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2 years ago

The cost of 8 kg of butter is $ 904. what will be the cost of 14 kg of butter

Mathematics

1 answer:

kvv77 [185]2 years ago

6 0

Answer:

$1582

Step-by-step explanation:

cost of each kg 904/8=113

cost of 14 kg 113x14=1582

the cost of 14 kg of butter is $1582

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Analyze the diagram to complete the statements. The m∠MXN is the m∠YZX. The m∠LZX is the m∠ZYX + m∠YXZ. The m∠MYL is 180° − m∠ZY

Xelga [282]

By using the picture that was provided below, you can see that

m<MXN is greater than m<YZX.

Angles m<LZX is equal to m<ZYX + m<YZX and

m<MYL is equal to 180 degrees minus M<ZYX.

I hope this helps! Have a good day

8 0

2 years ago

Read 2 more answers

Assuming no direct factory overhead costs (i.e., inventory carry costs) and $3 million dollars in combined promotion and sales b

alexira [117]

Answer:

Step-by-step explanation:

Given that we assume no direct factory overhead costs (i.e., inventory carry costs) and $3 million dollars in combined promotion and sales budget, the Deal product manager wishes to achieve a product contribution margin of 35%.

Sales - variable cost = Fixed cost + profit

Here fixed cost = 3 million dollars

Sales - variable = contribution = 35%

35% should atleast meet the fixed cost

i.e. 35% = 3 million

100% = 8.57 million can be cost

Since fixed cost will not change and remain 3 million these 5,57 million can be given to material and labor costs

So material and labor cost should be limited upto 5.57 million increase.

3 0

2 years ago

Craig has $1850 dollars in a bank account that he uses to make automatic payments of $400.73 on his car loan. If Craig stops mak

Ivenika [448]

Given:

Amount in the bank account = $1850

Monthly payment of can loan = $400.73

To find:

When would automatic payments make the value of the account zero?

Solution:

Craig stops making deposits to that account. So, amount $1850 in the bank account is used to make monthly payment of can loan.

On dividing the amount by monthly payment, we get

$\dfrac{1850}{400.73}=4.61657$

It means, the amount is sufficient for 4 payment but for the 5th payment the amount is not sufficient.

Therefore, the 5th automatic payments make the value of the account zero.

6 0

2 years ago

A police car's siren operates at a frequency of 2,000 Hz. What frequency would a stationary listener observe if the police car i

SpyIntel [72]

we can use formula

$f_L=f_s(\frac{v_a-v_p}{v_a+v_p} )$

where

fL is listener frequency

fs is siren frequency

va is the speed of sound in air

vs is police car speed

so, we have

$f_s=2000Hz$

$v_a=345m/s$

$v_p=65mph$

now, we can change it into m/s

$v_p=65*\frac{1609}{3600}m/s$

$v_p=29.05139m/s$

now, we can plug it

$f_L=2000(\frac{345-29.05139}{345+29.05139} )$

$f_L=1689.33263Hz$ ..............Answer

3 0

2 years ago

G identify the solution of the recurrence relation an = 6an − 1 – 8an − 2 for n ≥ 2 together with the initial conditions a0 = 4,

maksim [4K]

Via the generating function method, let

$G(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Then take the recurrence,

$a_n=6a_{n-1}-8a_{n-2}$

multiply everything by $x^n$ and sum over all $n\ge2$ :

$\displaystyle\sum_{n\ge2}a_nx^n=6\sum_{n\ge2}a_{n-1}x^n-8\sum_{n\ge2}a_{n-2}x^n$

Re-index the sums or add/remove terms as needed in order to be able to express them in terms of $G(x)$ :

$\displaystyle\sum_{n\ge2}a_nx^n=\sum_{n\ge0}a_nx^n-(a_0-a_1x)=G(x)-4-10x$

$\displaystyle\sum_{n\ge2}a_{n-1}x^n=\sum_{n\ge1}a_nx^{n+1}=x\sum_{n\ge1}a_nx^n=x\left(G(x)-a_0\right)=x(G(x)-4)$

$\displaystyle\sum_{n\ge2}a_{n-2}x^n=\sum_{n\ge0}a_nx^{n+2}=x^2\sum_{n\ge0}a_nx^n=x^2G(x)$

So the recurrence relation is transformed to

$G(x)-4-10x=6x(G(x)-4)-8x^2G(x)$
$(1-6x+8x^2)G(x)=4-14x$
$G(x)=\dfrac{4-14x}{1-6x+8x^2}=\dfrac{4-14x}{(1-4x)(1-2x)}=\dfrac1{1-4x}+\dfrac3{1-2x}$

For appropriate values of $x$ , we can express the RHS in terms of geometric power series:

$G(x)=\displaystyle\sum_{n\ge0}(4x)^n+3\sum_{n\ge0}(2x)^n=\sum_{n\ge0}\bigg(4^n+3\cdot2^n\bigg)x^n$

which tells us that

$a_n=4^n+3\cdot2^n$

3 0

2 years ago