All categories

3 years ago

Which equation, when solved, results in a different value of x than the other three?

Mathematics

2 answers:

pashok25 [27]3 years ago

8 0

Answer:

11.3 = 8.3 + 0.6x

Step-by-step explanation:

You would subtract the 8.3 from 11.3 so you would get 3 = 0.6x. Then you would divide both sides by 0.6 resulting in x = 5.

myrzilka [38]3 years ago

7 0

Answer:

Step-by-step explanation:

dont ask me why it just is B

You might be interested in

Help on this complete the square problem.<br><br><br><img src="https://tex.z-dn.net/?f=x%5E2%2B3x%2B7" id="TexFormula1" title="x

Lesechka [4]

X2+3x+7=0 that’s the answer

6 0

3 years ago

What are three ratios that are equivalent to 15:38

MatroZZZ [7]

equivalent ratio's to 15:38

15 x 2 = 30
38 x 2 = 76

30:76 is equivalent to 15:38

15:38 x 3

15 x 3 = 45
38 x 3 = 114

45:114 is equivalent to 15:38

15:38 x 4

15 x 4 = 60
38 x 4 = 152

60:152 is equivalent to 15:38

3 0

3 years ago

Read 2 more answers

Karen's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Karen $5.50 per pound, and type

Lady_Fox [76]

Answer:

32 lb

Step-by-step explanation:

Let a = number of pounds of type A coffee used.

Let b = number of pounds of type B coffee used.

From the statement:

"This month's blend used three times as many pounds of type B coffee as type A"

we get this equation:

b = 3a

"Type A coffee costs Karen $5.50 per pound, and type B coffee costs $4.25 per pound."

She spend (in dollars) 5.50a + 4.25b, but since b = 3a, we get

5.5a + 4.25(3a) = 5.5a + 12.75a.

She spent $584, so

5.5a + 12.75a = 584

18.25a = 584

a = 32

She used 32 lb of type A coffee.

4 0

3 years ago

An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen

Slav-nsk [51]

Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$