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3 years ago

I can buy 3 cans of soda for $1.05. How many cans can buy at the same rate for $21?

Mathematics

2 answers:

kakasveta [241]3 years ago

7 0

Answer:

60 cans of soda

Step-by-step explanation:

3 cans of soda for $1.05

$21 = ? cans of soda

21 / 1.05 = 20

so if 20 $1.05's fit in $21 then how many cans fit in $21

20 x 3 = 60

you can buy 60 cans of soda with $21

jek_recluse [69]3 years ago

3 0

Answer:

20 cans

Step-by-step explanation:

Divide $21 by $1.05 and you get 20.

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Two concentric circles have radii 6 and 14. Find the length of a chord of the larger circle that is tangent to the smaller circl

Vitek1552 [10]

Answer:

Step-by-step explanation:

let 2x be the length of chord tangent to first circle.

so x=√(14²-6²)=√(14+6)(14-6)=√(20×8)=√160=4√10

length of chord=2x=2×4√10=8√10

6 0

3 years ago

Factor the expression 9x^2 - 30x + 25 I need help

MrMuchimi

Answer:

(3x - 5)²

Step-by-step explanation:

When we expand (3x - 5)², we get:

(3x - 5) (3x - 5)

When we multiply, we get:

9x² - 30x + 25

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7 0

3 years ago

For f(x)= x +2 and g(x)= 1/x find (g o f)

lukranit [14]

Hello :
<span>f(x)= x +2 and g(x)= 1/x
</span><span>(g o f)(x) = g(f(x))=g(x+2) = 1/(x+2)</span>

8 0

3 years ago

Question 9 (1 point)

Romashka [77]

Answer:

Step-by-step explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(82.5-1.5\times 76.56,159.06+1.5\times 76.56)$

(-32.34,273.9)

So, There is no outlier

Maximum = 207

League B in ascending order :

24.93,136,145,157,163.25,178,256

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

$Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5$

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(140.5-1.5\times 30.125,170.625+1.5\times 30.125)$

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256

5 0

3 years ago

4/12 + 3/12 = ???????

iris [78.8K]

7/12

..............................

4 0

3 years ago

Read 2 more answers