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Neko [114]
2 years ago
9

Which is a step in the process of calculating successive discounts of 8% and 10% on a $50 item?

Mathematics
1 answer:
Vanyuwa [196]2 years ago
7 0
C is the right answer.
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Solve for t. You must write your answer in fully simplified form. <br><br> −19=7t
nikitadnepr [17]

Answer:

t=-19/7

Step-by-step explanation:

you need to isolate for t which means you need to get t all alone on one side of the equation. Do do that you can divide both sides by 7. This gives you -19/7=7t/7

we know that anything divided by itself is 1 so we get -19/7=t

This is the simplest for because 19 and 7 are both prime numbers, so you can't simplify them.

7 0
2 years ago
Read 2 more answers
A factory dumps an average of 2.43 tons of pollutants into a river every week. if the standard deviation is 0.88 tons, what is t
vovikov84 [41]
25.8%  
First, determine how many standard deviations from the norm that 3 tons are. So:
 (3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273  
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%  
So the probability that more than 3 tons will be dumped in a week is 25.8%
5 0
2 years ago
I need you to answer with a, b, c, d
solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

\begin{gathered} ax^2+bx+c=0 \\  \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

For the given function:

f(x)=2x^2-10x-3x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}x=\frac{10\pm\sqrt[]{100+24}}{4}\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\  \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\  \end{gathered}\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\  \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\  \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Then, the zeros of the given quadratic function are:

\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\  \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Answer: Third option

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liraira [26]

Answer:

mustard

Step-by-step explanation:

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5 0
2 years ago
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Write the quadratic equation whose roots are 2 and -4 and whose leading coefficient is 2
Goryan [66]

Answer:

2x^2+4x-16

Step-by-step explanation:

The quadratic can be written as

f(x) = a(x-z1)(x-z2) where z1 and z2 are the roots

f(x) = a (x-2)(x- -4)

a is the leading coefficient

f(x) = 2(x-2)(x+4)

     = 2(x^2 -2x+4x-8)

     = 2(x^2 +2x-8)

     = 2x^2 +4x-16

5 0
3 years ago
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