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3 years ago

Pip, Angad and Nick share some sweets in the ratio 4:4:1. Pip gets 20 sweets. How many

Mathematics

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

Step-by-step explanation:

The 4 part of the ratio represents Pip's 20 sweets

Dividing 20 by 4 gives the value of one part of the ratio

20 ÷ 4 = 5 sweets ← value of 1 part of the ratio

Thus

Pip gets 20 sweets

Angad gets 20 sweets ( 4 parts of the ratio )

Nick gets 5 sweets ( 1 part of the ratio )

Total number of sweets = 20 + 20 + 5 = 45

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State if the triangles in each pair are similar. If so, State how you know they are similar and complete the similarity statemen

yuradex [85]

Answer:

Answer is option 2

Step-by-step explanation:

We know that Angle M = Angle G (given in diagram)

We also know that Angle L in triangle LMN is equal to Angle L in triangle LGH

As two angles are equal in both triangles they are similar.

But why is it Triangle LGH instead of Triangle HGL?

As we know M=G therefore they should be in the same place in the name Of the triangle. In triangle LMN M is in the middle therefore Angle G should also be in the middle

7 0

3 years ago

Can someone help me with this please

pochemuha

Answer: No

Step-by-step explanation:

What we have to do is to substitute the values of x and y that we are given into the inequality:

x = 8

y = 7

y > 4x - 6

7 > 4(8) - 6

7 > 32 - 6

7 > 26

This is not true so it isn't a solution

3 0

3 years ago

Read 2 more answers

Suppose that prices of recently sold homes in one neighborhood have a mean of $265,000 with a standard deviation of $9300. Using

Hatshy [7]

Answer:

Range = (237100, 292900)

Step-by-step explanation:

Using Chebyshevs Inequality:

$P(|X - \mu | \le k \sigma )\ge 1 -\dfrac{1}{k^2}= 0.889$

$1 -\dfrac{1}{k^2}= 0.889$

$\dfrac{1}{k^2}= 1- 0.889$

$\dfrac{1}{k^2}=0.111$

$k = \sqrt{\dfrac{1}{0.111}}$

$k \simeq 3$

Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ ± kσ

where;

μ = 265000

σ = 9300

Range = 265000 ± 3(9300)

Range = 265000 ± 27900

Range = (265000 - 27900, 265000 + 27900)

Range = (237100, 292900)

3 0

2 years ago

The graph below describes the journey of a train between two cities.

Veseljchak [2.6K]

Answer:

[A] Acceleration in first 15 min = 200 km/h²

[B] Distance between two cities = 125 km

[C] Average Speed of journey = 62.5 km/hr

Step-by-step explanation:

To Solve:

Acceleration in first 15 min

Distance between two cities

Average speed of journey

Solve:

[A] Work out the acceleration, in km/h?, in the first 15 mins.

Each horizontal block is 1/8 hr = 7.5 min

Each vertical block is 10 km/hr

Time Velocity km/hr

0 Min ( 0 hr) 0

15 Min (1/4 hr) 50

45 Min (3/4 hr) 50

60 MIn ( 1 hr) 100

90 Min ( 3/2 hr) 100

120 Min ( 2hr) 0

Acceleration in first 15 min (1/4 hr) = (50 - 0)/(1/4 - 0) = 50/(1/4)

= 200 km/h²

Hence, Answer for [A] = 200 km/h²

[B] Work out the distance between the two cities. $60 Velocity (in km/h)

= (1/2)(0 + 50)(1/4 - 0) + 50 * (3/4 - 1/4) + (1/2)(50 + 100)(1 - 3/4) + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)

= 25/4 + 25 + 75/4 + 50 + 25

= 125

Distance between two cities = 125 km

[C] Work out the average speed of the train during 20 the journey.

Average Speed of journey = 125/2 = 62.5 km/hr

[RevyBreeze]

5 0

2 years ago

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ddd [48]

Answer:

141.4

Step-by-step explanation:

i think

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3 years ago