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3 years ago

11

You might need: Calculator Problem Zane is a dangerous fellow who likes to go rock climbing in active volcanoes. One time while

inside a volcano, he heard some rumbling, so he decided to climb up out of there as quickly as he could. He climbed up at a rate of 44 meters per second. After 33 seconds, he was 1313 meters below the edge of the volcano. How far was Zane below the edge of the volcano when he started climbing? How long did it take Zane to reach the edge of the volcano?

2 answers:

Ksenya-84 [330]3 years ago

7 0

Zane climb up in 33 seconds

44•33 = 1452 m.

so, Zane start from 1313 + 1452 = 2765 m below the edge.

to reach the edge with constant speed, Zane need 2765/44 = 62.84 seconds

we can round it to 63 seconds

Lesechka [4]3 years ago

6 0

Answer: Zane was 25 meters below the edge of the volcano when he started and he took 6.25 seconds to reach the edge of volcano.

Step-by-step explanation:

Since wee have given that

Speed at which he climb up = 4 meters per second

After 3 seconds, he was 13 meters below the edge of the volcano.

When he started climbing, Position of him below the edge of the volcano is given by

$3\times 4+13\\\\=12+13\\\\=25\ meters$

Time taken by Zane to reach the edge of the volcano is given by

$=\dfrac{Distance}{Speed}\\\\=\dfrac{25}{4}\\\\=6.25\ seconds$

Hence, Zane was 25 meters below the edge of the volcano when he started and he took 6.25 seconds to reach the edge of volcano.

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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

2)

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

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balu736 [363]

Answer:

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Step-by-step explanation:

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2 years ago

Alussa took a math test with 20 questions

anastassius [24]

Answer:

Step-by-step explanation: 100 divided by 20 is 5 therefore, each question is worth 5%. take how many questions she got right and multiply by 5. then determine if her score is less than or greater then 90%.

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3 years ago

In triangle ΔABC, ∠C is a right angle and CD is the height to AB . Find the angles in ΔCBD and ΔCAD if: Chapter Reference a m∠A

Strike441 [17]

Answer:

Given A triangle ABC in which

∠C =90°,∠A=20° and CD ⊥ AB.

In Δ ABC

⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]

⇒20° + ∠B + 90°=180°

⇒∠B+110° =180°

∠B =180° -110°

∠B = 70°

In Δ B DC

∠BDC =90°,∠B =70°,∠BC D=?

∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]

90° + 70°+∠BC D =180°

∠BC D=180°- 160°

∠BC D = 20°

In Δ AC D

∠A=20°, ∠ADC=90°,∠AC D=?

∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]

20°+90°+∠AC D=180°

110° +∠AC D=180°

∠AC D=180°-110°

∠AC D=70°

So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°

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4 years ago

A rectangular park of length 60 m breadth 50 m encloses w volleyball court of length 18 m and breadth 10 m. Find the area of the

Zanzabum

Given

A rectangular park of length 60 m breadth 50 m encloses with volleyball court of length 18 m and breadth 10 m.

To find:

The area of the park excluding the court at the rate of Rs 110 per square meter.

Solution:

Area of a rectangle is:

$Area=length \times breadth$

Area of whole park is:

$A_1=60 \times 50$

$A_1=3000$

Area of volleyball court is:

$A_2=18 \times 10$

$A_2=180$

Now, the area of the park excluding the court is:

$A=A_1-A_2$

$A=3000-180$

$A=2820$

Therefore, the area of the park excluding the court is 2820 square meter.

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