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3 years ago

Gary ordered 30 pizzas for a party. 60% of the pizzas have 12 slices each. The remaining 40% of the pizzas have 10 slices each.

Mathematics

1 answer:

LekaFEV [45]3 years ago

4 0

Answer:

336 total slices

Step-by-step explanation:

60% of 30 = 18

18 * 12 = 216

40% of 30 = 12

12 * 10 = 120

216 + 120 = 336

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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

Chnstopher earned $142.20. He saved $24.50. He spent the rest on 2 new pairs

pogonyaev

Answer:

$58.25

Step-by-step explanation:

142.20 - 24.50 = 117.50

117.50 / 2 = 58.25

6 0

3 years ago

Read 2 more answers

HELP ASAP ILL GIVE BRAINLIST

Mila [183]

Answer:

Part A) ΔCGB and ΔPBE are similiar

Part B) They are similar because they are Vertical Angles/Alternate Interior Angles

Part C) Distance from B to E is 125; Distance from P to E is 275.

Step-by-step explanation:

5 0

2 years ago

QUICK!!!

Svetlanka [38]

Answer:

Option B is the right choice.

Step-by-step explanation:

Given:

An are CE and an inscribed angle CBE.

Measure of inscribed angle CBE = 25 °

We have to find the measure of the arc CE.

Concept:

An inscribed angle is an angle with its vertex on the circle.
The measure of an inscribed angle is half the measure the intercepted arc.
Measure of inscribed angle = 1/2 × measure of intercepted arc .

To find the measure of arc CE.

⇒ $\angle CBE = \frac{1}{2} \times m\ (arc\ CE)$

⇒ $2\times \angle CBE = m\ (arc\ CE)$

⇒ $2\times 25 = m\ (arc\ CE)$

⇒ $50 = m\ (arc\ CE)$

Measure of intercepted arc CE = 50 degrees.

The measure of arc CE = 50° so, option B is the right choice.

4 0

3 years ago

A garden is designed in the shape of a rhombus formed

Usimov [2.4K]

Answer:

120ft.

Step-by-step explanation:

A garden is designed in the shape of a rhombus formed from 4 identical 30°-60°-90° triangles. The shorter distance across the middle of the garden measures 30 feet. What is the distance around the perimeter of the garden? 120ft.

4 0

2 years ago