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2 years ago

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A sixth grade class has 12 boys and 24 girls. Q: For every 2 boys, there are 4 girls, do you agree with this statement? Explain

2 answers:

dimaraw [331]2 years ago

7 0

Agree....There are 12 boys and every 2 = 6 groups 6x2
There are 24 girls and every 4 to the 2 =6 groups 6x4

nevsk [136]2 years ago

4 0

Agree, there are double the amount of girls total in the class, so logically, the groups of girls will always stay double the amount of boys

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The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2.Three scores extracted from the te

Luba_88 [7]

Answer:

102

Step-by-step explanation:

We have the mean (m) 128.5 and the standard deviation (sd) 8.2, we must calculate the value of z for each one and determine whether or not it is an outlier:

z = (x - m) / sd

In the first case x = 148:

z = (148 - 128.5) /8.2

z = 2.37

In the second case x = 102:

z = (102 - 128.5) /8.2

z = -3.23

In the first case x = 152:

z = (152 - 128.5) /8.2

z = 2.86

The value of this is usually between -3 and 3, therefore when x is 102 it goes outside the range of the value of z, which means that this is the outlier.

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2 years ago

Javier is cooling a liquid in a science lab. The liquid starts at 85 degrees, and he cools it 3.5 degrees every minute for 12 mi

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85- (3.5*12)

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2 years ago

For a sample variance of n = 36 that has a sample variance of 1,296, what is the estimated error for the sample?

lukranit [14]

Answer:

6

Step-by-step explanation:

Given :

Sample size, n = 36

Sample variance, s² = 1296

The estimated standard error can be obtained using the relation :

Standard Error, S. E = standard deviation / √n

Standard deviation, s = √1296 = 36

S.E = 36/√36

S.E = 36/6

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A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for

Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

$\implies h=\frac{1}{w^2}$

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

$= 2\times \frac{1}{w^2}(w+2w)$

$=\frac{6w}{w^2}$

$=\frac{6}{w}$

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

$C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}$

$=2w^2+4w^2+\frac{12}{w}$

$=6w^2+\frac{12}{w}$

Differentiating with respect to w,

$C'(w) = 12w -\frac{12}{w^2}$

Again differentiating with respect to w,

$C''(w) = 12 + \frac{24}{w^3}$

For maxima or minima,

C'(w) = 0

$\implies 12w -\frac{12}{w^2}=0$

$\implies 12w^3 - 12=0$

$w^3-1=0\implies w = 1$

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

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3 years ago