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gayaneshka [121]
3 years ago
11

the mean height of the 30 children in a room is 76 cm james whose height is 107 cm went into the room what is the new mean of th

e height of the children in the room?​
Mathematics
1 answer:
Snezhnost [94]3 years ago
4 0

Answer:

The new mean is 77cm

Step-by-step explanation:

If the mean height of the 30 children in a room is 76 cm, then the sum of their ages is

30 \times 76 = 2280cm

If James with height 107cm joins them, then the total number of children now becomes 30+1=31 and the total height becomes: 2280cm+107cm=2387cm

The new mean now becomes:

\bar x =  \frac{2387}{31}  = 77cm

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Answer:

7/10

Step-by-step explanation:

7/8 = .875

10/12 = .833

12/14 = .857

7/10 = .7

1/2 - .5

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Tom is buying topsoil for the flower bed shown below. One bag of topsoil covers 151515 square meters. How many bags of topsoil d
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Answer:

2 bags of topsoil.

Step-by-step explanation:

The attached figure shows a flower bed.

One bag of topsoil covers 15 square meters.

We need to find how many bags of topsoil does Tom need to cover his flower bed.

The area of the flower bed is :

A=\dfrac{1}{2}\times 12\times 5\\\\=30\ m^2  

Let he has to cover x bags of topsoil. So,

x = Area of flower bed/Area of 1 bag of top soil

x = 30/15

x = 2

Hence, he will need 2 bags of topsoil to cover his flower bed.

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3 years ago
a mother is five times old as her daughter. in 6 years, the mother will be three times as old as her daughter. how old is each n
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30,6

Step-by-step explanation:

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2 years ago
Okay heres another one for some of you lucky people who answer questions. Here is a ligit math question: What 2+2 I goota know s
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Answer:

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Step-by-step explanation:

5 0
3 years ago
EXAMPLE 1 (a) Find the derivative of r(t) = (2 + t3)i + te−tj + sin(6t)k. (b) Find the unit tangent vector at the point t = 0. S
Tatiana [17]

The correct question is:

(a) Find the derivative of r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(b) Find the unit tangent vector at the point t = 0.

Answer:

The derivative of r(t) is 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is (j/2 + 3k)

Step-by-step explanation:

Given

r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(a) To find the derivative of r(t), we differentiate r(t) with respect to t.

So, the derivative

r'(t) = 3t²i +[e^(-t) - te^(-t)]j + 6cos(6t)k

= 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).

Now,

r'(0) = 3t²i + (1 - t)e^(-t)j + 6cos(6t)k; at t = 0

= 3(0)²i + (1 - 0)e^(0)j + 6cos(0)k

= j + 6k (Because cos(0) = 1)

r'(0) = j + 6k

r(0) = (2 + t³)i + te^(−t)j + sin(6t)k; at t = 0

= (2 + 0³)i + (0)e^(0)j + sin(0)k

= 2i (Because sin(0) = 0)

r(0) = 2i

Note: Suppose A = xi +yj +zk

|A| = √(x² + y² + z²).

So |r(0)| = √(2²) = 2

And finally, we can obtain the unit tangent vector

r'(0)/|r(0)| = (j + 6k)/2

= j/2 + 3k

8 0
3 years ago
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