Tommy has 5 jars of marbles. Each jar is 2/3 filled with marbles. How many jars of marbles does Tommy have

A County Superintendent of Highways is interested in the numbers of different types of vehicles that regularly travel within his

Karolina [17]

Answer:

a) The ratio of passenger cars to pickup trucks is $\frac{5}{7}$

The ratio of pickup trucks to passenger cars is $\frac{7}{5}$

The ratio of passenger cars to total vehicles is $\frac{5}{12}$

The ratio of pickup trucks to total vehicles is $\frac{7}{12}$

b) The tape diagram is shown in the attached picture.

c) The tape diagram has 12 equal-sized parts.

d) The total quantity that the tape diagram represent is 192.

e) The value of each individual part of the tape is 16.

f) There were registered 80 passenger cars and 112 pickup trucks.

Step-by-step explanation:

a) According to the information in the problem, that for every 5 passenger cars registered, there were 7 pickup trucks registered.

So the ratios are:

Passenger cars to pickup trucks: $\frac{5}{7}$

Pickup trucks to passenger cars: $\frac{7}{5}$

If we calculate the total vehicles: 5+7=12

So, we can get two ratios more:

Passenger cars to total vehicles is $\frac{5}{12}$

Pickup trucks to total vehicles is $\frac{7}{12}$

b) We make a tape diagram where each box represents a vehicle so we draw 5 boxes for cars and 7 boxes for pickup trucks.

c) If we count the total quantity of boxes, there are 12.

d) This diagram represents the total quantity of registrations in August which is 192.

e) To get the representation of each part of the tape diagram we have to divide 192 by 12. 192÷12=16

f) Finally, we multiply 16 by the number of boxes of each type of vehicle:

Cars: 16×5=80

Pickup trucks: 16×7=112

6 0

2 years ago

Distinguish between rote and rational counting

sergejj [24]

Answer: There is a difference between rote counting and rational counting. Rote counting involves the memorization of numbers. Rational counting tells children "how many there are." For children to count rationally, they need to demonstrate one-to-one correspondence.

6 0

1 year ago

Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1

irina [24]

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}$

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)

5 0

2 years ago

Read 2 more answers

Find the average value of the function f(x, y, z) = 5x2z 5y2z over the region enclosed by the paraboloid z = 4 − x2 − y2 and the

S_A_V [24]

The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.

Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. 

If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV 

∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz 

= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 

∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz 

= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ 

= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ 

= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 

∴ F = 10935π/8 ÷ 81π/2 = 135/4

3 0

2 years ago

Show that b^2 is greater than equal to or less than ac, according as a, b, c are in A.P., G.P.

ValentinkaMS [17]

Answer/Step-by-step explanation (ac > b² or b² < ac. )

A/c to question, we have to show:-

b² >ac in A.P ........ (1)

b² = ac in G.P .....(2)

b² < ac in H.P. ..... (3)

b = a+c/2 (A.P)

b = √ac ( G.P)

b = 2ac/a+c (H.P)

In A.P :

b² > ac = b² - ac

= (a+c/2)² - ac

= (a²+2ac+c²/4) - ac = a² + 2ac + c² - 4ac / 4

= a² - 2ac + c² / 4 = ( a - c ) ² / 4 > 0 Hence, b²>ac

In G.P:-

b = √ac

Hence, b² = ac

In H.P :- b² < ac = ac > b² = ac - b² = ac - ( 2ac / a+c)

= ac(a+c) - 2ac / a+c

= a²c + ac² - 2ac / a+c

= ac(2ac - 2) / a+c > 0

Hence, ac > b² or b² < ac.

5 0

1 year ago