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3 years ago

Factor this expression: 12 + 20

Mathematics

1 answer:

liberstina [14]3 years ago

5 0

Answer:

Step-by-step explanation:

12 = 4 × 3

20 = 4 × 5

12 + 20 = 4 × 3 + 4 × 5 = 4 × (3 + 5)

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All employees at three stores of a large retail chain were asked to fill out a survey.a. Is it random b. Systematic c. Stratifie

aleksandrvk [35]

Answer:

d. Cluster

Step-by-step explanation:

Random: Random is asking a group of people from a population. For example, to estimate the proportion of Buffalo residents who are Bills fans, you ask 100 Buffalo residents and estimate to the entire population.

Systematic: Similar to random. For example, you want to estimate something about a population, and your sample is every 5th people you see on the street.

Cluster:Divides the population into groups, with geographic characteristics.. Each element is the groups is used. Suppose you want to study the voting choices of Buffalo Bills players. You can divide into offense, defense and special teams, and ask each player of these 3 groups.

Stratified: Done on a group of clusters, that is, from each cluster(group), a number of people are selected.

In this problem, we have that:

All employees at three stores of a large retail chain were asked to fill out a survey.

Divided by clusters(stores).

So the orrect answer is:

d. Cluster

5 0

3 years ago

At the district competition for 5th grade jump-roper of the year, the following number of jumps without stopping were recorded p

deff fn [24]

First, we arrange the data from least to greatest:
90, 100, 100, 200, 203, 224, 233, 237, 245, 257, 264, 265, 265, 266, 277, 279, 285, 288, 288, 290, 290, 291, 291, 300

There are 24 data all in all. The mean is calculated by dividing by 24 the sum of all the data.
mean = (5828)/24242.83

The median of the number of jumps is the mean of the 12th and 13th term.
265

5 0

2 years ago

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are cla

AnnyKZ [126]

Answer:

(a) Name: Multinomial distribution

Parameters: $p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

(b) Range: $\{(x,y,z)| x + y + z=20\}$

Parameters: $p_1 = 5\%$ $n = 20$

$(d)\ E(x) = 1$ $Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1) = 0.7359$

$(h)\ E(Y) = 17$

Step-by-step explanation:

Given

$p_1 = 5\%$

$p_2 = 85\%$

$p_3 = 10\%$

$n = 20$

$X \to$ High Slabs

$Y \to$ Medium Slabs

$Z \to$ Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

$X \to$ Number of voids considered as high slabs

$Y \to$ Number of voids considered as medium slabs

$Z \to$ Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

$p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

And the mass function is:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

$n = 20$

The number of voids (x, y and z) cannot be negative and they must be integers; So:

$x + y + z = n$

$x + y + z = 20$

Hence, the range is:

$\{(x,y,z)| x + y + z=20\}$

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

$p_1 = 5\%$ and $n = 20$

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

$p_1 = 5\%$ and $n = 20$

$E(x) = p_1* n$

$E(x) = 5\% * 20$

$E(x) = 1$

$Var(x) = E(x) * (1 - p_1)$

$Var(x) = 1 * (1 - 5\%)$

$Var(x) = 1 * 0.95$

$Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3)$

In (b), we have: $x + y + z = 20$

However, the given values of x in this question implies that:

$x + y + z = 1 + 17 + 3$

$x + y + z = 21$

Hence:

$P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P{X \le 1, Y = 17, Z = 3)$

This question implies that:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)$

Because

$0, 1 \le 1$ --- for x

In (e), we have:

$P(X = 1, Y = 17, Z = 3) = 0$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0$

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

In (a), we have:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

So:

$P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}$

Expand

$P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}$

Using a calculator, we have:

$P(X=0; Y=17; Z = 3) = 0.07195$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

$P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1)$

This implies that:

$P(X \le 1) = P(X = 0) + P(X = 1)$

In (c), we established that X is a binomial distribution with the following parameters:

$p_1 = 5\%$ $n = 20$

Such that:

$P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}$

So:

$P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}$

$P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}$

$P(X=0) = 1 * 1 * (95\%)^{20}$

$P(X=0) = 0.3585$

$P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}$

$P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}$

$P(X=1) = 0.3774$

So:

$P(X \le 1) = P(X = 0) + P(X = 1)$

$P(X \le 1) = 0.3585 + 0.3774$

$P(X \le 1) = 0.7359$

$(h)\ E(Y)$

Y has the following parameters

$p_2 = 85\%$ and $n = 20$

$E(Y) = p_2 * n$

$E(Y) = 85\% * 20$

$E(Y) = 17$

8 0

3 years ago

Given A = {x ∈ ℤ : 4|x} and B = {x ∈ ℤ : 2|x}. Prove A ⊆ B.

Rashid [163]

Let a ∈ A. Then a is some integer that is divisible by 4, so we can write a = 4k for some integer k.

We can simultaneously rewrite a as a = 2•2k, so 2 clearly divides a, which means a ∈ B as well.

Therefore A ⊆ B.

4 0

2 years ago

Can someone help ASAP I’ll give brainiest

olga nikolaevna [1]

Answer:

the second one is correct

Step-by-step explanation:

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7 0

2 years ago

Read 2 more answers