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3 years ago

14

What is the range of the data? 2,8,7,8,6,15,6,3,8

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

13

Step-by-step explanation:

The range of a data set is the difference between the greatest and least values in the set. In this case, the greatest value is 15 and the least value is 2 so the range is 15 - 2 = 13.

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MissTica

That is basically splitting the sentence so that it’s easier for you to solve it. It’s basically the same thing

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3 years ago

Solve for all values of x by factoring.<br> x^2 + 4x – 30 = -3x

vichka [17]

Answer:

${x}^{2} + 4x - 30 = - 3x \\{x}^{2} + 7x - 30 = 0 \\ {x}^{2} - 3x + 10x - 30 = 0 \\ x(x - 3) + 10(x - 3) = 0 \\ (x - 3)(x + 10) = 0 \\ \boxed{x = 3} \\ \boxed{x = - 10}$

<h3><em><u>x=3 or x=-10</u></em> is the right answer.</h3>

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3 years ago

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Answer
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Explanation:

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3 years ago

[98 POINTS]<br><br> Explain how you would calculate the volume of the figure shown.<br> Be detailed.

N76 [4]

Answer:

1282.81824016

Step-by-step explanation:

Find the volume of the cone (the formula is π×r²×h/3) and the volume is 359.19

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3 years ago

Read 2 more answers

A jeepney ride cost 11 pesos for the first 4 kilometers and each additional integer kilometers adds 3.50 pesos to the fare, use

Nookie1986 [14]

Answer:

$f(d)=\left\{ \begin{matrix} 11 & \text{for} & 0< d\leq 4 \\-3+3.5d & \text{for}& d > 4 \end$

Step-by-step explanation:

Cost of first 4 kilometers ride = 11 pesos.

So, for the first 4 km, the fair is constant i.e.

for 1 km the fare is 11 pesos,

for 2 km the fair is also 11 pesos,

similarly, for 3 km of 4 km, the fare is 11 pesos.

Hence, for the distance $0\geq d\geq 4$ , the fair function,

$f(d)=11\cdots(i)$

After 4 km, there is an increment of $ 3.50 for each kilometer.

So, the fare function up to 5 kilometers,

$f(d)=11+3.5=14.5$

So, the fare function up to 6 kilometers, i.e for the distance $5$ ,

$f(d)=11+2\times3.5=18$

This can be arranged as the fare function up to 6 km,

$f(d)=11+(6-4)\times3.5=18$

Similarly, the fare function up to $d$ kilometer $(n>4)$ ,

$f(d)=11+(d-4)\times3.5$

$\Rightarrrow f(d)=-3+3.5d\cdots(ii)$

Hence, from equations (i) and (ii),

$f(d)=\left\{ \begin{matrix} 11 & \text{for}\; 0< d \leq 4 \\-3+3.5d & \text{for}\; d > 4 \end$

8 0

3 years ago