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2 years ago

9

The range of the following data set is 9,5,14,24,12

1 answer:

dimaraw [331]2 years ago

4 0

Range would be 19 (24-5)

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Round 222702 to the nearest ten thousand

melomori [17]

220000 because 2 rounds down

4 0

2 years ago

Read 2 more answers

F(x)=18x+11, find f(5)

Natalija [7]

Answer:

$\boxed {\tt f(5)=101}$

Step-by-step explanation:

We are given the function:

$f(x)=18x+11$

and asked to find f(5). Therefore, we must substitute 5 in for each $x$ .

$f(5)=18(5)+11$

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

Multiply 18 and 5.

$f(5)=90+11$

Add 90 and 11.

$f(5)=101$

f(5) is equal to 101.

5 0

2 years ago

A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

mestny [16]

Answer:

The value is $W= 2640 \ ft \cdot lb$

Step-by-step explanation:

From the question we are told that

The weight of the bucket is $F = 5 lb$

The depth of the well is $x_1 = 60 \ ft$

The weight of the water is $W_w = 42 lb$

The rate at which the bucket with water is pulled is $v = 1.5 \ ft/s$

The rate of the leak is $r = 0.15 lb/s$

Generally the workdone is mathematically represented as

$W = \int\limits^{x_1}_{x_o} {G(x)} \, dx$ ]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

$G(x) = F + (W_w- Ix)$

Here I is the rate of water loss in lb/ft mathematically represented as

$I = \frac{r}{v}$

=> $I = \frac{0.15 }{1.5 }$

=> $I = 0.1$

So

$G(x) = 5 + (42- 0.1x)$

=> $G(x) = 47- 0.1x)$

So

$W = \int\limits^{60}_{0} {47- 0.1x} \, dx$ ]

=> $W = [47x - \frac{0.1x^2}{2} ]|\left 60} \atop {0}} \right.$

=> $W= [47(60) - 0.05(60)^2]$

=> $W= 2640 \ ft \cdot lb$

7 0

2 years ago

An account earns simple interest. Find the interest earned.<br>$700 at 3% for 4 years

earnstyle [38]

Answer:

$84

Step-by-step explanation:

Simple Interest = Interest rate*Principle*number of years

= (3÷100)*$700*4

=$84

3 0

2 years ago

Few math questions TIME IS RUNNING OUT!! PLEASE HELP!!!!

olya-2409 [2.1K]

1.

$|5+6x|\ \textless \ 13\\\\-13\ \textless \ 5+6x\ \textless \ 13$

so

$\begin{cases}5+6x\ \textless \ 13\\5+6x\ \textgreater \ -13\end{cases}$

Answers B and D.

2.

$|x+9|\geq11\\\\x+9\ \textgreater \ 11\qquad\vee\qquad x+9\leq-11\\\\
\boxed{x\geq2\qquad\vee\qquad x\leq-20}$

Answer D.

3.

The same as point 2. Answer D.

4.

Answer C.

3 0

3 years ago