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QveST [7]
3 years ago
6

Evaluate the expression for d = 3. 2(d +8)= ??

Mathematics
2 answers:
Pavel [41]3 years ago
7 0
The answer is: 22


Step by step explanation
cestrela7 [59]3 years ago
5 0

Answer:

22

Step-by-step explanation:

You might be interested in
One warehouse had 185 tons of coal, another one had 237 tons. The first warehouse delivered 15 tons of coal to its clients daily
lianna [129]

Answer:

In 9 days will the second warehouse have 1.5 times more coal than the first one

Step-by-step explanation:

The coal left in warehouses after x days can be found using the equation:

  • first warehouse: 185 - 15x
  • second warehouse= 237-18x

Let the second warehouse have 1.5 times more coal than the first one after n days then  

  1. 1.5 ×(185 - 15n)=237-18n  which gives:
  2. 277,5-22,5n=237-18n and
  3. 40,5=4,5n
  4. 9=n

7 0
3 years ago
Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav
Furkat [3]

Answer:

  • There is no significant evidence that p1 is different than p2 at 0.01 significance level.
  • 99% confidence interval for p1-p2 is  -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

H_{0}: p1-p2=0

H_{a}: p1-p2≠0

Test statistic can be found using the equation:

z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

  • p1 is the sample proportion of the common attribute in population1 (\frac{16}{30} =0.533)
  • p2 is the sample proportion of the common attribute in population2 (\frac{1337}{1900} =0.704)
  • p is the pool proportion of p1 and p2 (\frac{16+1337}{30+1900}=0.701)
  • n1 is the sample size of the people from population1 (30)
  • n2 is the sample size of the people from population2 (1900)

Then z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}} ≈ 2.03

p-value of the test statistic is  0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2±z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}} where

  • z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704±2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}} ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
A histogram is better for _____________ data and a frequency polygon is better for ________data.
Ne4ueva [31]

A histogram is better for "discrete" data and a frequency polygon is better for "continuous" data.

<h3>What is histogram?</h3>

A histogram is a data representation that looks like a bar graph and buckets a wide range of categories into columns all along horizontal x-axis.

  • The numeric count or percentage of happenings in the data for every column is represented by the vertical y-axis.
  • Columns can be employed to visualize data distribution patterns.
  • Technical analysts use the MACD histogram in trading to making data in momentum.
  • The MACD histogram columns could provide buy and sell signals earlier than the MACD and signal lines.
<h3>What is frequency polygon?</h3>

Frequency polygons are a graphs depiction of the distribution that aids in data comprehension by utilizing a specific shape.

  • Frequency polygons are comparable to histograms but are more beneficial when comparing two or more different data sets.
  • The graph primarily displays cumulative frequency distribution data as a line graph.
  • Frequency Polygons are a type of graph that deciphers information or data and are broadly used in statistics.
  • This graphic form of data representation aids in the depiction of the data's shape and trend in a systematic and organized manner.

To know more about the histogram, here

brainly.com/question/2962546

#SPJ4

5 0
1 year ago
An airline allows each passenger 33 kilograms of luggage. Carl has to lower the weight of his suitcase by 25% to stay within the
damaskus [11]
1- 0.25= 0.75 (multiplier)

33÷0.75=44

The suitcase originally weighed 44kg

Hope this helps!
5 0
3 years ago
Read 2 more answers
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