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VashaNatasha [74]
2 years ago
14

A company buys pens at the rate of $5.50 per box for the first 10 boxes, $4.50 per box for the next 10 boxes, and $3.50 per box

for any additional boxes. how many boxes of pens can be bought for $177.00?
Mathematics
1 answer:
marusya05 [52]2 years ago
4 0

So it is really easy to solve firstly we can see how much does the first 10 boxes make which makes around 55$ obviously. Secondly 45$ for the next 10 boxes.

So for now we can simply calculate that we have spent around 100$ which means 20 boxes. The remaining money left is 77$ so we can buy 77/3.5 = 22 only 22 boxes with that money. Hence a total of 42 boxes.

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W = 3x + 7y solve for y
mario62 [17]

Answer:

The value of the equation y=\frac{W-3x}{7}.

Step-by-step explanation:

Consider the provided equation.

W = 3x + 7y

We need to solve the provided equation for y.

Subtract 3x from both side.

W-3x= 3x-3x+ 7y

W-3x=7y

Divide both sides by 7.

\frac{7y}{7}=\frac{W-3x}{7}

y=\frac{W-3x}{7}

Hence, the value of the equation is y=\frac{W-3x}{7}.

8 0
2 years ago
Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0
2 years ago
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Mariulka [41]
All you have to do is substitute all the Xs to and get a final y output.

for example:
if we take the number x is -1 all you do is:

y=-4(-1)+2
y=4+2
y=6
thats the first one done

7 0
2 years ago
Budgeting for home maintenance early can save money in the long run. Why save early compared to later, especially if the home is
Kruka [31]
Its A. all the other answers are unreasonable

4 0
2 years ago
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Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not
tangare [24]

Answer:

a.

\bar p_1=0.05\\\bar p_2=0.067

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let p_1  \& p_2 denote processes 1 & 2.

For p_1: T1=10,n1=200

For p_2:T2=20,n2=300

Therefore

\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067

b. To test for hypothesis:-

i.

H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05

ii.For a two sample Proportion test

Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\

iii. for \frac{\alpha}{2}=(-1.96,+1.96) (0.5 alpha IS 0.025),

reject H_o if|Z|>1.96

iv. Do not reject H_o. The noncomforting proportions are not significantly different as calculated below:

z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}

z=-0.78

c.(1-\alpha).100\% for the p1-p2 is given as:

(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt   \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645  \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0
2 years ago
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