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3 years ago

PLEASE HELP I NEED HELP ASAP PLEASEE

Mathematics

1 answer:

Serhud [2]3 years ago

5 0

<h3>Answer: Choice C -- (0.6, 1.9)</h3>

Explanation:

Look through the table to see where f(x) and g(x) are the same value. Unfortunately this never happens for any row. However, we can get fairly close when we look at row two with f(x) = 1.8 and g(x) = 2; this input is x = 0.6

A good guess is that the midpoint of 1.8 and 2 is the y coordinate of the solution, so this is where the 1.9 comes from. You add 1.8 and 2 to get 3.8, then divide in half to get 1.9

So when x = 0.6 a good guess at the y value of the solution is y = 1.9, meaning that (0.6, 1.9) is an estimate of the solution

note: this is likely not the true solution, but it may be fairly close

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Step-by-step explanation:

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Which can be the first step in finding the equation of the line that passes through the points (5, -4) and (-1, 8) in slope-inte

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Answer:

y = -2x + 6

Step-by-step explanation:

Slope = (-4 - 8)/(5 + 1)

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2 years ago

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1.5m+1.25j

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m equals the number of muffins sold, and because each muffin sells for $1.50, 1.5m represents the income made from the muffins.

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2 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$