Ask question

Ask question

All categories

2 years ago

11

Viola is putting stamps in a book. She has 1,488 stamps. Each page in the book can hold 62 stamps. How many pages will she need

for all of her stamps.
A 12
B 18
C 24
D 32

2 answers:

Ostrovityanka [42]2 years ago

4 0

Answer:

Answer is C. 24

Step-by-step explanation:

1488 ÷ 62 = 24

ioda2 years ago

4 0

Answer:

Yes, I agree with other person it is 24

Step-by-step explanation: 1488/62=24

So, your answer is C

༼ つ ◕_◕ ༽つ

You might be interested in

I need all 4 help pls

slavikrds [6]

Answer:

I dont know the answer wait ima get it right now

Step-by-step explanation

i do not know the answer

3 0

3 years ago

Read 2 more answers

How to simplify 2mg x 5m²k

Damm [24]

hey, I'm not sure about this answer, but I think it's true, sorry

m(2g+5mk)

6 0

3 years ago

A student weighs a sample of the shale and

tia_tia [17]

interesting choice of pfp

4 0

3 years ago

A random sample of 10161016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity o

Korolek [52]

Answer:

0.509

Step-by-step explanation:

Given that a random sample of 1016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?"

Of the 1016 adults surveyed, 517 indicated that televisions are a luxury they could do without.

To find out the point estimate of population proportion of adults in the country who believe that televisions are a luxury they could do without.

We find that point estimate is nothing but sample proportion

Sample proportion = favourably voters/total people surveyed

= $\frac{517}{1016} \\=0.5088\\=0.509$

6 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago