describe a situation in which a formula could be use more easily if it were rearranged, include the formula in your description.

What is the solution to this system of linear equations?

GarryVolchara [31]

Answer:

I got (1,2).

Step-by-step explanation:

Process of elimination:

Multiply the second equation by 3 to get same terms, subtract and solve for Y to get 2.

Plug 2 into second equation for Y to get x=1

(1,2)

6 0

2 years ago

Amanda went to the store to purchase ink pens. She found three kinds of pens. The first cost $4 each; the price of the second ki

KonstantinChe [14]

Answer:

Amanda buy first kind of pen = 3

Amanda buy second kind of pen = 2

Amanda buy third kind of pen = 15

Step-by-step explanation:

Given - Amanda went to the store to purchase ink pens. She found three

kinds of pens. The first cost $4 each; the price of the second kind

was 4 for $1; and the cost for the third kind was 2 for $1. She bought

20 pens and she bought at least one of each kind. (It is possible to

buy only 1 of the pens that are "4 for $1" or "2 for $1".) The cost was

$20.

To find - When she got back to her office, Amanda decided to turn this into

a math problem for me. She asked: how many of each kind did I

buy?

Proof -

Let Amanda buy first kind of pen = x

second kind of pen = y

third kind of pen = z

As given,

She bought total pen = 20

⇒x + y + z = 20 ...............(1)

Now,

As given,

cost for first kind pen = $4 for 1 pen

As she bought x pens of first kind , so

Cost of x pens of first kind = $4x

Now,

The price of the second kind was 4 for $1

⇒Cost of second kind = $ $\frac{1}{4}$ for 1 pen

As she bought y pens of send kind , so

Cost of y pens of second kind = $ $\frac{1}{4}y$

Now,

The price of the third kind was 2 for $1

⇒Cost of third kind = $ $\frac{1}{2}$ for 1 pen

As she bought z pens of send kind , so

Cost of z pens of third kind = $ $\frac{1}{2}z$

Now,

As given, The cost was $20

⇒4x + $\frac{1}{4}y$ + $\frac{1}{2}z$ = 20

⇒16x + y + 2z = 80 .....................(2)

∴ we get 2 equations

x + y + z = 20 .....................(1)

16x + y + 2z = 80 .....................(2)

Now,

Subtract equation (1) from equation (2) , we get

16x + y + 2z - ( x+ y + z )= 80 - 20

⇒16x + y + 2z - x - y - z = 60

⇒15x + z = 60

⇒z = 60 - 15x

Now,

Put the value of z in equation (1) , we get

x + y + 60 - 15 x = 20

⇒ y - 14x = 20 - 60

⇒y - 14x = -40

⇒14x - y = 40

⇒y = 14x - 40

Now,

we get

z = 60 - 15x

y = 14x - 40

As given

she bought at least one of each kind

it means x > 1, y > 1, z > 1

Now,

If x = 1, then y = 14 - 40 = -26

Not possible

If x = 2 , then y = 14(2) - 40 = -12

Not possible

If x = 3, then y = 14(3) - 40 = 2 and z = 60 - 15(3) = 15

Possible.

If x = 4, then y = 14(4) - 40 = 16 and z = 60 - 15(4) = 0

Not Possible.

If x = 5, then y = 14(5) - 40 = 30 and z = 60 - 15(5) = -15

Not Possible.

∴ we get

x = 3, y = 2, z = 15

Amanda buy first kind of pen = x = 3

Amanda buy second kind of pen = y = 2

Amanda buy third kind of pen = z = 15

7 0

3 years ago

In a football tournament, the Bees scored 9 less than three times as many points as the Hornets. The Wasps scored 28 more points

Elina [12.6K]

Given:

Bees scored 9 less than three times as many points as the Hornets.

Wasps scored 28 more points than the Hornets.

Together the three teams scored 184 points.

To find:

The required equation for this scenario.

Step-by-step explanation:

Let x represent the number of points scored by the Hornets.

Bees scored 9 less than three times as many points as the Hornets.

Bees score = 3x-9

Wasps scored 28 more points than the Hornets.

Wasps score = x+28

Together the three teams scored 184 points.

$(3x-9)+x+(x+28)=184$

Therefore, the required equation is $(3x-9)+x+(x+28)=184$ .

3 0

3 years ago

ID: A

snow_tiger [21]

Answer:

13.63 miles per hour

Step-by-step explanation:

100 yards / 15 sec converted to miles per hour

100 yards x 60 sec x 60 min x 1 mile = 13.63 miles/hr

15 sec 1 min 1 hr 1760 yards

the mistake on Mrs Dukes is the conversion from miles to yards.

1 mile = 1760 yards NOT 5280 yards

7 0

3 years ago

A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

Tems11 [23]

An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
$P(x)=C(n,x)p^x(1-p)^{n-x}$ where
p=probability of success
n=number of trials
x=number of successes $C(n,x)=\frac{n!}{x!(n-x)!}$

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.

8 0

3 years ago