00075

%error

0.00075/0.00125 × 100

60%

4 0

3 years ago

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago

Please help my finale math exam is do to night

Misha Larkins [42]

Answer:

Dimension volume

6 cups (1.5 L) 8 x 2 inch (20 x 5 cm) Round Cake Pan

6 cups (1.5 L) 9 x 1.5 inch (23 x 4 cm) Round Cake Pan

6 cups (1.5 L) 8 x 8 x 1.5 inch (20 x 20 x 4 cm) Square Cake Pan

8 cups (2 L) 9 x 2 inch (23 x 5 cm) Round Cake Pan

here are four outcomes that sum to seven out of 24 possible outcomes. Therefore the probability of values of four sided die and six sided

Step-by-step explanation:

4 0

2 years ago

The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3

RSB [31]

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability of a defective calculator is 10 percent.

This means that $p = 0.1$

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243$

There is a 24.3% probability that one of the calculators will be defective.

3 0

3 years ago