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3 years ago

Which equation best represents the relationship?

Mathematics

1 answer:

Agata [3.3K]3 years ago

4 0

Answer: (C) $C=\frac{5}{9}(F-32)$

Explanation:

The difference between the degrees in Fahrenheit and the constant 32 corresponds to the expression in brackets (F-32). The fraction 5/9 of that difference completes the expression: (5/9)(F-32) and gives the degrees Celsius.

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If 11 (n) equals 143 what is n?

UkoKoshka [18]

The value of n that will make the expression equal is 13

<h3>Equations and expression</h3>

Equations are expressions separated by equal sign

Given the equation

11n = 143

Divide both sides by 11 to have;

11n/11 = 143/11

n = 13

Hence the value of n that will make the expression equal is 13

Learn more on linear equation here; brainly.com/question/14323743

#SPJ1

4 0

1 year ago

Please help. Need to find the equation of the tangent at point P. Give your answer in the form y=ax+b where a and b are both fra

SVETLANKA909090 [29]

Answer:

y=7/24x+625/24

Step-by-step explanation:

-7x+24y=-25"2

-7x+24y=625

+7x +7x

24y=7x+625

divide 24

y=7/24x+625/24

Got this for my mathswatch too

6 0

3 years ago

What is the biggest number less than 1,000 that has only 2s in its prime factorization?

Contact [7]

Answer:

2^9 = 512

Step-by-step explanation:

2^9 = 512 is the largest number less than 1000 that has only 2s in its prime factorization.

8 0

3 years ago

A sample of 13 sheets of cardstock is randomly selected and the following thicknesses are measured in millimeters. Give a point

valina [46]

<h3>Answer: 0.061</h3>

===============================================

Explanation:

Add up the values to get

1.96+1.81+1.97+1.83+1.87+1.84+1.85+1.94+1.96+1.81+1.86+1.95+1.89= 24.54

Then divide over 13 (the number of values) to get 24.54/13 = 1.8876923 which is approximate.

So the mean is approximately 1.8876923

---------------------

Now make a spreadsheet as shown below

We have the first column as the x values, which are the original numbers your teacher provided. The second column is of the form (x-M)^2, where M is the mean we computed earlier. We subtract off the mean and square the result.

After we compute that column of (x-M)^2 values, we add them up to get what is shown in the highlighted yellow cell at the bottom of the column.

That sum is approximately 0.04403076924

Next, we divide that over n-1 = 13-1 = 12

0.04403076924 /12 = 0.00366923077

That is the sample variance. Apply the square root to this to get the sample standard deviation. This is the point estimate of the population standard deviation. As the name implies, it works for samples that estimate population parameters.

sqrt(0.00366923077) = 0.06057417576822

This rounds to 0.061 which is the final answer.

5 0

2 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use of one particular cust

Fiesta28 [93]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

$321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320$

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}$

STEP 3: Calculate the mean

$\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}$

STEP 4: Calculate the Standard deviation

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}$

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

$\begin{gathered} \text{IQR}=Q_3-Q_1 \\ \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\ \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}$

STEP 6: Find the Interquartile Range

$\begin{gathered} IQR=Q_3-Q_1 \\ \text{IQR}=479.5-363.5 \\ \text{IQR}=116 \end{gathered}$

Hence, the interquartile range of the data is 116

3 0

1 year ago