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1 year ago

5

Mrs. Melanie Lazo pays Php 15.75 for each hair clip to its supplier. She decided to adds markup of Php 3.15. What is the markup

rate?

1 answer:

malfutka [58]1 year ago

8 0

Answer:

the answer to the problem is 20.8%

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If ∆ABC and ∆GEF are congruent, the value of x is___? and the value of y is ___?

attashe74 [19]

<u>Answer-</u>

<em>The value of y is </em><em>6 units</em><em>.</em>

<u>Solution-</u>

Given ∆ABC ≅ ∆GEF

If two triangles are congruent, then the corresponding sides are also congruent.

So,

EF = BC, GF = AC

Putting the values,

$\Rightarrow x-y=4$ --------------1

$\Rightarrow x-8=2$

$\Rightarrow x=10$ --------------------2

Putting the values of x in equation 1,

$\Rightarrow 10-y=4$

$\Rightarrow y=10-4=6$

Therefore, the value of y is 6 units.

7 0

1 year ago

Read 2 more answers

The area of the figure

STALIN [3.7K]

Rectangle:

4*8=32

Semicircle:

3.14*4^2= 50.24
50.24/2=25.12

Add:

25.12+32= 57.12

Final answer: A

5 0

1 year ago

Read 2 more answers

Question: 2. Musah Stands At The Centre Of A Rectangular Field. He First Takes 50 Steps North, Then 25 Steps West And Finally 50

Iteru [2.4K]

Answer:

60.36 steps West from centre

85.36 steps North from centre

Step-by-step explanation:

<em>Refer to attached</em>

Musah start point and movement is captured in the picture.

1. He moves 50 steps to North,
2. Then 25 steps to West,

3. Then 50 steps on a bearing of 315°. We now North is measured 0°

or 360°, so bearing of 315° is same as North-West 45°.

<em />

<em>Note. According to Pythagorean theorem, 45° right triangle with hypotenuse of a has legs equal to a/√2.</em>

<u />

<u>How far West Is Musah's final point from the centre?</u>

25 + 50/√2 ≈ 60.36 steps

<u>How far North Is Musah's final point from the centre?</u>

50 + 50/√2 ≈ 85.36 steps

7 0

1 year ago

Train A and train B stops at Swindon at 10:30 . Train A stops every twelve minutes and train B stops every 14 Mins , when do the

Nata [24]

Answer: at 11:54

Step-by-step explanation:

Let's define the 10:30 as our t = 0 min.

We know that Train A stops every 12 mins, and Train B stops every 14 mins, they will stop at the same time in the least common multiple of 12 and 14.

To find the least common multiple of two numbers, we must do:

LCM(a,b) = a*b/GCD(a,b)

Where GCD(a, b) is the greatest common divisor of a and b.

In this case the only common divisior of 12 and 14 is 2.

So we have:

LCM(12, 14) = 12*14/2 = 84.

Then the both trains will stop 84 minutes after 10:30

one hour has 60 mins, so we can write 84 minutes as:

1 hour and 24 minutes = 1:24

Then they will stop at the same time at 10:30 + 1:24 = 11:54

4 0

1 year ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

1 year ago