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photoshop1234 [79]
3 years ago
13

A sweater is on sale for 40 percent off its origanal price of 29.95 what is the amount of savings

Mathematics
2 answers:
horrorfan [7]3 years ago
8 0

0.4×29.95

=11.98 so your answer is 11 dollars and 98 cents

worty [1.4K]3 years ago
8 0

Hey There!

So what you need to do is <em>29.95 - 40% </em>which is equal to <em><u>11.98.</u></em>

So the amount of savings is <u><em>11.98! :D</em></u>

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4vir4ik [10]

Answer:

x=\frac{32}{3}

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4 0
3 years ago
Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production
Lena [83]

The question supplied is incomplete :

The x parameters aren't given ; "Units with weights less than or greater than ounces will be classified as defects"

Assume the unit weights Given are ;

Units with weights less than 10.8 or greater than 11.2 ounces will be classified as defects

Just follow the procedure in the solution for any value of unit weight interval given.

Answer:

0.3173

Step-by-step explanation:

Mean weight, m = 11 ounces

Standard deviation, s = 0.2 ounces

Find the Zscore for each unit weight :

Z = (x - mean) / standard deviation

P(x < 10.8) :

Z = (10.8 - 11) / 0.2 = - 1

P(Z < - 1) = 0.15866

P(x > 11.2) :

Z = (11.2 - 11) / 0.2 = 1

P(Z > 1) = 0.84134

P(x < 11.2) - P(x < 10.8) = 1 - (P(Z < 1) - P(Z < - 1)) = 1 - 1 - (0.84134 - 0.15866) = 1 - 0.68268 = 0.31732

Hence, Probability of a defect is 0.3173

7 0
3 years ago
PLeAsE hElP mE wItH mY MaTh PlEaSe LiNk AttAcHeD.
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The answer is going to be A) 9:14
7 0
4 years ago
Equations of Proportional Relationships
Alla [95]

Answer:

y=5x is the correct answer

4 0
3 years ago
Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0
Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is x^2 - 36 = 0

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

x^{2}-12 x+36=0  --- eqn 1

x^{2}+12 x-36=0 -- eqn 2

x^{2}-36=0  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is x^2 - 36 = 0

3 0
3 years ago
Read 2 more answers
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