Ask question

Ask question

All categories

4 years ago

11

Your name is included in a list of 6 names. One name is selected at random. Write the probability, as a decimal, that it is your

name.

1 answer:

zvonat [6]4 years ago

6 0

The probability of my name being chosen is 1/6. As a decimal, that's 0.16666...

You might be interested in

Write and solve an equation for the total number of possible combinations from flipping a coin 10 times.

vladimir2022 [97]

Step-by-step explanation:

every flip has 2 possible outcomes.

1 flip has 2.

2 flips have 2×2 = 2² = 4

3 flips have 2×2×2 = 2³ = 8

and so on.

the number of possible combinations flipping a coin n times is

C(n) = 2^n

for 10 flips

C(10) = 2¹⁰ = 1024

possible outcomes or combinations.

5 0

1 year ago

Read 2 more answers

Is the quotient of two rational numbers always a rational number? Explain.

Natasha_Volkova [10]

Answer:

Yes,

Step-by-step explananation

The quotient of two rational numbers is always rational, and the reason for this lies in the fact that the product of two integers is always an rational number.

6 0

4 years ago

Which of the following is the equation of a line in a slope= 5 and y-intercept at (0,-3)?

asambeis [7]

Answer:y=5x-3

Step-by-step explanation:

7 0

3 years ago

Elizabeth likes to trek with her friends during the weekend, which is also a great way to socialize and stay fit. What type is t

tresset_1 [31]

Answer:

D. a complex sentence

Step-by-step explanation:

A. is incorrect because it does not include a conjunction to connect the two sentences.

B. is incorrect because the sentence is not asking a question.

C. is incorrect because the sentence includes two sentences.

D. is correct because a complex sentence includes and independent clause(elizabeth likes to trek with her friends during the weekend) and a dependent clause( which is also a great way to socialize and stay fit)

6 0

3 years ago

Read 2 more answers

A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals

Sophie [7]

Answer:

a) $w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

b) $202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case $\epsilon = \pm 1$ since that's the tolerance 1C

d) $\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smalles value on this case $\delta =0.113$

e) For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

$T(w)= 0.1 w^2 +2.156 w +20$

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

$203= 0.1w^2 +2.156 w +20$

And we can rewrite the expression like this:

$0.1w^2 +2.156 w-183=$

And we can solve this using the quadratic formula given by:

$w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a =0.1, b =2.156 and c=-183. If we replace we got:

$w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

$202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case $\epsilon = \pm 1$ since that's the tolerance 1C

Part d

For this case we can do this:

$\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smallest value on this case $\delta =0.113$

Part e

For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

8 0

4 years ago