All categories

3 years ago

In Kailah's class the ratio of girls to boys is 13 to 8. If there were 16 boys, how many girls will there be?

Mathematics

2 answers:

VladimirAG [237]3 years ago

5 0

Answer:

I believe that the answer is 26

Step-by-step explanation:

The ratio of girls to boys is 13 to 8 so there are 13 girls for every 8 boys so if there are 16 boys (8X2) then there are 26 girls (13X2)

Lady_Fox [76]3 years ago

5 0

I think 22.
there are 5 more girls than boy with 13:8

You might be interested in

A multiplication table. In the row labeled 3, the numbers 3, 6, 9, 12, 15, 18, 21, 23, and 27 are highlighted. In the row labele

kondaur [170]

Answer:

24:27

Step-by-step explanation:

i got it right on my unit test :)

3 0

3 years ago

The _______ represents half the distance between the maximum of a sine or cosine graph

ziro4ka [17]

Answer: amplitude

Step-by-step explanation: The amplitude represents half the distance between the maximum of a sine or cosine graph.

3 0

3 years ago

Which expression has value of 15 when = 12?

frosja888 [35]

Answer:

I think its 3p not sure but u can try that

3 0

3 years ago

K divided by 11 is 7.<br>

sp2606 [1]

Answer:

Step-by-step explanation:

3 0

3 years ago

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.

7 0

3 years ago