All categories

3 years ago

30-[11+{18-(48÷8×2)}+7]+15

Mathematics

1 answer:

Inessa05 [86]3 years ago

3 0

30-(11+(18-(48/8×2))+7)+15

= 30-(11+(18-12)+7)+15

= 30-(11+6+7)+15

= 30-24+15

= 21

You might be interested in

The Lee family buys a new car that has a sales price of $35,000. If the car dealer requires a minimum down payment of 15%, what

Klio2033 [76]

Answer:

Si un electrón no consume, ni libera energía el átomo se encuentra en

Step-by-step explanationSi un electrón no consume, ni libera energía el átomo se encuentra en:

6 0

3 years ago

Robin can mow a lawn in 3 hours, while Brady can mow the same lawn in 4 hours. How many hours would it take for them to mow the

Sauron [17]

The answer to your question is B.
Please mark as Brainiest Answer.

5 0

3 years ago

Read 2 more answers

A box contains 5 plain pencils and 5 pens. A second box contains 3 color pencils and 7 crayons. One item from each box is chosen

kotegsom [21]

The events 'select a plain pencil' and 'select a color pencil' are independent. therefore the probability of both occurring is the product of their probabilities:
$P(plain\ pencil\ and\ color\ pencil)=\frac{5}{10}\times\frac{3}{10}=\frac{3}{20}$
The answer is 3/20 or 0.15.

8 0

4 years ago

Listed below are amounts (in millions of dollars) collected from parking meters by a security company in a certain city. A lar

Sever21 [200]

Answer:

a i.) The mean for the security company is $ 1.47 million

a ii.) The median for the security company is $ 1.55 million

a iii.) The mean for the other companies is $ 1.96 million.

a iv.) The mean for the other companies is $ 2.0 million.

b) B. The mean and the median for the security company are both lower than the mean and the median for the collections performed by other companies.

c) B. Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees.

Step-by-step explanation:

By mean, we sum all the observation and divide by the sample size (n). Where sample size is the number of observations.

Mathematically,

$\bar{x} = \frac{\sum\limits_{i=1}^{n}{x_{i}}}{n}$ , where xi is the observations or values given.

By median, we mean middle number. So, we sort the values in ascending or descending order and take the value(s) in the middle. If the sample size is even, we take the 2 middle values and obtain the average.

See below, R programming codes for the solution.

##########...... R code

s = c(1.6,1.8,1.6,1.8,1.7,1.2,1.1,1.2,1.2,1.5)

c = c(1.5,2.1,1.9,2.2,1.9,1.7,2.1,2.2,2.2,1.8)

length(s)

mean(s); median(s)

mean(c);median(c)

t = (mean(s)-mean(c))/sqrt((var(s) + var(c))/length(s))

pnorm(t)

##############################################

For question C, we compute the test statistics, and obtain the p-value, see the last two lines of the R codes.

And since the p-value is less than level of significance, the test is significant. Thus, we conclude that:

<em>Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees.</em>

6 0

4 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