Ask question

Ask question

All categories

Katyanochek1 [597]

4 years ago

12

assume that when adults with smartphones are randomly selected, 45% use them in meetings or classes. If 6 adult smartphone users

are randomly selected, find the probability that exactly 3 of them use their smartphones in meetins or classes.

1 answer:

Ilya [14]4 years ago

6 0

Answer:

0.3032

Step-by-step explanation:

Use binomial probability.

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

P = ₆C₃ (0.45)³ (0.55)³

P = 0.3032

You might be interested in

me.Walker needs 22 cups of flour to bake 12 cakes.how many cups of. flour will she need to bake 30 cakes?

lara [203]

We can infer that more cups of flour are needed as the cakes are increased, so we can apply a rule of three, simple with direct proportion to find the answer:
12 cakes ----> 22 cups flour
30 cakes ----> x

x = (30)(22)/12
x = 55
therefore 55 cups of flour are needed to bake 30 cakes

5 0

4 years ago

!HELP! SHOW WORK PLEASE SO I CAN UDNERSTAND!

Nata [24]

Ok.

The volume of a cone is the volume of a cylinder over 3.

The volume of a cone is:
$\frac{1}{3} \pi r^{2}h$

This problem gives us the diameter, which is twice of the radius. So, we'll just half it.
Radius (me) = 4 cm.

Substitute our given values into the equation.
$\frac{1}{3} \pi (4)^{2}h \ or \ \frac{1}{3} \pi (4)^{2}(10) \ or \ \frac{1}{3} \pi (16)(10)$

Multiply
$\frac{1}{3} \pi(160) \ or \ \frac{1}{3} (3.14)(160) \ or \ 1.047(160)$

Multiply
$167.52 \ cm^{3}$

This would make $167 \ cm^{3}$ the closest option.

4 0

4 years ago

Read 2 more answers

To create a confidence interval from a bootstrap distribution using percentiles, we keep the middle values and chop off a certai

Nataliya [291]

Answer:

a

$\frac{\alpha }{2} = 2.5 \%$

b

$N = 25$

Step-by-step explanation:

From the question we are told that

The number of bootstrap samples is n = 1000

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally the percentage of values that must be chopped off from each tail for a 95% confidence interval is mathematically evaluated as

$\frac{\alpha }{2} = \frac{0.05}{2} = 0.025 = 2.5 \%$

=> $\frac{\alpha }{2} = 2.5 \%$

Generally the number of the bootstrap sample that must be chopped off to produce a 95% confidence interval is

$N = 1000 * \frac{\alpha }{2}$

=> $N = 1000 * 0.025$

=> $N = 25$

4 0

4 years ago

Which inequality correctly compares two of these times? A) 54.05 > 54.15 B) 54.5 < 54.1 C) 54.15 > 54.05 D) 54.1 < 5

koban [17]

c is right answer because 54.15 is bigger than 54.05

3 0

3 years ago

The graph of y = 4x2 - 4x – 1 is shown.

11Alexandr11 [23.1K]

Answer:

i) The approximate solutions are: $x_{1} \approx -0.207$ , $x_{2} \approx 1.207$ .

ii) The approximate solutions are: $x_{1}\approx -0.5$ , $x_{2} \approx 1.5$ .

Step-by-step explanation:

i) The best approach to estimate graphically the solution of $4\cdot x^{2} - 4\cdot x - 1 = 0$ is graphing the following system of equations:

$y = 4\cdot x^{2}-4\cdot x - 1$ (1)

$y = 0$ (2)

And labeling the points in which both intersects each other. We include the result in the image 'solution-i'. The approximate solutions are: $x_{1} \approx -0.207$ , $x_{2} \approx 1.207$ .

ii) The best approach to estimate graphically the solution of $4\cdot x^{2} - 4\cdot x - 1 = 2$ is graphing the following system of equations:

$y = 4\cdot x^{2}-4\cdot x - 1$ (1)

$y = 2$ (2)

And labeling the points in which both intersects each other. We include the result in the image 'solution-ii'. The approximate solutions are: $x_{1}\approx -0.5$ , $x_{2} \approx 1.5$ .

5 0

4 years ago