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2 years ago

5

At a school party there are 40 cookies. The ratio of chocolate cookies to oatmeal cookies is shown in the tape diagram. Chocolat

e: 5 Oatmeal: 3 How many oatmeal cookies were at the party?

2 answers:

katen-ka-za [31]2 years ago

8 0

5x + 3x = 40

8x = 40

x = 5

3 x 5 = 15 oatmeal cookies

Sholpan [36]2 years ago

4 0

<h2>Answer:</h2>

The number of Oatmeal cookies at the party were:

15

<h2>Step-by-step explanation:</h2>

It is given that:

The total number of cookies at the school party were: 40 cookies

Also the ratio of chocolate cookies to Oatmeal cookies are: 5: 3

So, let the number of chocolate cookies= 5x

and the number of Oatmeal cookies= 3x

Hence, we have:

5x+3x=40

( since number of chocolate and oatmeal cookies is equal to the total number of cookies)

i.e.

8x=40

on dividing both side of the equation by 8 we have:

x=5

Hence, Number of chocolate cookies= 5×5=25 cookies.

and Number of Oatmeal cookies= 5×3= 15 cookies.

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Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

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A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the number

serg [7]

Answer:

The probability distribution is

X : 0 1 2

f(X) : 2/7 4/7 1/7

Step-by-step explanation:

Let x is the number of defective sets purchased by the hotel.

Total number of television sets = 7

Total number of defective television sets = 2

Since there are 2 defective television sets and hotel purchase 3 sets, So, X=0,1,2.

If X=0,

$P(X=0)=\frac{^5C_3}{^7C_3}=\frac{10}{35}=\frac{2}{7}$

If X=1,

$P(X=1)=\frac{^5C_2\cdot ^2C_1}{^7C_3}=\frac{10\cdot 2}{35}=\frac{4}{7}$

If X=2,

$P(X=2)=\frac{^5C_1\cdot ^2C_2}{^7C_3}=\frac{5\cdot 1}{35}=\frac{1}{7}$

The probability distribution is

X : 0 1 2

f(X) : 2/7 4/7 1/7

The probability histogram is shown below.

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What is 4.66666666667 as a fraction

ss7ja [257]

Step-by-step explanation:

466666666667/100000000000

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Ava wants to figure out the average speed she is driving. She starts checking her car’s clock at mile marker 0. It takes her 4 m

VLD [36.1K]

<h2>Answer:</h2>

The average speed of car is:

0.75 miles per minutes.

The equation of line is:

$n=\dfrac{3}{4}t$

<h2>Step-by-step solution:</h2>

The table that describes the time and number of miles marked is given by:

Time Miles

0 0

4 3

8 6

Clearly we could observe that with the increase in time by every 4 minutes the number of mile increases by 3.

i.e. the average speed is calculated as the ratio of change in miles to the change in time.

Hence,

$Average\ Speed=\dfrac{3-0}{4-0}\\\\\\i.e.\\\\\\Average\ Speed=\dfrac{3}{4}=0.75$

Hence, the average speed=0.75 miles per minutes.

Also we will find the equation of the lines that related the miles and time by using the two-point formula.

i.e. any line passing through (a,b) and (c,d) is calculated by using the formula:

$y-b=\dfrac{d-b}{c-a}\times (x-a)$

Here (a,b)=(0,0) and (c,d)=(4,3)

i.e. the equation is given by:

$n-0=\dfrac{3-0}{4-0}\times (t-0)\\\\i.e.\\\\n=\dfrac{3}{4}t$

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