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ss7ja [257]
2 years ago
8

The image below represents a 12 x 16 room with an 8 x 10 piece of linoleum centered in the room. The yellow and blue rectangles

extend the length of their respective sides. Where these two rectangles overlap, there is a green rectangle. In the paragraph box below, explain why the total colored area is 62 square feet.

Mathematics
2 answers:
n200080 [17]2 years ago
7 0
Since the rectangles are colored in, we can take the area of each color, then add them together. To find all the lengths and widths, subtract as needed: Since the center rectangle's length is 10 feet, subtract 10 from the length of the room which is 16 feet. Those 6 units must be divided in half to give 3 feet for the width of the blue rectangle. Similarly, subtract 8 from 12 and divide the 4 in half to give 2 feet for the width of the yellow rectangle. This also reveals the length and width of the green rectangle. The length is 3 and the width 2, so the area is found 3x2=6. The area of the blue rectangle is found 3x10=30. The area of the yellow rectangle is found 13x2=26. Add these areas up: 6+30+26 = 62 square feet.
Andrei [34K]2 years ago
3 0

Answer:for the people here to find out the area of the green rectangle the answer is 6

Step-by-step explanation:

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Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0
1 year ago
Two equal groups of seedlings, and equal in height, were selected for an experiment. One group of seedlings was fed Fertilizer A
ser-zykov [4K]

Answer:

Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0

Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0

Here to test Fertilizer A height is greater than Fertilizer B

Two Sample T Test:

t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}

Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}

S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154

t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524

P value for Test Statistic of P(3.3524,26) = 0.0012

df = n1+n2-2 = 26

Critical value of P : t_{0.025,26}=2.05553

We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.

6 0
1 year ago
One morning, eight buses arrive at a bus stop.
Semenov [28]

Answer:

0 and 5 minutes

Step-by-step explanation:

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

We have given 8 numbers and we have to find the other two numbers a minute late for each bus to arrive at the bus stop.

Also, we have given the median and mode of all 10 buses i.e. 3.5 and 0 respectively.

Since the mode is 0, so the frequency of 0 must be greater than 2 times and less than or equal to 4 times.

The frequency of 0 is 4 is not possible as then we didn't get a median 3.5 then.

So the frequency of 0 is 3.

Hence one of number is 0 in two unknown number of a minute late for each bus arrive at the bus stop.

Now, we have the number of minutes late for each bus after arranging them in ascending order is: 0 0 0 2 2 6 8 8 9

For getting a median 3.5 we must have a number between 2 and 6. Let it be x then:

\frac{2+x}{2} = 3.5

⇒x = 5

Hence the number of minutes late for each bus is: 0 0 0 2 2 5 6 8 8 9

Thus, 0 and 5 are minutes late for the two-afternoon buses.

6 0
2 years ago
Read 2 more answers
In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
Sergeeva-Olga [200]

Answer:

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0
1 year ago
The type of part being manufactured today should be 1.27 inches wide, with a tolerance of 0.04 inches. If a part does not meet t
olasank [31]
The range of widths is {1.23,1.31} so none will be rejected, answer A.

3 0
2 years ago
Read 2 more answers
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