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3 years ago

Can someone explain this please???

Mathematics

1 answer:

xeze [42]3 years ago

4 0

A function is a rule that assigns exactly one output to a given input. The input is taken from a set called the domain, and the corresponding output belongs to a set called the range.

1. In this exercise, we're calling the pool of patients 1-8 the domain, and the pool of nurses A-D the range. The given table describes a function because any patient is assigned to only one nurse.

2. This wouldn't be a function if at least one patient was assigned to more than one nurse. If this were to happen in practice, the patient could be, say, given the same dose of some medicine twice if the nurses aren't careful.

3. Making the nurse pool the domain and the patient pool the range would give a relation that is not a function, since more than one patient is assigned to one nurse.

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Help asap I will give u huge amount of brainly points

Minchanka [31]

Answer:

$\boxed{\sf -\dfrac{5}{8}}$

Use BODMAS rule,

where:

brackets
order
division
multiplication
addition
subtraction

Given expression:

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-14\times\left(\dfrac{1}{4} \right)^2]$

simplify exponent

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-14\times\left(\dfrac{1}{16} \right)]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-\(\dfrac{14}{16}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-\(\dfrac{7}{8}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:\dfrac{32}{8} -\(\dfrac{7}{8}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:\dfrac{25}{8}]$

multiply fractions

$\rightarrow \sf -\dfrac{25}{40}$

$\rightarrow \sf -\dfrac{5}{8}$

8 0

1 year ago

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Solve : 11 (9-v)=0 a) -9 b) 9 c) 99 d)-99

garri49 [273]

B. V=9 Multiply it out, then subtract 99 from both sides, then divide by 11

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3 years ago

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Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

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

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

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

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

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3 years ago

(20 PTS!!)

adell [148]

The answer would be "both groups show about the same average time"

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Step2247 [10]

The graph is at a slope:1 with a y-intercept:-4

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