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3 years ago

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The mean of the values in a data set is c. If each of the values in the data set were multiplied by 6, what would be the mean of

the resulting data?

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

Mean of the data would be 6c

Step-by-step explanation:

<u>Rule to Remember:</u> Whenever each term of the data is multiplied by the same number, the new mean is previous mean multiplied by that number.

In this case, original value of mean was c. Each value of the data is multiplied with 6 so a result the new mean would be 6 times the original mean i.e. 6c

Let us consider that original values in the data set are x, y and z. Their mean would be:

$Mean=\frac{x+y+z}{3}$

If each value of the data is multiplied with 6, the new values would be 6x, 6y and 6z. The mean in this case will be:

$Mean=\frac{6x+6y+6z}{3}\\Mean=\frac{6(x+y+z)}{3}\\Mean=6\times\frac{x+y+z}{3}\\Mean=6c$

We can see from above that the new mean is 6 multiplied to the original mean.

Therefore, the answer to this question is 6c.

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3 years ago

COMPUTE<br><br> 3 ( 2 1/2 - 1 ) + 3/10

Juli2301 [7.4K]

Answer:

<h3> $\boxed{ \frac{24}{5} }$ </h3>

Step-by-step explanation:

$\mathsf{3(2 \frac{1}{2} - 1) + \frac{3}{10} }$

Convert mixed number to improper fraction

$\mathrm{3( \frac{5}{2} - 1) + \frac{3}{10} }$

Calculate the difference

⇒ $\mathrm{3( \frac{5 \times 1}{2 \times 1} - \frac{1 \times 2}{1 \times 2} }) + \frac{3}{10}$

⇒ $\mathrm{ 3 \times( \frac{5}{2} - \frac{2}{2}) } + \frac{3}{10}$

⇒ $\mathrm{3 \times ( \frac{5 - 2}{2} ) + \frac{3}{10} }$

⇒ $\mathrm{3 \times \frac{3}{2} + \frac{3}{10} }$

Calculate the product

⇒ $\mathrm{ \frac{3 \times 3}{1 \times 2} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{9}{2} + \frac{3}{10}}$

Add the fractions

⇒ $\mathsf{ \frac{9 \times 5}{2 \times 5} + \frac{3 \times 1}{10 \times 1} }$

⇒ $\mathrm{ \frac{45}{10} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{45 + 3}{10 } }$

⇒ $\mathrm{ \frac{48}{10} }$

Reduce the numerator and denominator by 2

⇒ $\mathrm{ \frac{24}{5} }$

Further more explanation:

<u>Addition </u><u>and </u><u>Subtraction</u><u> </u><u>of </u><u>like </u><u>fractions</u>

While performing the addition and subtraction of like fractions, you just have to add or subtract the numerator respectively in which the denominator is retained same.

For example :

Add : $\mathsf{ \frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} } = \frac{4}{5}$

Subtract : $\mathsf{ \frac{5}{7} - \frac{4}{7} = \frac{5 - 4}{7} = \frac{3}{7} }$

So, sum of like fractions : $\mathsf{ = \frac{sum \: of \: their \: number}{common \: denominator} }$

Difference of like fractions : $\mathsf{ \frac{difference \: of \: their \: numerator}{common \: denominator} }$

<u>Addition </u><u>and </u><u>subtraction</u><u> </u><u>of </u><u>unlike </u><u>fractions</u>

While performing the addition and subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fractions. Thus, obtained fractions should be reduced into lowest terms if there are any common on numerator and denominator.

For example:

$\mathsf{add \: \frac{1}{2} \: and \: \frac{1}{3} }$

L.C.M of 2 and 3 = 6

So, ⇒ $\mathsf{ \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} }$

⇒ $\mathsf{ \frac{3}{6} + \frac{2}{6} }$

⇒ $\frac{5}{6}$

Multiplication of fractions

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term.

When any number or fraction is divided by a fraction, we multiply the dividend by reciprocal of the divisor. Let's consider a multiplication of a whole number by a fraction:

$\mathsf{4 \times \frac{3}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6}$

Multiplication for $\mathsf{ \frac{6}{5} \: and \: \frac{25}{3} }$ is done by the similar process

$\mathsf{ = \frac{6}{5} \times \frac{25}{3} = 2 \times 5 \times 10}$

Hope I helped!

Best regards!

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The function f(x)= square root of x is translated using the rule (x, y) → (x – 6, y + 9) to create A(x). Which expression descri

Alex73 [517]

The function $f(x)=\sqrt{x}$ is translated according to the rule

$(x,y)\rightarrow (x-6,y+9).$

This translation is translation 6 units to the left and 9 units up.

1. Translation of the function $f(x)=\sqrt{x}$ 6 units to the left gives you the function $g(x)=\sqrt{x+6}.$

2. Translation of the function $g(x)=\sqrt{x+6}$ 9 units up gives you the function $A(x)=\sqrt{x+6}+9.$

The domain of the function $A(x)$ is $[-6,\infty),$ then the range of the function $A(x)$ is $[9,\infty).$

Answer: $[9,\infty).$

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3 years ago