Consider the figure.

Can someone solve this its probably easy but i dont know

Rzqust [24]

Answer:

Step-by-step explanation:

A={26,30,32,44,48}

B={22,26,34}

A∩B=26

number of elements=1

elements in ξ are {22,24,26,28,30,32,34,36,38,40,42,44,46,48}

number of elements in ξ=14

P(A∩B)=1/14

I can't draw venn diagram but i give hint

A={30,32,44,48}+{26}

B={22,34}+{26}

A∩B={26}

6 0

3 years ago

A population of values has a normal distribution with μ = 155.4 and σ = 49.5 . You intend to draw a random sample of size n = 24

xz_007 [3.2K]

Answer:

(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.

(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.

Step-by-step explanation:

Let the random variable X follow a Normal distribution with parameters μ = 155.4 and σ = 49.5.

(a)

Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:

$P(158.6 < X$

*Use a standard normal table.

Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.

(b)

A sample of n = 246 is selected.

Compute the probability that a sample mean is between 158.6 and 159.2 as follows:

$P(158.6 < \bar X$

*Use a standard normal table.

Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.

4 0

3 years ago

Read 2 more answers

This is a Differential Equations problem, I only need help on part 1 from question five. I need steps as well, thank you.

jolli1 [7]

Answer:

a) $y(t) = \sqrt{2t + 1 }$

b) 1.5 hours after thickness will be 2 inches.

Step-by-step explanation:

$\frac{dy}{dt} = \frac{1}{y} \\ \\ ydy = dt \\ \\ integrating \: both \: sides \\ \\ \int y \: dy = \int 1 \: dt \\ \\ \frac{ {y}^{2} }{2} = t + c \\ \\ {y}^{2} = 2t + 2c \\ \\ y (t)= \sqrt{2t + 2c} ...(1) \\ \\ a) \: \: \: plug \: t = 0 \: in \: (1) \\ \\ y (0)= \sqrt{2(0) +2 c} \\ \\ y (0)= \sqrt{0 +2 c} \\ \\ 1 = \sqrt{2c} \: \: ( \because \: y (0)=1) \\ \\ 2c = 1 \: \implies \: c = \frac{1}{2} \\ \\ plug \: c = \frac{1}{2} \: in \: (1) \\ \\ y(t) = \sqrt{2t + 2 \times \frac{1}{2} } \\ \\ \huge \: \red {y(t) = \sqrt{2t + 1 } } \\ \\b) \: \: plug \: y(t) = 2 \: in \: above \: equation \\ \\ 2 = \sqrt{2t + 1} \\ \\ 4 = 2t + 1 \: \\ (squaring \: both \: sides) \\ \\ 4 - 1 = 2t \\ \\ 2t = 3 \\ \\ t = \frac{3}{2} \\ \\ t = 1.5 \: hours \\$